Common Ground question for Dual (+/-) Power Supply

[FenderBender]

Hello,

Inspired by the PSU videos, I am trying to design a dual power supply based on the LT3080. it will use 4 of these regulators to provide current limiting and voltage adjustment: 2 per rail.

Since I would like to use this supply to power various projects including op amps, power amps, and other devices requiring dual supply, I need to be able to have a common ground between the two of them.

In order to achieve this, I thought about some previous concepts I understand. If I have two 9V batteries in series, you create a center tap ground between the two batteries and each of the other two terminals will have a positive and negative potential in reference to the center tap.

My question is: Will this work the same if I have two of these identical power supplies and I connect the + of one to the - of the other. Will that work?

Thanks.

[sacherjj]

Yes.  As long as the ground for the PSU is not tied to the earth ground.  (That is sometimes done, but not a good idea for this reason.)

Assuming they were set to +/- 6V.  Ground for one might be -6 V in reference to the other power supply.  That supply's ground would look like +6 V to the first supply.

I would encourage you to look at doing current limiting and voltage limiting with the same regulator.  That isn't much more difficult and saves money in regulator and increases the possible voltage swing.

[Mechatrommer]

your concept about +/- power is about right. except the 4 x LT3080 which i'm not sure. i believe LT3080 is "positive regulator" ie source current, not sink (for -ve rail), but then i'm not so sure, havent look at the datasheet.

[sacherjj]

your concept about +/- power is about right. except the 4 x LT3080 which i'm not sure. i believe LT3080 is "positive regulator" ie source current, not sink (for -ve rail), but then i'm not so sure, havent look at the datasheet.

He is talking about making two positive regulated supplies.  Then when you connect them to each other and set one +/- junction as ground, you have dual voltage supplies.  This works, since the transformers allow all the power supply to float until you define a reference ground.

[FenderBender]

Thanks. Will I need two transformers? I have a feeling I will...

Heavy influence from the PSU videos. Thinking of driving the 3080 set pin with a LM317LZ (100ma) adjustable regulator and an op amp with a feedback loop. No digital for me..yet. still trying to study Dave's schematic and figure out what resistors I'm using.

[Psi]

Will I need two transformers? I have a feeling I will...

You can use one transformer if it has two separate output windings.
That way there is no electrical connection between the two until the point after the regulators when you create a common.

[FenderBender]

That's what I thought. Might as well just use two unless I can find a cheap-ish toroid.

[sacherjj]

That's what I thought. Might as well just use two unless I can find a cheap-ish toroid.

You are most likely going to be cheaper for two EI styles over a dual secondary EI or toroid. 

[FenderBender]

Righto.

After reviewing my thoughts, it seems rather...foolish to use the LT3080 at all. It's no doubt a good reg but the cost of 4 of them is close to $20. Plus all this nonsense with driving pins, while there are perfectly acceptable products that do not require this and work fine with simple voltage dividers. I'm not trying to drive it with an MCU.

Instead, I am revisiting the L200 regulator which can use an external op-amp to limit current. The L200 is not perfect; I understand. It may need a little extra love to get good results, but it's still a good robust and versatile reg.

Datasheet is here: http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00000053.pdf

Figure 23 shows the adjustable voltage and current limit set up. For the feedback loop on the 741, can I change the values of the resistor and potentiometer on the divider? Can I make 1k -> 100R and the 100k pot -> 10k. It's surprisingly difficult to find 100k pots with the specs I want for whatever reason...

In the same manner, I also happen to have a pretty expensive 5k 10turn pot that I was going to use for another project.
I'm going to change the 10k pot they have speced for volt-adjust to a 5k pot. Besides, that would give me a max voltage of around 15V versus ~30V with the 10kpot. Less useless turns on the pot that do nothing. Better resolution also?

Does this seem a more viable solution?

Thanks.

[Psi]

It's easier to use a positive and a negative regulator IC. Then you can use a center-tapped transformer which will be easier to source.

[FenderBender]

It's easier to use a positive and a negative regulator IC. Then you can use a center-tapped transformer which will be easier to source.

Perhaps, but then I must discern a simple way to apply current limiting to each rail. I'm sure it is not hard, but my experience is limited, though I would like to design my own.

[sacherjj]

The one advantage of using two separate positive regulators is that you have the option of series or parallel connection, to give you double voltage (plus +/- supply) or double current in a single supply. 

The one big advantage of the LT3080 is that is can go down to 0V.  This can be faked with cheaper regulators, by setting up the reference to be -1.2V off of a center tapped rectifier circuit or something similar.

[FenderBender]

Thank you very much everyone. I think I will stick with the dual L200s and the external op-amps...I've been debating for months about how I should go about building this. Perhaps you've seen me talk about it before...but anyway...

Can someone help me with my question about the feedback loop resistors above?

Thanks.

[TerminalJack505]

Figure 23 shows the adjustable voltage and current limit set up. For the feedback loop on the 741, can I change the values of the resistor and potentiometer on the divider? Can I make 1k -> 100R and the 100k pot -> 10k. It's surprisingly difficult to find 100k pots with the specs I want for whatever reason...

I took a look at the datasheet and from what I can tell, here's what is going on with that feedback loop...

Take all of this with a large grain of salt.  My analysis may be completely wrong.  You would definitely want to double-check my analysis and breadboard this up and test it out.

First, I don't really know what R1 is there for, honestly.  I'm pretty sure its value can remain as-is, though.  It doesn't factor-in when determining the maximum current.  From what I can tell, it is just some kind of surrogate for the current sense resistor that is normally between pins 5 and 2.

The op amp along with R2 and P1 are what set the voltage at pin 2--the current limit pin.

The op amp's output voltage--pin 2--is,

  V₂ = V_o - I_o*R3*P1/R2

or

  V₂ = V_o - V_sense*P1/R2

where,  V_sense is the voltage drop across the current sense resistor, R3.  V_o is the regulated output voltage, as labeled in the schematic.  I_o is the output current.

Note that V₅, the voltage at pin 5 is just,

  V₅ = V_o + V_sense

Voltage limiting kicks in when V₂ is 0.45 volts below V₅.  V₅ is at most 0.45V above V₂, in other words.

The maximum output current is when V₅ - V₂ is 0.45V.  So,

  I_o(max) = 0.45V / (R3 * (1 + P1/R2) )

You can use this equation to determine the size for R2 when the values of P1 and R3 are already set.

The larger the (user set) value of the potentiometer P1, the smaller I_o(max) will be.  So if you plug P1's maximum value into the equation above you will solve for the smallest possible I_o(max).

When P1 is zero ohms then I_o(max) will basically be:

  I_o(max) = 0.45V / R3

Which for the schematic in figure 23 would be 4.5A.  It says 1.5A so I don't know where that comes from.  This may indicate a problem with my analysis.

To set an upper limit for I_o(max) you could simply put a resistor in series with P1--call it R5.  Essentially, this gives a minimum value for P1. 

The equation for I_o(max) in this case would be:

  I_o(max) = 0.45V / (R3 * (1 + (P1 + R5)/R2) )

So when P1 is zero R5 would determine I_o(max)'s upper limit.  R5 will affect the lower limit as well, obviously.

If you choose upper and lower bounds for I_o(max), set R3 to 0.1 (as in the schematic) and P1 to 10K then you can solve for R2 and R5.

For a lower bound of 20mA and an upper bound of 1.5A...

  0.020A = 0.45V / (0.1 * (1 + (10K + R5) / R2))
  1.5A     = 0.45V / (0.1 * (1 + (  0   + R5) / R2))

Re-writing this as a linear system gives:

  0.448*R2 - 0.002*R5 = 20
  0.3*R2     - 0.15*R5   = 0

Solving this system gives the values:

  R2 = 45.045
  R5 = 90.0901

, which will give you ballpark values for R2 and R5.  You could use E96 values of 45.3 and 90.9, for example.  If you plug these values back into the equations you can figure out the exact upper and lower limits for I_o(max).

Note that this is probably why they used a 470 ohm R2 and a 100K P1 in their example.  Remember, they didn't have a resistor (which we've called R5) in series with P1.  That will limit the worst-case current through R2 and P1 to just less than 1 mA (0.45V / 470). 

The worst case current for a R2 of 45.3 and a P1 of 10K in series with 90.9 would be 3.3 mA (0.45V / 136.2).  Which shouldn't be a problem.  But if you didn't have R5 then this would be about 10 mA (0.45V / 45.3).

Hopefully that gives you some idea about what seems to be going on and how to select a value for R2.  R1 should be left as-is at 1K.

[TerminalJack505]

I think I may have solved the mystery of R1.  Like I said in my earlier post, I didn't know what purpose it was supposed to serve.

It looks like the schematic is likely wrong.  R1 most likely is supposed to go in series with P1--like I proposed doing with the resistor I named R5.

This will make my analysis jive with the figures shown on the schematic.

Namely, the maximum shown output current of 1.5A.

  I_o(max) = 0.45V / (0.1 * (1 + 1000/470))

Yields ~1.44A.

[FenderBender]

Figure 23 shows the adjustable voltage and current limit set up. For the feedback loop on the 741, can I change the values of the resistor and potentiometer on the divider? Can I make 1k -> 100R and the 100k pot -> 10k. It's surprisingly difficult to find 100k pots with the specs I want for whatever reason...

I took a look at the datasheet and from what I can tell, here's what is going on with that feedback loop...

Take all of this with a large grain of salt.  My analysis may be completely wrong.  You would definitely want to double-check my analysis and breadboard this up and test it out.

First, I don't really know what R1 is there for, honestly.  I'm pretty sure its value can remain as-is, though.  It doesn't factor-in when determining the maximum current.  From what I can tell, it is just some kind of surrogate for the current sense resistor that is normally between pins 5 and 2.

The op amp along with R2 and P1 are what set the voltage at pin 2--the current limit pin.

The op amp's output voltage--pin 2--is,

  V₂ = V_o - I_o*R3*P1/R2

or

  V₂ = V_o - V_sense*P1/R2

where,  V_sense is the voltage drop across the current sense resistor, R3.  V_o is the regulated output voltage, as labeled in the schematic.  I_o is the output current.

Note that V₅, the voltage at pin 5 is just,

  V₅ = V_o + V_sense

Voltage limiting kicks in when V₂ is 0.45 volts below V₅.  V₅ is at most 0.45V above V₂, in other words.

The maximum output current is when V₅ - V₂ is 0.45V.  So,

  I_o(max) = 0.45V / (R3 * (1 + P1/R2) )

You can use this equation to determine the size for R2 when the values of P1 and R3 are already set.

The larger the (user set) value of the potentiometer P1, the smaller I_o(max) will be.  So if you plug P1's maximum value into the equation above you will solve for the smallest possible I_o(max).

When P1 is zero ohms then I_o(max) will basically be:

  I_o(max) = 0.45V / R3

Which for the schematic in figure 23 would be 4.5A.  It says 1.5A so I don't know where that comes from.  This may indicate a problem with my analysis.

To set an upper limit for I_o(max) you could simply put a resistor in series with P1--call it R5.  Essentially, this gives a minimum value for P1. 

The equation for I_o(max) in this case would be:

  I_o(max) = 0.45V / (R3 * (1 + (P1 + R5)/R2) )

So when P1 is zero R5 would determine I_o(max)'s upper limit.  R5 will affect the lower limit as well, obviously.

If you choose upper and lower bounds for I_o(max), set R3 to 0.1 (as in the schematic) and P1 to 10K then you can solve for R2 and R5.

For a lower bound of 20mA and an upper bound of 1.5A...

  0.020A = 0.45V / (0.1 * (1 + (10K + R5) / R2))
  1.5A     = 0.45V / (0.1 * (1 + (  0   + R5) / R2))

Re-writing this as a linear system gives:

  0.448*R2 - 0.002*R5 = 20
  0.3*R2     - 0.15*R5   = 0

Solving this system gives the values:

  R2 = 45.045
  R5 = 90.0901

, which will give you ballpark values for R2 and R5.  You could use E96 values of 45.3 and 90.9, for example.  If you plug these values back into the equations you can figure out the exact upper and lower limits for I_o(max).

Note that this is probably why they used a 470 ohm R2 and a 100K P1 in their example.  Remember, they didn't have a resistor (which we've called R5) in series with P1.  That will limit the worst-case current through R2 and P1 to just less than 1 mA (0.45V / 470). 

The worst case current for a R2 of 45.3 and a P1 of 10K in series with 90.9 would be 3.3 mA (0.45V / 136.2).  Which shouldn't be a problem.  But if you didn't have R5 then this would be about 10 mA (0.45V / 45.3).

Hopefully that gives you some idea about what seems to be going on and how to select a value for R2.  R1 should be left as-is at 1K.

Thank you so much for writing this. Indeed, it can be hard to analyze stuff like this. I'll take your word for it and we'll see what we can come up with.

Thanks.

[FenderBender]

I think I may have solved the mystery of R1.  Like I said in my earlier post, I didn't know what purpose it was supposed to serve.

It looks like the schematic is likely wrong.  R1 most likely is supposed to go in series with P1--like I proposed doing with the resistor I named R5.

This will make my analysis jive with the figures shown on the schematic.

Namely, the maximum shown output current of 1.5A.

  I_o(max) = 0.45V / (0.1 * (1 + 1000/470))

Yields ~1.44A.

Quite interesting. I found this article: http://www.flyelectric.ukgateway.net/lithium-charger.htm

It is supposed to be a revised version of the L200 datasheet version with better linearity. Configuration seems to be very similar in regard to R1 and he has verified both work (L200 datasheet and his own). I'm sorry I can't be of any more help!

But that's what I've seen.

[TerminalJack505]

If you have an L200 and you are going to base your circuit on the datasheet figure 23 example then I would definitely breadboard it and make sure it behaves properly at the upper limit of the current limit.  I think that you will find that the circuit as shown will actually allow for more than the 1.5A that they say. 

That may not be a big deal, though.  The regulator has an internal limiter at 2A.

[TerminalJack505]

I played around with this circuit using a math application (SMath Studio) that Amspire pointed out in another thread.  (Nice software, by the way.) 

I also double-checked my analysis with a simulator.  The simulation agrees with my analysis.  The simulation also verifies that R1 has no affect on the circuit's behavior.  Which, if you think about it, how could it?  It is just sitting there between two low impedance voltage sources.  Although there may be some start-up condition that it is being used for.

The first image is a graph of the maximum allowed current, I_o(max), as a function of the setting of a 10-turn potentiometer.  This is in-line with the poor linearity mentioned by the the website you found.

The red line is a graph of the circuit as it is in the datasheet.  I cropped it out of the graph but it allows a maximum I_o(max) of 4.5A.  The blue line is a graph of the circuit with the changes I suggested.

As you can see, the circuit as-is (or the circuit with my suggested changes) is going to be far from ideal.

The second image is just the function of the potentiometer and a sample plot of a ten turn 10k pot.

I don't have any suggestions so far as making it more linear.  I just want to make sure you are aware of what you can expect.

[FenderBender]

I see...Quite unlinear. You didn't happen to simulate the other one I linked, did you? http://www.flyelectric.ukgateway.net/lithium-charger.htm


Thanks for the help.

[TerminalJack505]

I see...Quite unlinear. You didn't happen to simulate the other one I linked, did you? http://www.flyelectric.ukgateway.net/lithium-charger.htm


Thanks for the help.

No, I sure didn't.  When I first looked at it I thought it was basically the same circuit but with different component values.  After looking at it again I notice that the op amp's inverting input is taken from the 10k pot's wiper so that may make a bigger difference than I thought.

I'll see what I can come up with and let you know what I find out.

[FenderBender]

Thanks a bunch.

[TerminalJack505]

Here are plots for how the 'Fly Electric' circuit will respond as the two pots are varied. 

The x axis represents the setting of the 250k potentiometer.  This is similar to the earlier graph.

Each of the different lines represent different settings of the 10k potentiometer R6 (named P2 in the function definition.)  For comparison I also plotted the graph for the circuit from the datasheet.  (This is the very bottom, dark red line.)

If you refer to the second image the lines on the graph appear in the same order on the graph.  (Reading from the top of the graph and preceding downward.)

The P2 variable of the function represents the resistance that appears on the left-hand side of the wiper.  This resistance will be in series with R7 (aka P1.)  The remaining resistance appears on the right-hand side and in series with R2.

[FenderBender]

Wow Terminal- Thanks!

So I thought I understood the graph before but now I'm a bit confused.

On the x-axis- What units is that in? Ohms? Because you said it was the setting of the 250k pot, but I only see up to ~10?

The y-axis is the current limit? So at f(2) for the top blue line, the current limit is 2.25A? Does that make sense?

And the rest of them seem to almost go on forever upwards...What does that mean?

[TerminalJack505]

On the x-axis- What units is that in? Ohms? Because you said it was the setting of the 250k pot, but I only see up to ~10?

The x-axis is the setting of a 10-turn potentiometer.  (The number of turns from the leftmost position.)  I used this in the original graph since you mentioned you were wanting to use a 10k, 10 turn pot.  I used the same units in this graph just so we'd be comparing apples-to-apples.

The y-axis is the current limit? So at f(2) for the top blue line, the current limit is 2.25A? Does that make sense?

Yes, the y-axis is the maximum allowed current: I_o(max).  The top blue line would have a minimum I_o(max) of around 1.85A and go upward as the 250k pot is changed.

And the rest of them seem to almost go on forever upwards...What does that mean?

They will all max out when they cross the x=10 axis since the pot can't be turned any further. 

What it really means though is those settings to the far left are pretty useless.  Particularly for a power supply since 1) the part already maxs-out at 2A and 2) the minimums are too high.  The circuit was meant for charging batteries so it may have been okay for that.

If you are set on using the L200 then you are probably going to have to introduced another op amp for the current limiting feature to get both the range and linearity that you would need.