How to split single DC powersupply in + and - volts for opamp rails

[robsims]

Hi, i'm a beginner and i have read many articles about opamps. What i still don't understand is how to split my single dc powersupply in +16 volts and -16 Volts to connect to the rails of a LM324N opamp. Can someone please explain it to me, or if you know a very good article, please post the link. Many thanks in advance

[james_s]

In a practical sense you can't really split a single power supply. There are tricks you can use to create a virtual ground though and run an op-amp off a single supply. A pair of 9V batteries in series is a cheap and easy way of producing a split supply for experimenting with op amps.

[themadhippy]

if its not mission critical ,or to much current 2 resistor of the same value,  in series,1 end to your +16 volts the other end to the -16 Volts and your "ground"  is the middle point  were the 2 resistors join

[EPAIII]

A SINGLE DC supply with +16 and -16 Volts?? Do you mean it does not have a common ground? Just the two + and - outputs? I have never seen a power supply like that. I can understand a 32 Volt supply, but that could be relabeled in any number of ways (+20 and -12 V or +1 and -31 Volts or -10 and -42 Volts, and many, many more) without some common reference point. In general these alternate forms of labeling the outputs are not used and it would be labeled as a 32 Volt supply.

So, I must assume you have a single power supply that has only ONE 16 Volt output and it is floating (neither terminal is tied to any external point, including ground). This is the only way your statement makes any sense. Going on that assumption, you then need a common (ground) point that is at the mid point between those two (+16 and -16) Voltages. This would give you +8, Gnd or 0V, and -8V.

Your OP amp seems to come in a quad package: four OP amps in one package. Chances are you will have one or more unused ones in your design. Also, most OP amp circuits only need the common or ground to carry a very small current. Now, IF your circuit only consists of OP amp circuitry that needs that common or ground then an unused section of the four OP amp package can be used to supply that Voltage. A Voltage divider between your +16 and -16 Volts, with two equal resistors (10Ks would work) would have it's center point, which would be at the mid point Voltage would be connected to the + input and the output would be connected to the - input. The output would provide your needed mid-point or ground. This only works if you keep within the rated output current of the OP amp which seems to be guaranteed to be +/-10 mA.

If you need a greater common/ground current, then there are Voltage follower circuits and ICs that can be added on the OP amp's output to produce it.

So yes, you can split a single power supply.

Hi, i'm a beginner and i have read many articles about opamps. What i still don't understand is how to split my single dc powersupply in +16 volts and -16 Volts to connect to the rails of a LM324N opamp. Can someone please explain it to me, or if you know a very good article, please post the link. Many thanks in advance

[rdl]

Can someone please explain it to me, or if you know a very good article, please post the link.

The Art of Electronics Third Edition Chapter 9 - Voltage Regulation and Power Conversion, provided free by the authors.

https://artofelectronics.net/wp-content/uploads/2016/02/AoE3_chapter9.pdf

Not a lot actually pertains to what you ask, but you will learn a lot about power supplies. If you are any good at building and can get the parts, there is a schematic supplied which would solve your problem.

[robsims]

Yes EPAIII, i have a single power supply that has only ONE 30 Volt output and it is floating (neither terminal is tied to any external point, including ground).

So if i have two resistors in series, the top/beginning of the first resistor will be connected to the plus post of the powersupply, The plus rail of the opamp will also be connected to the plus post of the powersupply.  Mid point of the two resistors would be connected to the + input and the output would be connected to the - input. Where do i connect the negative - rail of the opamp?

[rdl]

Using just a couple of resistors is one way to split a voltage, and it might work depending on what you're doing with the op amp, but overall it's probably the worst way. Trying going to google or wherever and search for "rail splitter circuit".

[pqass]

Rail splitters have been discussed here before. 
See: https://www.eevblog.com/forum/beginners/rail-splitter-circuit-question/

[EPAIII]

Here is a schematic. This was my first attempt to draw a schematic with KiCAD so please forgive any rough spots. I think it conveys the idea.

I have divided your single 30 Volts into a +15 and a -15 Volt labels. This is just your power supply with labels that are more meaningful for your op amp circuit. The LM324 is powered in the normal way from your power supply, with plus to pin 4 and minus to pin 11. The section of that op amp in the schematic is generating a Voltage that is half way between the two power supply rails (+ and - 15V) so I have labeled it as 0 Volts.

The connection between pins 1 and 2 turns it into a Voltage follower with high gain. So the output, pin 1 follows the Voltage at the non-inverting (+) input, pin 3, limited only by the current available at that output. And since the input impedance at both of the input pins of the op amp is quite high, it does not upset the Voltage value determined by the two resistors.

Yes EPAIII, i have a single power supply that has only ONE 30 Volt output and it is floating (neither terminal is tied to any external point, including ground).

So if i have two resistors in series, the top/beginning of the first resistor will be connected to the plus post of the powersupply, The plus rail of the opamp will also be connected to the plus post of the powersupply.  Mid point of the two resistors would be connected to the + input and the output would be connected to the - input. Where do i connect the negative - rail of the opamp?

[EPAIII]

If you are going to play with op amps, I strongly recommend you get a copy of "IC Op Amp Cookbook" by Walter G. Jung. It is an old book and has at least three versions. It can be found on the internet from a multitude of sellers. Well worth the price.

[rstofer]

Chapter 4 of "Op Amps For Everyone" gets into single supply op amp circuits.  Basically, the op amp doesn't care what you think ground is all about, it will work just as well on a single supply except that there needs to be some kind of rail splitter (2 resistors) to provide a reference ground for the signal inputs and outputs.

https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

There are single supply topics all over the Internet.

https://www.ti.com/lit/an/sboa059/sboa059.pdf

Kirchhoff's Laws come up quite a bit

https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/a/ee-kirchhoffs-laws

[TimFox]

"Rail splitters" are a reasonable way to convert a single ungrounded power supply into a split power supply, positive and negative with respect to a ground terminal, if the current drawn from the positive and negative terminals by the load are roughly equal, so that the "common" output does not have to deal with a large current.
I have a very old Wiltron 350 phase meter that did exactly that, since almost all the analog electronics were op amps, running off +/- 15V, with very low output currents from the individual amplifiers.

[robsims]

Many thanks guys. I will analyze the info you have given. If i have more questions, i'll post them

[rdl]

I will add one thing more, this very good, short and clear video by w2aew (Alan) about how op amps and their power supplies interact.

#79: Op Amp Power Supply Considerations: split, single, virtual ground, etc. - a tutorial

https://youtu.be/MtccB9K09ck

[robsims]

Here is a schematic. This was my first attempt to draw a schematic with KiCAD so please forgive any rough spots. I think it conveys the idea.

I have divided your single 30 Volts into a +15 and a -15 Volt labels. This is just your power supply with labels that are more meaningful for your op amp circuit. The LM324 is powered in the normal way from your power supply, with plus to pin 4 and minus to pin 11. The section of that op amp in the schematic is generating a Voltage that is half way between the two power supply rails (+ and - 15V) so I have labeled it as 0 Volts.

The connection between pins 1 and 2 turns it into a Voltage follower with high gain. So the output, pin 1 follows the Voltage at the non-inverting (+) input, pin 3, limited only by the current available at that output. And since the input impedance at both of the input pins of the op amp is quite high, it does not upset the Voltage value determined by the two resistors.

Yes EPAIII, i have a single power supply that has only ONE 30 Volt output and it is floating (neither terminal is tied to any external point, including ground).

So if i have two resistors in series, the top/beginning of the first resistor will be connected to the plus post of the powersupply, The plus rail of the opamp will also be connected to the plus post of the powersupply.  Mid point of the two resistors would be connected to the + input and the output would be connected to the - input. Where do i connect the negative - rail of the opamp?

Hi EPAIII,  i made some modifications, see attached picture. Are they good?

[TimFox]

For your rail splitter, the 30 V power supply should not be grounded.
Instead, the op-amp output becomes ground, and the two ends of the 30 V power supply become +15 V and -15 V with respect to that ground.
As I said above, the difference in load currents on the +15 and -15 V terminals needs to flow to the op-amp output, which limits how much imbalance can be tolerated.
Since there will be decoupling capacitors from +15 and -15 V to the new ground, the op amp will see a capacitive load, and that must be considered in the feedback design.

[RJSV]

   You could do it, similar to Thermahippy's advice, only just use 1 supply battery.  I would take 2 resistors, maybe 3.3 kohms each, in series, with the middle point call it: " AC ground".
So, casually approached, you've then got 12 VDC supply, as + and -, and you've created an AC ground, that could be attached to any metal case, and any sheild grounds.
   For a better ground point, (DC) I would consider using a pair of 9 V batteries, that being less power, if (and when) something on the bench (accidently) contacts places it shouldn't.  So you would have one of the 9V as a (-9V rail), )with it's positive connected to ground, and the other 9V battery as a (+9V rail), having it's (-) connected to ground. Then that's a good time to check your work; before hooking up any IC.
With that full plus and minus rail set, and valid ground, you won't need to create the 2 resistors divider, or the cap.  Suggested put a capacitor from ground to each rail (observing correct + and minus on any polarized capacitor.)
   So that's 2 batteries and 2 capacitors, like maybe 100 uF each should be adequate.  Many designs feature putting a second capacitor, tiny mica type, for absorbing the really tiny bit high frequency noise spikes, as the little cap.s are faster acting, than a power capacitor, like 100 uF.
  Keep in mind; Standard 9V batteries don't have rediculously high capacity, for long running times.

[RJSV]

   Since you didn't mention 'Voltage Follower' explicitly, I wanted to also mention that is what you are describing.  The voltage control action gets active when that (differential) input strays from the value presented at the positive input.  If plus input gets 'too big', meaning that it exceeds the negative differential input, then that input difference is amplified. So the resulting control 'swing' tends to oppose and try to counter any trend there.  Similarly, but opposite effect, of classic 'op amp negative feedback' is a control output action opposing any trend to be much LESS voltage, than the '+' differential input.
   Just saying; that's a classic ' Voltage Follower', in that it equals the (feeble) input, but with that 20 mA or so maximum the 'follower' action can drive a light load.
It's really an interesting, and ZEN-like quality.
(Try Google 'Ron Widlar, INTEL Corp.' sometime )

[RJSV]

Oops.
   Got my + and - mixed up...
   The '+' differential input is the pin the circuit is 'following', and so that's the voltage input (to follow at 1X or unity gain.)

[RJSV]

   My preferred single 9 V battery supply, for LM-324 OP Amp projects (see schematic picture) was usually connected to any metal chassis or box, at the negative point, of battery, then calling that ground.  By creating a virtual AC ground, with the 2 resistors and large capacitor, that doesn't conflict as you could always think of having two batteries, 4.5 V each, with ground in the middle, of the series connected batteries.
There's a difference, that being that your virtual ground is for supplying very small currents, whereas 'regular' ground can sink more current, and thus be part of power, rather than just a reference voltage.
   A whole can of worms can arise, when you move up to experimenting with higher bandwidth OP amps,
where you have to take into account things like inductance of capacitor leads, things like that.  But that's way up there, in frequency.

[Zero999]

Is it just amplifying AC?
If so, use a potential divider to bias the output to half the supply and use AC coupling capacitors.

Credit to audioguru for the original schematics.

[robsims]

After reading the last posts i am confused now. Remember,  i'm a beginner. I only want to amplify a small DC + or - voltage. My concern is how to connect the opamp rails with one 30 volt power supply and how to connect the load. I made some modifications on the drawing. I attached it again. Is the drawing good? Many thanks guys.

[RJSV]

   Drawings looks close, I would need an input capacitor, connected to that junction of the 2 divider resistors.  Also, I would use slightly high resistor value, like 10 k each, as your input has to drive the voltage at that DC mid-point, with small AC input voltages.
   Set your gain, so, with equal resistors (going to differential 'negative' input), AC gain will be 1.  With a two to one resistor pair, like 20k with 10 k, on the op AMP negative input,AC  gain is X2.
Thanks.

[EPAIII]

You are missing the idea. The op amp in my circuit is not driving a load. It is CREATING a point that is half way between the positive and negative supply lines.

If you were using a dual power supply, one for a positive Voltage and the other for a negative Voltage, then these two separate power supplies would be CONNECTED TOGETHER, minus to plus. And this connection point would be what is generally called GROUND in an op amp circuit. This type of connection of TWO power supplies produces THREE output terminals which would be labeled V+, Gnd, and V-. And if the Gnd terminal is considered 0 Volts, then V+ would be a positive Voltage and V- would be a negative Voltage IN REFERENCE TO THAT GND OR 0 VOLTAGE.

But you want to operate your op amp circuit with only ONE power supply. So, that power supply's terminals become the V+ and V- terminals AND my circuit provides a Voltage point half way between them for YOUR GND point. And in my circuit that CREATED GND point is at the OUTPUT pin of the op amp (pin 1). You do not connect a load between that point and your power supply's negative terminal (which you labeled GND).

That point (pin 1) is, for all intents and purposes, the circuit's ground. Your input signal's ground would be connected to that point. All points in any op amp circuit that are labeled GND are connected to that point. And any load which is connected to any other op amp in your circuit would use that point as it's reference GND.

So, where you drew +30 Volts, change it to +15 Volts. Where you added a ground symbol, change it to -15 Volts. And where you have a load resistor, REMOVE that load and connect the output pin (#1) of the op amp to your circuit GROUND.

Then create your circuit using these THREE Voltage supply points for power and also for your signal ground.

Here is a schematic. This was my first attempt to draw a schematic with KiCAD so please forgive any rough spots. I think it conveys the idea.

I have divided your single 30 Volts into a +15 and a -15 Volt labels. This is just your power supply with labels that are more meaningful for your op amp circuit. The LM324 is powered in the normal way from your power supply, with plus to pin 4 and minus to pin 11. The section of that op amp in the schematic is generating a Voltage that is half way between the two power supply rails (+ and - 15V) so I have labeled it as 0 Volts.

The connection between pins 1 and 2 turns it into a Voltage follower with high gain. So the output, pin 1 follows the Voltage at the non-inverting (+) input, pin 3, limited only by the current available at that output. And since the input impedance at both of the input pins of the op amp is quite high, it does not upset the Voltage value determined by the two resistors.

Yes EPAIII, i have a single power supply that has only ONE 30 Volt output and it is floating (neither terminal is tied to any external point, including ground).

So if i have two resistors in series, the top/beginning of the first resistor will be connected to the plus post of the powersupply, The plus rail of the opamp will also be connected to the plus post of the powersupply.  Mid point of the two resistors would be connected to the + input and the output would be connected to the - input. Where do i connect the negative - rail of the opamp?

Hi EPAIII,  i made some modifications, see attached picture. Are they good?

[EPAIII]

Yes, it is only good for the rated, output current of the op amp used. In this case, the LM324 is rated for about 20 mA so if more than that is needed some additional circuitry would be needed.

The times I have used this type of power splitter it has proven itself to be up to the task. Normally the ground in an op amp circuit is mostly used as a bias point in the input circuits. And the inputs of the op amp are, by definition high impedance so they draw very little current. I did warn about this when I first suggested it.

...see attached picture. Are they good?

that opamp based splitter only good for few mA... if you want to put more load, use push pull power transistor on its output.

[MathWizard]

I made a split rail supply , that just splits a voltage source, with pos/neg LM317's. It worked ok for basic little circuits that hardly pull any current on a breadboard. But I barely used it, once I got a real PSU.

[Zero999]

After reading the last posts i am confused now. Remember,  i'm a beginner. I only want to amplify a small DC + or - voltage. My concern is how to connect the opamp rails with one 30 volt power supply and how to connect the load. I made some modifications on the drawing. I attached it again. Is the drawing good? Many thanks guys.

What's the input voltage and required gain? The circuits I posted will work for AC only.

Are you using all of the op-amps in the LM324?

The idea is to bias the output of the op-amp at half the supply and treat its output as the 0V/earth/ground node and the +/- side of the 30V power supply as +15V and -15V. 0V is just a reference from where all voltages are measured. Here's a demonstration of a DC amplifier with a gain of 10, running off a split supply. The input is swept from -1V to +1V and the op-amp amplifies it to -10V to +10V.

[RJSV]

   Yes, zero999, your two schematics help to see the meaning, of 'supplied ground' (supplied by left side portion of schematics).  And, with that supplied artificial ground being the 'ground' symbols shown in the right-side half of those 2 schematics.
I can see how that is designated as a low output current, only really for supplying an artificial ground.
Sorry to be repeating what's been said already, but zero999 that pair of diagrams helped a lot!
Was pleased to see a simple demo of AC gain,

HOWEVER:  Why not just use 2 resistors, and large capacitor, to supply a low current ground reference ?

[TimFox]

Note that the output current of the op amps powered by the rail splitter is part of the current that must be supplied by the rail-splitter output (U1 in Zero999's drawing).
This may require too large a current through the two resistors (i.e., too small resistance):  it's a quantitative question.

[Zero999]

   Yes, zero999, your two schematics help to see the meaning, of 'supplied ground' (supplied by left side portion of schematics).  And, with that supplied artificial ground being the 'ground' symbols shown in the right-side half of those 2 schematics.
I can see how that is designated as a low output current, only really for supplying an artificial ground.
Sorry to be repeating what's been said already, but zero999 that pair of diagrams helped a lot!
Was pleased to see a simple demo of AC gain,

HOWEVER:  Why not just use 2 resistors, and large capacitor, to supply a low current ground reference ?

The resistor values would have to be very low to avoid affecting the circuit. Bypassing with capacitors would only work for AC and I've already demonstrated that. The op-amp configured as a unity gain buffer has a very low output impedance, so it will keep the output voltage close to middle, as long as the current through the 0V rail doesn't exceed the output's current rating.

Note that the output current of the op amps powered by the rail splitter is part of the current that must be supplied by the rail-splitter output (U1 in Zero999's drawing).
This may require too large a current through the two resistors (i.e., too small resistance):  it's a quantitative question.

Yes, transistors can be added, or op-amps connected in parallel, to increase the output current, but there's little point if it's only powering one op-amp, or even two and the output current isn't that high. If U1 and U2 are both the LM324, then if U1's output current is too high, the same would be true for U2, as all of U2's output current flows through U1's output. If anything, U1 can be pushed a little harder since its output voltage sits at half the supply voltage, whist other op-amps might go nearer the rails where the transistors get biased closer to saturation. If it's powering multiple op-amps and one op-amp is sourcing and the other sinking, the currents cancel to some degree.

[Zero999]

Here's a simulation showing the effect of using a potential divider. With 4k7 resistors, which would continuously draw 30/(4.7k+4.7k) = 30/9.4k = 3.2mA, 96mW from 30V, and a 10k load on the op-amp, the supply rails shift by nearly 2.5V relative to 15V, as the op-amp's output swings from -10V to +10V, supplying -/+1mA to the load resistor.

The impedance of the 0V node is equal to R1||R2, which is 4k7/2 = 2k35 in this case. The op-amp supplies 10V/10k = 1mA to the load resistor and a smaller current to R4 and R3, which have a total value of 20k + 180k = 200k and will draw an additional 10/200k = 0.05mA when the op-amp's output is at 10V, giving a total output current of 1.05mA. 1.05m*2.35k = 2.47V, which is seen on +V and -V, as 0V is the fixed reference point and can't be changed. Note the difference between +V and -V is always 30V, assuming V1 is well-regulated. Both +V and -V rise and fall by 2.5V simultaneously.

[robsims]

You are missing the idea. The op amp in my circuit is not driving a load. It is CREATING a point that is half way between the positive and negative supply lines.

If you were using a dual power supply, one for a positive Voltage and the other for a negative Voltage, then these two separate power supplies would be CONNECTED TOGETHER, minus to plus. And this connection point would be what is generally called GROUND in an op amp circuit. This type of connection of TWO power supplies produces THREE output terminals which would be labeled V+, Gnd, and V-. And if the Gnd terminal is considered 0 Volts, then V+ would be a positive Voltage and V- would be a negative Voltage IN REFERENCE TO THAT GND OR 0 VOLTAGE.

But you want to operate your op amp circuit with only ONE power supply. So, that power supply's terminals become the V+ and V- terminals AND my circuit provides a Voltage point half way between them for YOUR GND point. And in my circuit that CREATED GND point is at the OUTPUT pin of the op amp (pin 1). You do not connect a load between that point and your power supply's negative terminal (which you labeled GND).

That point (pin 1) is, for all intents and purposes, the circuit's ground. Your input signal's ground would be connected to that point. All points in any op amp circuit that are labeled GND are connected to that point. And any load which is connected to any other op amp in your circuit would use that point as it's reference GND.

So, where you drew +30 Volts, change it to +15 Volts. Where you added a ground symbol, change it to -15 Volts. And where you have a load resistor, REMOVE that load and connect the output pin (#1) of the op amp to your circuit GROUND.

Then create your circuit using these THREE Voltage supply points for power and also for your signal ground.

Here is a schematic. This was my first attempt to draw a schematic with KiCAD so please forgive any rough spots. I think it conveys the idea.

I have divided your single 30 Volts into a +15 and a -15 Volt labels. This is just your power supply with labels that are more meaningful for your op amp circuit. The LM324 is powered in the normal way from your power supply, with plus to pin 4 and minus to pin 11. The section of that op amp in the schematic is generating a Voltage that is half way between the two power supply rails (+ and - 15V) so I have labeled it as 0 Volts.

The connection between pins 1 and 2 turns it into a Voltage follower with high gain. So the output, pin 1 follows the Voltage at the non-inverting (+) input, pin 3, limited only by the current available at that output. And since the input impedance at both of the input pins of the op amp is quite high, it does not upset the Voltage value determined by the two resistors.

Yes EPAIII, i have a single power supply that has only ONE 30 Volt output and it is floating (neither terminal is tied to any external point, including ground).

So if i have two resistors in series, the top/beginning of the first resistor will be connected to the plus post of the powersupply, The plus rail of the opamp will also be connected to the plus post of the powersupply.  Mid point of the two resistors would be connected to the + input and the output would be connected to the - input. Where do i connect the negative - rail of the opamp?

Hi EPAIII,  i made some modifications, see attached picture. Are they good?

yes EPAIII, i missed the point. Thought you're drawing was how to connect the opamp on a single powersupply. But ok, can you or someone else make a simple drawing of the a opamp with all the connections?

I have a LM324N IC. I only want to use one opamp in that IC package, so i only need a drawing of one opamp. I have a single 30 volt DC powersupply and i want to connect it to the opamp. I want to use the opamp as a non-inverting amplifier for a positive Dc input voltage. The drawing must have the + and - inputs of the opamp connected. The drawing must also have the connections from the powersupply to the opamp rails.  I also want a load resistor connected to the opamp output. I hope i'm not asking stupid questions. I'm a beginner and i want to see how all the connections on all the pins of the opamp are made in detail (with all ground symbols etc) in order to make it work perfectly. Many thanks  for helping me out

[RJSV]

   Zero999 supplied a good diagram featuring 1 opamp to create a virtual ground, that won't react to the circuit it feeds.
All you need to do is view those ground symbols, on that gain producing output, on the right side of the schematic view; those symbols are where the virtual ground connects.  Yeah, both opamp inputs are.shown connected.
Sorry, you might have noticed, some of us are learning too, but keep asking if not clear.

[rdl]

I have a LM324N IC. I only want to use one opamp in that IC package, so i only need a drawing of one opamp. I have a single 30 volt DC powersupply and i want to connect it to the opamp. I want to use the opamp as a non-inverting amplifier for a positive Dc input voltage.

If you're only wanting to amplify a DC voltage maybe you don't need the virtual ground. The LM324 is made for use with single supply DC. When operated this way it will work down close to 0 volts. There should be a suitable circuit shown in the data sheet

[robsims]

   Zero999 supplied a good diagram featuring 1 opamp to create a virtual ground, that won't react to the circuit it feeds.
All you need to do is view those ground symbols, on that gain producing output, on the right side of the schematic view; those symbols are where the virtual ground connects.  Yeah, both opamp inputs are.shown connected.
Sorry, you might have noticed, some of us are learning too, but keep asking if not clear.

Now i got it. Thanks. Yes, i understand the schematic of Zero999

[robsims]

Here's a simulation showing the effect of using a potential divider. With 4k7 resistors, which would continuously draw 30/(4.7k+4.7k) = 30/9.4k = 3.2mA, 96mW from 30V, and a 10k load on the op-amp, the supply rails shift by nearly 2.5V relative to 15V, as the op-amp's output swings from -10V to +10V, supplying -/+1mA to the load resistor.

The impedance of the 0V node is equal to R1||R2, which is 4k7/2 = 2k35 in this case. The op-amp supplies 10V/10k = 1mA to the load resistor and a smaller current to R4 and R3, which have a total value of 20k + 180k = 200k and will draw an additional 10/200k = 0.05mA when the op-amp's output is at 10V, giving a total output current of 1.05mA. 1.05m*2.35k = 2.47V, which is seen on +V and -V, as 0V is the fixed reference point and can't be changed. Note the difference between +V and -V is always 30V, assuming V1 is well-regulated. Both +V and -V rise and fall by 2.5V simultaneously.

All the ground symbols are virtual grounds and are connected together. Very good drawing. It shows how to connect everything.  Yes, i got it now. Thanks Zero999.

[Zero999]

Here's a simulation showing the effect of using a potential divider. With 4k7 resistors, which would continuously draw 30/(4.7k+4.7k) = 30/9.4k = 3.2mA, 96mW from 30V, and a 10k load on the op-amp, the supply rails shift by nearly 2.5V relative to 15V, as the op-amp's output swings from -10V to +10V, supplying -/+1mA to the load resistor.

The impedance of the 0V node is equal to R1||R2, which is 4k7/2 = 2k35 in this case. The op-amp supplies 10V/10k = 1mA to the load resistor and a smaller current to R4 and R3, which have a total value of 20k + 180k = 200k and will draw an additional 10/200k = 0.05mA when the op-amp's output is at 10V, giving a total output current of 1.05mA. 1.05m*2.35k = 2.47V, which is seen on +V and -V, as 0V is the fixed reference point and can't be changed. Note the difference between +V and -V is always 30V, assuming V1 is well-regulated. Both +V and -V rise and fall by 2.5V simultaneously.

All the ground symbols are virtual grounds and are connected together. Very good drawing. It shows how to connect everything.  Yes, i got it now. Thanks Zero999.

Yes, but I wouldn't recommend that circuit, because +V and -V vary with the load current. It won't work, if the load on the op-amp is higher. Use the one with the extra op-amp, which is free, as you have unused op-amps in the LM324 package.


For completeness, here's how to connect two or more op-amps in parallel to increase the current capacity of the 0V rail. U1 is the master and U2 the slave. To add another op-amp, connect it as per U2, with the +input to U2's +input and the output connected to the -input and 0V, via a 100R resistor. Negative feedback reduces the effective value of the output resistance to much lower than the output resistors in parallel.

[robsims]

Here's a simulation showing the effect of using a potential divider. With 4k7 resistors, which would continuously draw 30/(4.7k+4.7k) = 30/9.4k = 3.2mA, 96mW from 30V, and a 10k load on the op-amp, the supply rails shift by nearly 2.5V relative to 15V, as the op-amp's output swings from -10V to +10V, supplying -/+1mA to the load resistor.

The impedance of the 0V node is equal to R1||R2, which is 4k7/2 = 2k35 in this case. The op-amp supplies 10V/10k = 1mA to the load resistor and a smaller current to R4 and R3, which have a total value of 20k + 180k = 200k and will draw an additional 10/200k = 0.05mA when the op-amp's output is at 10V, giving a total output current of 1.05mA. 1.05m*2.35k = 2.47V, which is seen on +V and -V, as 0V is the fixed reference point and can't be changed. Note the difference between +V and -V is always 30V, assuming V1 is well-regulated. Both +V and -V rise and fall by 2.5V simultaneously.

All the ground symbols are virtual grounds and are connected together. Very good drawing. It shows how to connect everything.  Yes, i got it now. Thanks Zero999.

Yes, but I wouldn't recommend that circuit, because +V and -V vary with the load current. It won't work, if the load on the op-amp is higher. Use the one with the extra op-amp, which is free, as you have unused op-amps in the LM324 package.


For completeness, here's how to connect two or more op-amps in parallel to increase the current capacity of the 0V rail. U1 is the master and U2 the slave. To add another op-amp, connect it as per U2, with the +input to U2's +input and the output connected to the -input and 0V, via a 100R resistor. Negative feedback reduces the effective value of the output resistance to much lower than the output resistors in parallel.

Here is a picture out of the LM324N datasheet . The first circuit is quite simple.  i can amplify a small signal with it. Then i will connect the output  to a transistor to increase the current. Think this is good too

[RJSV]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

[robsims]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

[Zero999]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

Then you need a comparator circuit, not an amplifier.

The easiest way to do this is to buy an undervoltage protection relay.

[robsims]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

Then you need a comparator circuit, not an amplifier.

The easiest way to do this is to buy an undervoltage protection relay.

 Many thanks Zero999. I'll read a few articles about comparator circuits

[robsims]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

Then you need a comparator circuit, not an amplifier.

The easiest way to do this is to buy an undervoltage protection relay.

Look at this link   
 I can use use one of the four opamps on my LM324 IC as a comparator. With some other components (zener diode etc)  i can make my mains AC undervoltage protection device. Nice

[EPAIII]

This reply shows what I was talking about. One section of the op amp for the creation of the needed ground reference point and ANOTHER section of the op amp to do the job you want done.

If you try to use just a pair of resistors (with or without the filter capacitors) for the creation of the ground reference point, then ALL of the other circuit connections that are made to that resistor junction will be trying to pull the Voltage at that point up or down. This could lead to instability. For just one example, notice that the load is connected to this point. So, if the output to the load causes enough current through that load, it can pull the created ground point up or down in the same direction. That would then pull the + input of the op amp in that same direction. And this would be a positive feedback situation. The output could easily lock up at either the positive or negative supply Voltage. This may or may not happen, depending on the values of components in the circuit. But even if it does not lock up, this can also change the gain of the op amp, perhaps giving an incorrect output.

There is no point in my making another drawing because it would be like the one that Zero999 made with two sections of the op amp. This is what I would recommend.

PS: Comparator ICs are a lot like op amps: + and - high impedance inputs, high gain, etc. They are just optimized for different purposes. So a section of a multiple op amp chip can be used as a comparator in most cases.

Here's a simulation showing the effect of using a potential divider. With 4k7 resistors, which would continuously draw 30/(4.7k+4.7k) = 30/9.4k = 3.2mA, 96mW from 30V, and a 10k load on the op-amp, the supply rails shift by nearly 2.5V relative to 15V, as the op-amp's output swings from -10V to +10V, supplying -/+1mA to the load resistor.

The impedance of the 0V node is equal to R1||R2, which is 4k7/2 = 2k35 in this case. The op-amp supplies 10V/10k = 1mA to the load resistor and a smaller current to R4 and R3, which have a total value of 20k + 180k = 200k and will draw an additional 10/200k = 0.05mA when the op-amp's output is at 10V, giving a total output current of 1.05mA. 1.05m*2.35k = 2.47V, which is seen on +V and -V, as 0V is the fixed reference point and can't be changed. Note the difference between +V and -V is always 30V, assuming V1 is well-regulated. Both +V and -V rise and fall by 2.5V simultaneously.

All the ground symbols are virtual grounds and are connected together. Very good drawing. It shows how to connect everything.  Yes, i got it now. Thanks Zero999.

Yes, but I wouldn't recommend that circuit, because +V and -V vary with the load current. It won't work, if the load on the op-amp is higher. Use the one with the extra op-amp, which is free, as you have unused op-amps in the LM324 package.


For completeness, here's how to connect two or more op-amps in parallel to increase the current capacity of the 0V rail. U1 is the master and U2 the slave. To add another op-amp, connect it as per U2, with the +input to U2's +input and the output connected to the -input and 0V, via a 100R resistor. Negative feedback reduces the effective value of the output resistance to much lower than the output resistors in parallel.

[radiolistener]

it's very hard to split PSU output from unipolar to bipolar, you can't do it without energy loss for transformation.

You can do unipolar to bipolar DC transformation by implementing powerful sine generator powered from unipolar DC, then put that sine into primary coil of the transformer and use two secondary coils with rectifiers to power bipolar regulator.

By using higher frequency you can reduce transformer size and weight. So, it's better to use 400 Hz for transformation instead of usual 50/60 Hz which is used for mains.

If possible, it's better to replace unipolar PSU with bipolar one. It will be much easy, cheaper and more efficient than adding unipolar to bipolar transformation module. If your device is powered from battery, you can add second battery to get bipolar power source.

[Zero999]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

Then you need a comparator circuit, not an amplifier.

The easiest way to do this is to buy an undervoltage protection relay.

Look at this link   
 I can use use one of the four opamps on my LM324 IC as a comparator. With some other components (zener diode etc)  i can make my mains AC undervoltage protection device. Nice

I haven't watched the video and don't have time to at the moment.

Don't mess around with the mains. Run the whole circuit off a small transformer, with a bridge rectifier and smoothing capacitor. The secondary voltage will vary, depending on the mains voltage. The ratio won't be as per the transformer's primary to secondary voltage because it will be specified for the rated output voltage, at the rated maximum current and will be higher with small loads.

The comparator circuit can monitor the rectified voltage on the smoothing capacitor, but it might be better to use a transformer with two secondary windings and use one to power the circuit and the other to monitor the voltage.

The LM324 will work as a comparator, but there are more suitable ICs. The LM311 is good because its output can directly sink 50mA, which is enough to drive a small relay, without an external transistor.

[robsims]

Watching this play out, I'm hoping you can ask any needed questions about GAIN and negative feedback, which has some subtle aspects, non-intuitive, like: assuming OPAMP has 'infinite' gain, by itself (open loop), actually might be 20,000 X gain, but open-ended circuits are too sensitive to ambient temperature.
Also, the input currents are considered as 'zero', while the input resistance is...infinite.  Those exaggerations aren't true of course, but negative feedback can make it so the circuit surrounding an OPAMP solely sets the gain.  Like, if output is reduced by 10X for the feedback, then, looking towards output, you can say it the other way; The output is 10X the input.  That all works out, using ohms law and current flow results, in the 2 feedback resistors.
   A lot of history, on how those OPAMP characteristics got set up, please search on
   'Ron Widler' and the 'current mirror',
   Fairchild Electronics Co.
Thanks, -Rick

Many thanks RJHayward, i have read  a few articles on opamp "rules", pos and negative feedback, i also looked at youtube vids explaining opamps such as dave jones. I'm going to setup the LM324 opamp on a breadboard and experiment with it. Ultimate goal is to make a switching device to disconnect  mains AC when there is an undervoltage. Mains AC is 120 volts and the device has to disconnect mains AC when the voltage is around 90 volts

Then you need a comparator circuit, not an amplifier.

The easiest way to do this is to buy an undervoltage protection relay.

Look at this link   
 I can use use one of the four opamps on my LM324 IC as a comparator. With some other components (zener diode etc)  i can make my mains AC undervoltage protection device. Nice

I haven't watched the video and don't have time to at the moment.

Don't mess around with the mains. Run the whole circuit off a small transformer, with a bridge rectifier and smoothing capacitor. The secondary voltage will vary, depending on the mains voltage. The ratio won't be as per the transformer's primary to secondary voltage because it will be specified for the rated output voltage, at the rated maximum current and will be higher with small loads.

The comparator circuit can monitor the rectified voltage on the smoothing capacitor, but it might be better to use a transformer with two secondary windings and use one to power the circuit and the other to monitor the voltage.

The LM324 will work as a comparator, but there are more suitable ICs. The LM311 is good because its output can directly sink 50mA, which is enough to drive a small relay, without an external transistor.

Many thanks for the advice Zero999

[RJSV]

   (I'm learning things, just by skimming through the various responses).
   I did amateur OpAmp circuit explorations using the LM-324 and the LM-339 comparator.  The LM-339 is open collector output, so you simply need to connect a pull-up, like 2.2 kohms, between output and the (+) supply point.
   I wouldn't read the SPEC as "32 V supply required", but for me, more conveinient  to just buy a 9V battery, although they only supply moderate power / life.
Plus, any boo-boos (tech term, eh), any accidental shorts won't be too much drama.
   I do remember something about, in the LM-324 Spec, something about the IC output voltage can't go up real close to the supply...like only up to supply minus 1.5 v.
I'll try check that soon.

   Nice country, I looked it up, thinking it was maybe near Cambodia or Australia.  Florida climate there ?

[robsims]

   (I'm learning things, just by skimming through the various responses).
   I did amateur OpAmp circuit explorations using the LM-324 and the LM-339 comparator.  The LM-339 is open collector output, so you simply need to connect a pull-up, like 2.2 kohms, between output and the (+) supply point.
   I wouldn't read the SPEC as "32 V supply required", but for me, more conveinient  to just buy a 9V battery, although they only supply moderate power / life.
Plus, any boo-boos (tech term, eh), any accidental shorts won't be too much drama.
   I do remember something about, in the LM-324 Spec, something about the IC output voltage can't go up real close to the supply...like only up to supply minus 1.5 v.
I'll try check that soon.

   Nice country, I looked it up, thinking it was maybe near Cambodia or Australia.  Florida climate there ?

Many thanks for the info. Yes Suriname is beautiful and the climate here is just as in Florida, minus the storms

[robsims]

I have some BA10339 IC's lying on scrap boards. I've read the datasheets and these are real comparators. I think i can use these to make the undervoltage protection device. Who thinks otherwise or has some additional information i can use?

[TimFox]

it's very hard to split PSU output from unipolar to bipolar, you can't do it without energy loss for transformation.

You can do unipolar to bipolar DC transformation by implementing powerful sine generator powered from unipolar DC, then put that sine into primary coil of the transformer and use two secondary coils with rectifiers to power bipolar regulator.

By using higher frequency you can reduce transformer size and weight. So, it's better to use 400 Hz for transformation instead of usual 50/60 Hz which is used for mains.

If possible, it's better to replace unipolar PSU with bipolar one. It will be much easy, cheaper and more efficient than adding unipolar to bipolar transformation module. If your device is powered from battery, you can add second battery to get bipolar power source.

It's actually very easy to use a rail splitter to obtain bipolar DC voltages from a (floating) unipolar supply.
It will be inefficient if there is a substantial unbalanced load, such as a high current from the output of an op amp powered by the bipolar supply.
In the case I cited of a Wiltron analog phase meter, where there was a string of op amps operating from +/- 15 V, AC coupled from one to the next, where the DC output current of each stage was low, it was a sensible design decision to use a 30 V supply and a simple rail splitter.
If you use a high-frequency generator, you may have to contend with noise from that circuitry.
Again, this is a quantitative question.

[Zero999]

it's very hard to split PSU output from unipolar to bipolar, you can't do it without energy loss for transformation.

You can do unipolar to bipolar DC transformation by implementing powerful sine generator powered from unipolar DC, then put that sine into primary coil of the transformer and use two secondary coils with rectifiers to power bipolar regulator.

By using higher frequency you can reduce transformer size and weight. So, it's better to use 400 Hz for transformation instead of usual 50/60 Hz which is used for mains.

If possible, it's better to replace unipolar PSU with bipolar one. It will be much easy, cheaper and more efficient than adding unipolar to bipolar transformation module. If your device is powered from battery, you can add second battery to get bipolar power source.

It's actually very easy to use a rail splitter to obtain bipolar DC voltages from a (floating) unipolar supply.
It will be inefficient if there is a substantial unbalanced load, such as a high current from the output of an op amp powered by the bipolar supply.
In the case I cited of a Wiltron analog phase meter, where there was a string of op amps operating from +/- 15 V, AC coupled from one to the next, where the DC output current of each stage was low, it was a sensible design decision to use a 30 V supply and a simple rail splitter.
If you use a high-frequency generator, you may have to contend with noise from that circuitry.
Again, this is a quantitative question.

It's possible to do it with high efficiency with a switched mode power supply.

[TimFox]

Or, as one of my German-citizen physics professors replied to a different suggestion:  "You could also pull up your pants with tongs."