How Current Limitation is happening in the circuit???

[ommsiva]

""[6] 680 Ohms is probably the right value to limit current through the LED which may be just 20ma. ""

Sir,

if current taken by load is more, inverting input will be high. Making the output of opamp -6V.
This -6v forward biases the LED because it is been connected to cathode and Anode is at ZERO. the adj pin and output pin has a loop with opamp which is open because D1 is short and I adj current  flows in D1 , D2 and to opamp.

I=(6-3)/680--------------(3v is considered cut in potential of LED)
I=4.41mA

How it is 20mA?

Correct me if i am wrong..

[MrAl]

""[6] 680 Ohms is probably the right value to limit current through the LED which may be just 20ma. ""

Sir,

if current taken by load is more, inverting input will be high. Making the output of opamp -6V.
This -6v forward biases the LED because it is been connected to cathode and Anode is at ZERO. the adj pin and output pin has a loop with opamp which is open because D1 is short and I adj current  flows in D1 , D2 and to opamp.

I=(6-3)/680--------------(3v is considered cut in potential of LED)
I=4.41mA

How it is 20mA?

Correct me if i am wrong..

Hi,

No that was just a quick guess about what they wanted to limit it to not that it would always be that level.
With a 20ma LED you may even want to limit it to 10ma, but that does not mean it cannot be just 2ma, which it would probably be sometimes or maybe most of the time.  So that limit is just to protect the LED just in case it tries to go higher.  It may never go higher though, and the limit could even be much lower.
I sometimes have limited my 20ma LEDs to just 5ma as that provides a lot longer life for the LED.  At 20ma the life is shorter, but even at 10ma the life is much longer, and at 5ma the half-life will be much, much longer.  If you need a lot of brightness (as in a flashlight) you may not be able to do this, but if not, it makes sense to limit the current to a lower level.

BTW I should mention that this circuit is a very old circuit probably appearing in application books as far back as 1980 or even before that.  A more modern circuit would most likely limit current in a different way, conserving energy.  The 0.2 Ohm resistor at the full load of 5 amps will consume 5 watts, which is a lot by today's standards, and would really require a 10 watt rated resistor which is fairly large.  For a sarcastic view, imagine trying to install a 10 watt ceramic resistor inside a cell phone.
You could look at other ways to do a current limit circuit which would use less power.  There are IC chips made today that are dedicated just for sensing current.  If you wanted to, you could look into that.  If you only use the power supply once in a while though or for short time periods, it may not be worth it to change anything though.  It's always up to you.
We also have to keep in mind this is a LINEAR power supply, so at low to mid output voltages and higher current output the dissipation in the transistor will be quite high, which requires a larger heat sink.  With 30v input and 10v output at 5 amps, the power in the series pass transistor will be 20*5=100 watts.  That's really high.  At 5v output and 5amps, 125 watts, which is also very high. The heat sink has to be huge.  If you look up some heat sinks you can do a few calculations to find out how big it has to be.
Back in the 1980's I worked on a program for Sandia Labs that had really bit heatsinks, about 12 inches wide and 18 inches long and about 4 inches high.  They are not that cheap, and they also take up a lot of room and need free air flow, so it's quite a job to get something like that to be practical.

[ommsiva]

Sir,

This circuit contain voltage regulator where the output voltage can be fixed by P2 and R4.
current control is done by opamp 741 ,R1,R2 and P1  and it forms a closed loop with L200.

My doubt is

1) non inverting input (Pin3) is connected to output which is the second end of current sense resistor and inverting input(pin2) is connected to first end of current sense resistor?

        so always my inverting input will be greater than non inverting input. so results in pin 6 always moves towards ground potential

2)Why we want R1 in the circuit?


Thank you

[MrAl]

Sir,

This circuit contain voltage regulator where the output voltage can be fixed by P2 and R4.
current control is done by opamp 741 ,R1,R2 and P1  and it forms a closed loop with L200.

My doubt is

1) non inverting input (Pin3) is connected to output which is the second end of current sense resistor and inverting input(pin2) is connected to first end of current sense resistor?

        so always my inverting input will be greater than non inverting input. so results in pin 6 always moves towards ground potential

2)Why we want R1 in the circuit?


Thank you

Hi,

It is unclear why they still included R1 in the circuit when they already decided to go with the op amp.  Perhaps there is some transient or startup condition requirements.

As to the output of the 741 going negative, that is easy to explain.
As the voltage across pin 5 and pin 2 of the L200 becomes greater, the output current will be more limited.  This would mean that one way or another we would want to increase the voltage difference when we want the circuit to go into current limit.  Since pin 5 is somewhat constant when compared to pin 2, driving pin 2 lower would accomplish this, and thus the 741 would take over the current limit function.

This must be a very old circuit though.  There are better choices for the replacement of the 741 op amp, most notably the LM358 assuming the response using the 741 was fast enough.  If the 741 was not really fast enough, then the LM358 will not be fast enough either, so an op amp with a faster slew rate might be used instead.  Of course stability is always an issue and has to be checked when using a faster component.

[ommsiva]

How to analyse this circuit?

We have to start from output or from input end. where to start?

[MrAl]

How to analyse this circuit?

We have to start from output or from input end. where to start?

Hello,

You would start by learning to analyze much simpler power supply circuits first.
You would also have to learn how to analyze circuits with bipolar transistors, such as in a differential amplifier.
Also good to know would be how a basic ua723 voltage regulator works.
Another point is to understand how power supply rectifier circuits work.

You probably want to start with rectifier circuits.

[ommsiva]

Sir,

If my transformer is 24v-0-24v @10 ampere, then after bridge rectification and filtering I will get around 24 square root 2 which is 34V.

My question is, I will get plus and minus 34V , How much current can I take from each line?

It's a bipolar regulated power supply. Suggest some methods to get high current negative voltage regulation.
 

Thank you all.

[Jwillis]

When you calculate the total out put power of a linear power supply, calculate it from the VRMS of the transformer and not the peak voltage after rectification. Although the peak voltage is 34V, upon load that voltage will drop significantly down to the VRMS of the transformer. And depending on the transformers own regulation it will drop more at full load. So you will not get the expected 34V output but maybe 23V at full load.
So if you want  to have say a 30V output then you would need at least a transformer above 30V taking in account its regulation. So for example to get 30V output at full load with a transformer with a 4% regulation you would need at least a 32V transformer. Assuming you can find a transformer with 4% regulation below 300 - 400VA. Most fall around 8 - 15% regulation some as much as 30%. Lower VA transformers will have poorer regulation compared to High VA transformers. 800 - 1000 VA transformers will have around 2 to 5% regulation.
So when you look at common linear power supplies on the market you will find that the transformer may be a high as 10V or higher above the power supplies maximum output.
I find that for every amp of output you need a transformer that has that many volts higher then the required output voltage. So for a 30V 10A power supply you would need a 40V transformer to hold regulation.

[ommsiva]

Hi to all,

In internet, they are two ways of increasing current handling capability of LM317 IC.

first circuit: 10 ohm current sensing resistor is present to develop base emitter potential of the transistor which is going to supply extra current.

Second circuit: 22 ohm current sensing resistor is present to develop base emitter potential of 2N2905 ,when collector current flows through 470 ohm resistor and voltage drop across it exceed the barrier potential of TIP73 , it will turn on.

Which circuit is better for practical working condition?

[MrAl]

Hi to all,

In internet, they are two ways of increasing current handling capability of LM317 IC.

first circuit: 10 ohm current sensing resistor is present to develop base emitter potential of the transistor which is going to supply extra current.

Second circuit: 22 ohm current sensing resistor is present to develop base emitter potential of 2N2905 ,when collector current flows through 470 ohm resistor and voltage drop across it exceed the barrier potential of TIP73 , it will turn on.

Which circuit is better for practical working condition?

Hi,

Probably neither unless you like really big heatsinks and pre-switcher era and pre-green era voltage regulator circuits 

Either of those could waste a lot of power, something that is frowned upon in this day and age, and also gets hot which is undesirable as well.
For short term use though it may make more sense.

[ommsiva]

Dear sir,

1)I found this circuit on web, TIP36C and TIP35C is going to do current boosting. Whether the 0.1 ohm /5W resistance should be in collector or emitter terminal?(to share equal current in all transistor)

2) D2 and D3 are 1 amp diode but RPS supplies up to 10 amps of current. Is it correct to use 1AMP diode?

3) Whether current limiting function of LM317 will happen when current drawn by base of TIP35C's exceed  at 2.2Amps?


Thank you

[MrAl]

Dear sir,

1)I found this circuit on web, TIP36C and TIP35C is going to do current boosting. Whether the 0.1 ohm /5W resistance should be in collector or emitter terminal?(to share equal current in all transistor)

2) D2 and D3 are 1 amp diode but RPS supplies up to 10 amps of current. Is it correct to use 1AMP diode?

3) Whether current limiting function of LM317 will happen when current drawn by base of TIP35C's exceed  at 2.2Amps?


Thank you

To understand how this circuit works and answer those questions it would help a lot to do a simulation that's what simulators are made for.  They help understand what happens and what could happen when things vary during the actual operation of the circuit.
Without doing a simulation here are some ideas.

[1]  It's a little difficult to say if the 0.1 Ohm resistors would work better in the emitters than in the collectors, but there does seem to be a strong possibility that idea would make the circuit work better.  For one, the collectors act more like constant currents which means the resistors will not do as much, and also if moved to the emitters a little more voltage drop in the emitter circuits may allow the smaller transistor to work better for limiting the total output current because that would loosen up the saturation voltage requirement of the smaller transistor.
A more easily noticed problem though is the rating of the 0.06 Ohm resistors.  5 watts isnt enough because at some point they could be working too close to that power level.  10 watts (or even higher) would be better, and with free air flow all around.

[2]
Yes, I would say those two diodes should be at least of the 1N5400 series which are 3 amp diodes not 1 amp diodes.  The higher the rating the better as they are for reverse polarity protection against an external active load like a battery.  With the 1N5400 series, the protection relies partly on the behavior of the diode when it blows out.  They are banking on it shorting out.  If it happens to open up, the protection is lost.
There are better ways to do this too but they are more complicated.  For example, the higher power diodes plus a series fuse.  If a reversed active load is connected to the output and the diode conducts, the fuse blows, and the load is disconnected.

[3]
The current to the base of the larger transistors is hard to predict exactly, but if we estimate using a minimum beta of 10 and with the LM317 drawing 1 amp, that means each base current is about 0.3 amps, ideally.  The problem then seems to be that R1 is not the right value if the goal is to get 10 amps  out of the output of this circuit.  With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less.
This means R1 value would have to be decreased.  If it was 0.7 Ohms and with 2 amps through it, the power would be 2.8 watts, so a minimum of 5 watts would be a good starting point for the power rating of that resistor, with free air flow.

These ideas should be verified with a simulation.  This circuit isn't too complicated so it's not hard to use with a simulator.  I might try this myself if I get a chance later today or tomorrow morning.  I'll update with any additional info I might find out.
However, because of the more obvious problems lurking here and there in this circuit, I would study it out carefully before building a prototype.  A simulator would help a lot, followed by a bench test.  One of the things to pay attention to is the power dissipation in the transistors as well as the resistors.  I'd check every resistor power rating to make sure it can handle the required power.  I do not see any way around using somewhat larger heatsinks either for the transistors, and one for each LM3xx device.

Linear power supplies are great for use for short term testing of other circuits and devices.  Not as good for long term use as they eat up too much energy in most cases.


This idea should be studied in more depth with a simulation.

[ommsiva]

Sir, I don't understand this calculation

"With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less"

if emitter takes 1 amp then base current is 0.1A, then three base current add together and makes 0.3A.
Assume VEB of power transistor =1V @ 1A, VE=(Vin-V_0.06)=20-(0.06*3)=19.82V then Base voltage VB should be 1V less, VB=18.82V

[MrAl]

Sir, I don't understand this calculation

"With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less"

if emitter takes 1 amp then base current is 0.1A, then three base current add together and makes 0.3A.
Assume VEB of power transistor =1V @ 1A, VE=(Vin-V_0.06)=20-(0.06*3)=19.82V then Base voltage VB should be 1V less, VB=18.82V

Hi,

Yes, I'm sorry the actual current would be 1/10 of the emitter current so if the emitter current was 1 amp the base current would be 0.1 amps and the total current 0.3 amps (not each base has 0.3 amps).  However, with 9 amps output each emitter would pass roughly 3 amps each, and 3/10 is 0.3 and the total would be close to 1 amp.
This means if the current limit of the LMxxx device was 1 amp with 9 amps total through the emitters, that's 10 amps output, and that's 2 amps through resistor R1 approximately.

If the current limit of the LMxxx device was 1.5 amps, then the emitter currents would total 8.5 amps, with total base currents 0.85 amps, which means current through R1 would be about 2.4 amps.

If the current limit of the LMxxx device was 2 amps, then the emitter currents would total 8 amps, and that means about 0.8 amps into the bases, which is a total of 2.8 amps through the resistor R1. Thus R1 has to be sized based on something like that.

Does this make more sense now?

The main idea though is that the point of current limit is reached when the LMxxx device goes into current limit as that limits the current through the bases as well because the voltage drop across R1 will be limited.
Because we are dealing with two base emitter drops though this could change significantly with temperature rise, and the temperature will rise.  This means it is a current limit but not a precision current limit at least until the whole circuit comes up to operating temperature and the ambient temperature is constant.

I also recommended checking this circuit over very carefully because there were obvious errors.  Even with only 1 amp through that 2 Ohm resistor we see 2 watts, and the resistor is rated for 2 watts.  That means even with that unreal lower current the resistor would get way too hot.

[ommsiva]

good morning to all,

for past 2 months i was involved in studying power supply design and now I started building Lab power supply with LM317 and 337.
Positive supply was adjustable from 1.2 to 28V, but negative was starting from -8 to -24v.

Initially when i soldered and connected the circuit, my adjust pin of 337 was grounded and IC was heated, i have corrected that now and output was adjustable from -8 to -24V.

My question,

1) I have 7805 IC for powering peripheral circuit, after capacitive filtering the voltage was 35V.the output of 7805 is 10V instead of 5V, why?
2) why my negative supply was from -8 to 24V?
3) how to check LM337IC regulator?

Thanks you all

[xavier60]

The outputs should be 27V and -27V with the values shown.
Best way to check them is by function in a simple test circuit, https://uk.farnell.com/lm317-resistor-and-output-voltage-calculator
If R2 is zero ohms, the output should regulate at close to 1.25V
Note that the LM317 and LM337 pinouts are different.

For the LM337, https://www.datasheets.com/de/reference-design/typical-application-circuit-for-lm337-1-5-a--adjustable-output--negative-voltage-regulator-125315

[ommsiva]

Sir,

Please suggest simple method of deriving High current negative adjustable regulator.

[xavier60]

The design in Reply #39 would be ok with current limiting added.
Have you tested the LM337?

[ommsiva]

Yes. Two times. LM337.


Last time 22 ohm current sensing resistor had smoke.


Any body can propose simple method of obtaining high current adjustable negative voltage

[MrAl]

Yes. Two times. LM337.


Last time 22 ohm current sensing resistor had smoke.


Any body can propose simple method of obtaining high current adjustable negative voltage

Hi,

You probably have to learn how to troubleshoot these circuits as well as build them.
You use a meter and measure voltages and currents and try to figure out what is not right.
Most of the time you have to just use a little power at first to test.  That would mean something like if you need 30 volts input you start with maybe 5 to 10 volts input and a light load, then try to figure out what currents are flowing and what voltages are present and see if anything looks wrong as you go.

I say all this because you got the LM317 version going so you should be able to get the LM337 version going too.
One of the big differences is the external transistors being used for the LM337 and the LM317.
As you power up, compare the voltages and currents you measure in the LM317 circuit to the LM337 circuit and see what is different about the LM337 circuit, other than the polarities.

[ommsiva]

Sir,

please find the attached circuit, I connected this in bread board. The 120 ohm of 1Watt was fired two times. I cant find the reason for its breakdown. got frustrated after building this circuit.

[xavier60]

That circuit is correct. It should work properly if wired correctly and the LM337 and transistors aren't damaged. These parts can be unforgiving to wiring mistakes.
Also, breadboard isnt good for this kind of circuitry. It's best to have the LM337 secured to a heatsink and the passive components soldered directly to it, including bypass capacitors to prevent possible oscillation.
 As suggested earlier, start simple. Wire up a test circuit with a new LM337 without the transistors and run it from a low voltage low power source.
And show us some pictures.

[ommsiva]

Sir,

how much current can be obtained if I use 24v-0-24v , 10 ampere centre tapped transformer for positive and negative output each


According to theory output would be maximum 23v@ 6ampere on positive and negative each regulated outputs.

[MrAl]

Sir,

how much current can be obtained if I use 24v-0-24v , 10 ampere centre tapped transformer for positive and negative output each


According to theory output would be maximum 23v@ 6ampere on positive and negative each regulated outputs.

Hi,

What theory is that?

You have to know the Volt Ampere rating of the transformer.  It sounds like it could be a 240VA transformer but that's better obtained from the manufacturer.

[ommsiva]

Sir,

Rms power= 48*10=480va

Average power=(.9*Vac)(.6*Iac)
=(.9*24)(0.6*10)
=21.6*6
=129.6 watts
On end of coil I could take=129.6 watts

Overall I could take 260watt from 480va transformer.