How Current Limitation is happening in the circuit???

[MrAl]

Sir,

1)Normally Base emitter voltage V_BE=V_B-V_E , But you have taken V_BE=V_B+V_E.
What is the reason for that?

2)What is the contribution of these terms interms of current limiting from these equation, Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)-Io*R5 ?

for example
when load current is 10 amps,V_R5=10*0.055=0.55V, then what is the contribution of Vref*(R6+Rd))/(R6+R2+Rp) term?

                           the first term should be 1.25V (0.7V+0.55V) and second term should be 0.55V. is that correct?

Thank you

Hello again,

[1]
I do not see Vb+Ve anywhere in my replies so you will have to explain what gave you that idea.
The Vbe voltage is obviously the Vb minus the Ve, but those are implied in my equations.  If you calculate Vbe from my equations you will see that Ve does not have to be mentioned because Ve is at absolute ground (0v) and so calculating Vbe would be the same as calculating Vb because Vb-0v is Vbe.
If you still think there is a mistake, just explain how you got the idea that I was adding Vb and Ve and I'll look it over.

[2]
The contribution from the other resistors and Vref come in as an addition to the voltage across the current sense resistor, as can be seen in the final formula.  In other words, we are not just sensing Io (Iout) we are biasing that with a small voltage in order to get an adjustment range using a rather larger value potentiometer.  Had we just used a potentiometer, it would have to be on the order of 0.1 Ohms at some 10 watts or something.  By using a biasing voltage, we can use a larger value potentiometer which is more common and more easily obtained.  We might even be able to up that to 1k instead of 500 Ohms if we change R2 to around 12k or something like that, and R6 up in value as well.  I'll leave this idea for later though, and of course all this has to be tested on the bench.
When we add that bias however, there is the chance that the current regulation will become less stable, so that has to be tested in real life on the bench.
As you said, the output current Io at 10 amps creates a voltage of 0.55v across the sense resistors, and so the bias only has to be about 0.15 volts (roughly) in order for the current limit to start to kick in (with Vbe 0.7 volts).  Had we adjusted the bias voltage to 0.425 volts with P1, then Vbe would reach 0.7 volts with only 0.275 volts across the sense resistors, meaning that the current would now be limited to just 5 amps instead of 10 amps.  So you can see the benefit of using a bias voltage.
To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps.
There are other ways to try too though, such as with a 1 Ohm power resistor in parallel with a 100 Ohm (or so) potentiometer, but we lose more power that way because the power resistor has to be larger in order to get to the lower end of the adjustment range.

[ommsiva]

Dear sir,

          1)now i understood that voltage developed across current sense resistor is applied to the base and emitter is at input common.

          2)"To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps ", similar to the first circuit of this thread.
         
          3) But this 10Amp Load current will develop 0.55V at the bottom end of R6 and already we have calculated 0.13V top of R6 , P1 top will be .7V respectively.   How this 0.55V will affect my initial bias arrangement done using Vref ? then i have to apply Superposition Theorem?

thanks a lot.

[Andy Chee]

Since this topic has drifted.

Please start from the beginning. What are you trying to do?

From what I can tell, they are a 3rd year university student, attempting to build an adjustable current limited laboratory power supply (or possibly a CC-CV lithium battery charger).

However as a university project, they need to document all the fundamental theory and formulas that underpin its design.

In a sense, this has the feel of a "homework" type question, but applied to a final year graduation/keystone project!

[MrAl]

Dear sir,

          1)now i understood that voltage developed across current sense resistor is applied to the base and emitter is at input common.

          2)"To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps ", similar to the first circuit of this thread.
         
          3) But this 10Amp Load current will develop 0.55V at the bottom end of R6 and already we have calculated 0.13V top of R6 , P1 top will be .7V respectively.   How this 0.55V will affect my initial bias arrangement done using Vref ? then i have to apply Superposition Theorem?

thanks a lot.

Hi,

I guess you could say that, but I thought it was simpler to just realize that the voltage at the bottom of R6 depends on the output current because of the drop across R5, and that means that the top of R6 changes as a result of the output current to such that it will increase as the output current increases.
But yes, you can look at it as a superposition too because if we consider the voltage across R6 separately from the voltage across R5, then we have to add them to get the final voltage at the top of R6 and add that to the voltage across the bottom section of the potentiometer to get the voltage at the arm of the potentiometer and thus the Vbe voltage which is a major part of figuring out what the current limit set points will be.

We also have the small contributions from other parts of the circuit which we ignore because the currents are so small as compared to the output current, but we could add them into the equation as well if desired.

[ommsiva]

Hello All,

I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
again i changed to 470ohm,2Watts , again it started smoking.


I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?

How to find my transistor has went to breakdown?

Thank you all.

[MrAl]

Hello All,

I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
again i changed to 470ohm,2Watts , again it started smoking.


I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?

How to find my transistor has went to breakdown?

Thank you all.

Hello there,

You are probably going to have to learn to troubleshoot these circuits yourself.  That means making measurements with different loads, and possibly running the input voltage up a little at a time to see when the power dissipation starts to get too high in the resistor(s).  You keep having this same problem with just about every circuit you work on like this.  A resistor smokes in every circuit you posted so far.
To find out the problem you often have to take voltage and/or current measurements before anything smokes, and that means possibly with a light load and slowly increasing that load.  If it happens even with light load, then you might have to turn the input voltage up slowly until the problem shows itself but does not blow anything out.

So to start see if it runs with no load.  You don't seem to say what the load is when this happens, or even if there is a load or not.

[Xena E]

Hello All,

I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
again i changed to 470ohm, 2Watts , again it started smoking.


I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?

How to find my transistor has went to breakdown?

Thank you all.

Incredible!

Taking 470 ohms as the value of the resistor, you have to have over 30 Volts across it to even max out its design dissipation of 2 Watts.

Now, assuming that the base emitter junctions of the pass transistors are intact that means that the output would have to be shorted to ground and in that case almost all of that hypothetical voltage has to appear across the 0.1 Ohm emitter degeneration resistors, that will cause them to try to dissipate over 8.5kW each.

So as this isn't possible, (the transformer isn't that big is it?), it's either ignorance or Bullshit.

The circuit diagram as drawn won't work, in fact if the practical circuit is wired as drawn it would be the transformer that has smoke coming out of it, (you show the output of the rectifier on the 5.1Volt supply shorted to ground for starters).

If you really want help you need to furnish facts and details.

Regards,
Xena.

[ommsiva]

It is not an auxiliary coil that is used for deriving 5V for volt amp meter.


From the main transformer after full wave rectification,  I have used it.

I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.

[Xena E]

It is not an auxiliary coil that is used for deriving 5V for volt amp meter.


From the main transformer after full wave rectification,  I have used it.

 

No, you haven't. Not connected like that, its a dead short.

If you want help, please provide detailed photographs of your work so far. If you are a student, a tutor would throw out what you have provided so far as being inadequate.

I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.

The regulator chip you are using is ubiquitous, why not just copy an application note from one of the manufacturers?

I repeat, if you want help post photos and details.

Regards,
Xena.

[ommsiva]

MAM,

PLEASE FIND THE ATTACHMENT

[Xena E]

Is that stripboard?

 

Where's the rest of it?

(Transformer, load, volt amp meter, pass devices, as you have it connected).

I don't want to be rude, however I or anyone else here haven't got a frigging crystal ball! If a circuit is burning up components, there is either something fundamentally wrong, with the design or you have mis-connected the circuit.

Vague comments about things burning up and asking what's wrong as if we're at the bench with you is a waste of time.

OK, here's a stab in the dark. If I am to understand from your previous post that the meter you are using is connected to that 5.1V stabiliser and that is being powered from the same transformer winding/rectifier as the main circuit....

It is not an auxiliary coil that is used for deriving 5V for volt amp meter...
...From the main transformer after full wave rectification,  I have used it.

...if there is no form of galvanic isolation between the two there is a possible source of your burning up. The circuit as drawn does not support that though...

If you want something that just works buy one. If this is a project that demonstrates your skill level you have already achieved that.

REGARDS,
XENA.

[MrAl]

It is not an auxiliary coil that is used for deriving 5V for volt amp meter.


From the main transformer after full wave rectification,  I have used it.

I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.

Here is a cleaned up drawing.  You should try to post drawings that are very clear so people can help easier.

As mentioned by Xena, there is a short in the diagram indicated here by a red "X".

Also, you seem to be suggesting that there is only one transformer.  In the new diagram the way you have it connected, are the transformers T1 and T2 actually two different transformers or are they the same?  Alternately, is T2 just a separate winding on T1 perhaps?

[ommsiva]

Hi to all,

Mam,

Its my mistake. now I have attached, and they contain

1) 6 number of current sharing series pass transistor.

2) two potentiometer, one for voltage and other for current

3) two fuse, one for main 30V adjustable @ 5 amps and other for 5V.

MY thanks to good hearts present here and  name a few to mention their kindness

1) Mr Xavier who himself has bought LM337 and Power transistor(build circuits and given results) and shared his design ideas for building High current Negative regulator and given valuable feedback for the circuits.

2) Mr AI ,my Guru who always inspired me by his ideas and  way of thinking.

Thank you all.

[Xena E]

I'm very sorry Ommsiva your pictures seem deliberately not to show any detail, at all.

Pictures of the tops of T03 transistors may be useful in identifying the type numbers but don't convey anything about the connections made to them.

Two front panel photos ... of similar value.

A general internal shot of a disassembled rats nest of wiring.

All of the above at a resolution such that even if it were tidy construction and complete it couldn't be of any less use in assisting with your query.

As you seem to not like answering direct questions about your project, then the purpose of this thread is clearly intended as a piss-take of the kind people who have tried to help you.

Have a nice life.

X

[ommsiva]

Mam,

Sorry,   absolutely you are correct.  It's my mistake, each picture were nearly 6mb and our site allows maximum of 8mb totally for a reply.so i reduced the image size .Now I have disconnected power transistor from heat sink out of frustration.

[MrAl]

Hi to all,

Mam,

Its my mistake. now I have attached, and they contain

1) 6 number of current sharing series pass transistor.

2) two potentiometer, one for voltage and other for current

3) two fuse, one for main 30V adjustable @ 5 amps and other for 5V.

MY thanks to good hearts present here and  name a few to mention their kindness

1) Mr Xavier who himself has bought LM337 and Power transistor(build circuits and given results) and shared his design ideas for building High current Negative regulator and given valuable feedback for the circuits.

2) Mr AI ,my Guru who always inspired me by his ideas and  way of thinking.

Thank you all.

Hi,

I meant to ask, was this circuit working before this or did you just build it for the first time?

[ommsiva]

Sir,

Second time.

It was working well with 0.33 ohm 10watt.

Now 470 ohm got smoke.


Thank you

[xavier60]

Sir,

Second time.

It was working well with 0.33 ohm 10watt.

Now 470 ohm got smoke.


Thank you

Is the 470Ω the resistor in parallel with the CS resistor?
The 0.33Ω resistor you mention is the CS resistor?
What load were you testing with?

[ommsiva]

Sir,

I will check whether they are parallel.

i had 10ohm 50watt. i had an idea of keeping 20V output and check.

thank you.

[MrAl]

Sir,

Second time.

It was working well with 0.33 ohm 10watt.

Now 470 ohm got smoke.


Thank you

Hi,

If you are going to change values it would be best to update the schematic.  Please use the cleaner diagram I posted so we can read it better.

[MrAl]

Hello again,

Here is an updated schematic with some of the part numbers enumerated so it would be easier to discuss this circuit.

There are also two test conditions, with the values shown in green.  The pots are shown to have a fixed value for the two tests.
The two tests are with RL=4 and with RL=1.8, nothing else is changed.
The results are show on the diagram.
When RL=4 the circuit is not in current limit and puts out the set voltage level around 20 volts or so, and about 5 amps out.
When RL=1.8 the circuit is just starting to current limit so the output voltage level goes down a little to 18.6 volts and the current goes up to the set limit point set by R9 and R10.  The limit is set to be around 10 amps.

This means the circuit looks like it works OK so maybe it is a simpler problem that came up such as a blown out part or bad or wrong connection somewhere.  Note however that I did not check the LM317 part of the circuit.
Oh there is one problem, the 1000uf cap has to be made a lot bigger.

See the attached diagram for details.

Added later:
About that 1000uf cap, even 10000uf (10 times higher than 1000uf) might not be enough.  There would be a lot of ripple voltage with 10 amps output.  The problem is, if the ripple is too high then you need a higher input voltage in order to get a smooth output voltage of a fairly high value like 30 volts DC.  You can not go too high because of the limitations of the 723, so the burden becomes making the input filter capacitor value large enough.  20000uf would be better for example.

[xavier60]

Sir,

I will check whether they are parallel.

i had 10ohm 50watt. i had an idea of keeping 20V output and check.

thank you.

Also include what test you might have in your profile or signature. It will save anyone having to ask.

[Zero999]

I admit. I don't have time to read the entire thread in detail.

Some notes about the '723

The op-amp only works with its inputs 2V above the negative rail. This means that, in the usual configurations, it will only work down to a minimum output voltage of 2V. There are two possible ways round this:

1) Use a negative supply of at least 2V. This adds some complexity, but if it's run off a transformer, a simple voltage doubler circuit will do. Another downside is it effectively increases the total supply voltage to the '723 by 2V.

2) Use a potential divider to scale the control voltage, so it starts above 2V and design the op-amp circuit so it subtracts 2V off the input. This is more complicated, but doesn't increase the supply voltage to the '723.

The current limit is approximate because it depends on the base-emitter voltage of a transistor. It's only meant to protect the circuit, not as a constant current supply. The solution is to add an op-amp error amplifier and current sense resistor.

In another thread I sketched out a design for accurate current limiting an operation with an output voltage down to 5V. It's not quite finished and requires some frequency compensation, to prevent oscillation.

https://www.eevblog.com/forum/projects/lm-723-opamp-addon/msg5403488/#msg5403488

[MrAl]

Here are some charts that illustrate the effect of the capacitor value and the total input impedance Rs.



All of these are with an input sine wave peak of 30 volts and two diode drops 1.15 volts each,
frequency was 50Hz, Rload=3 Ohms.

Units: Volts, Ohms, Farads.


C=0.001000
Rs=0.00  Vmax=19.00  Vmin= 6.60  Vpp=12.398
Rs=0.10  Vmax=18.38  Vmin= 6.60  Vpp=11.782
Rs=0.20  Vmax=17.78  Vmin= 6.58  Vpp=11.200
Rs=0.30  Vmax=17.21  Vmin= 6.56  Vpp=10.651
Rs=0.40  Vmax=16.66  Vmin= 6.53  Vpp=10.134
Rs=0.50  Vmax=16.14  Vmin= 6.49  Vpp= 9.648
Rs=0.60  Vmax=15.64  Vmin= 6.45  Vpp= 9.191
Rs=0.70  Vmax=15.17  Vmin= 6.40  Vpp= 8.762
Rs=0.80  Vmax=14.71  Vmin= 6.35  Vpp= 8.358
Rs=0.90  Vmax=14.28  Vmin= 6.30  Vpp= 7.979
Rs=1.00  Vmax=13.87  Vmin= 6.25  Vpp= 7.623

C=0.005000
Rs=0.00  Vmax=27.10  Vmin=17.55  Vpp= 9.550
Rs=0.10  Vmax=25.92  Vmin=17.37  Vpp= 8.560
Rs=0.20  Vmax=24.37  Vmin=16.88  Vpp= 7.489
Rs=0.30  Vmax=22.64  Vmin=16.19  Vpp= 6.454
Rs=0.40  Vmax=20.91  Vmin=15.39  Vpp= 5.521
Rs=0.50  Vmax=19.27  Vmin=14.55  Vpp= 4.714
Rs=0.60  Vmax=17.76  Vmin=13.72  Vpp= 4.034
Rs=0.70  Vmax=16.40  Vmin=12.93  Vpp= 3.467
Rs=0.80  Vmax=15.19  Vmin=12.19  Vpp= 2.997
Rs=0.90  Vmax=14.11  Vmin=11.50  Vpp= 2.607
Rs=1.00  Vmax=13.15  Vmin=10.87  Vpp= 2.282

C=0.010000
Rs=0.00  Vmax=27.55  Vmin=21.46  Vpp= 6.082
Rs=0.10  Vmax=25.50  Vmin=20.61  Vpp= 4.896
Rs=0.20  Vmax=22.25  Vmin=18.64  Vpp= 3.614
Rs=0.30  Vmax=19.02  Vmin=16.41  Vpp= 2.609
Rs=0.40  Vmax=16.28  Vmin=14.37  Vpp= 1.906
Rs=0.50  Vmax=14.08  Vmin=12.65  Vpp= 1.427
Rs=0.60  Vmax=12.33  Vmin=11.23  Vpp= 1.097
Rs=0.70  Vmax=10.92  Vmin=10.06  Vpp= 0.865
Rs=0.80  Vmax= 9.79  Vmin= 9.09  Vpp= 0.696
Rs=0.90  Vmax= 8.85  Vmin= 8.28  Vpp= 0.571
Rs=1.00  Vmax= 8.07  Vmin= 7.59  Vpp= 0.476

C=0.015000
Rs=0.00  Vmax=27.63  Vmin=23.17  Vpp= 4.464
Rs=0.10  Vmax=24.33  Vmin=21.20  Vpp= 3.126
Rs=0.20  Vmax=19.41  Vmin=17.51  Vpp= 1.906
Rs=0.30  Vmax=15.43  Vmin=14.24  Vpp= 1.186
Rs=0.40  Vmax=12.56  Vmin=11.78  Vpp= 0.783
Rs=0.50  Vmax=10.51  Vmin= 9.96  Vpp= 0.548
Rs=0.60  Vmax= 9.00  Vmin= 8.59  Vpp= 0.402
Rs=0.70  Vmax= 7.85  Vmin= 7.54  Vpp= 0.306
Rs=0.80  Vmax= 6.95  Vmin= 6.71  Vpp= 0.241
Rs=0.90  Vmax= 6.23  Vmin= 6.04  Vpp= 0.194
Rs=1.00  Vmax= 5.64  Vmin= 5.48  Vpp= 0.159

C=0.020000
Rs=0.00  Vmax=27.66  Vmin=24.13  Vpp= 3.532
Rs=0.10  Vmax=22.87  Vmin=20.77  Vpp= 2.104
Rs=0.20  Vmax=16.78  Vmin=15.70  Vpp= 1.080
Rs=0.30  Vmax=12.67  Vmin=12.07  Vpp= 0.608
Rs=0.40  Vmax=10.03  Vmin= 9.65  Vpp= 0.379
Rs=0.50  Vmax= 8.25  Vmin= 8.00  Vpp= 0.257
Rs=0.60  Vmax= 6.99  Vmin= 6.81  Vpp= 0.184
Rs=0.70  Vmax= 6.06  Vmin= 5.92  Vpp= 0.138
Rs=0.80  Vmax= 5.34  Vmin= 5.23  Vpp= 0.108
Rs=0.90  Vmax= 4.77  Vmin= 4.68  Vpp= 0.086
Rs=1.00  Vmax= 4.31  Vmin= 4.24  Vpp= 0.070

C=0.025000
Rs=0.00  Vmax=27.68  Vmin=24.75  Vpp= 2.926
Rs=0.10  Vmax=21.32  Vmin=19.86  Vpp= 1.465
Rs=0.20  Vmax=14.58  Vmin=13.92  Vpp= 0.653
Rs=0.30  Vmax=10.64  Vmin=10.30  Vpp= 0.345
Rs=0.40  Vmax= 8.29  Vmin= 8.08  Vpp= 0.208
Rs=0.50  Vmax= 6.76  Vmin= 6.62  Vpp= 0.138
Rs=0.60  Vmax= 5.69  Vmin= 5.59  Vpp= 0.098
Rs=0.70  Vmax= 4.91  Vmin= 4.84  Vpp= 0.073
Rs=0.80  Vmax= 4.32  Vmin= 4.26  Vpp= 0.057
Rs=0.90  Vmax= 3.85  Vmin= 3.81  Vpp= 0.045
Rs=1.00  Vmax= 3.47  Vmin= 3.44  Vpp= 0.037

C=0.030000
Rs=0.00  Vmax=27.68  Vmin=25.18  Vpp= 2.500
Rs=0.10  Vmax=19.79  Vmin=18.74  Vpp= 1.049
Rs=0.20  Vmax=12.78  Vmin=12.36  Vpp= 0.419
Rs=0.30  Vmax= 9.12  Vmin= 8.91  Vpp= 0.212
Rs=0.40  Vmax= 7.03  Vmin= 6.91  Vpp= 0.125
Rs=0.50  Vmax= 5.70  Vmin= 5.62  Vpp= 0.082
Rs=0.60  Vmax= 4.79  Vmin= 4.73  Vpp= 0.058
Rs=0.70  Vmax= 4.12  Vmin= 4.08  Vpp= 0.043
Rs=0.80  Vmax= 3.62  Vmin= 3.59  Vpp= 0.033
Rs=0.90  Vmax= 3.23  Vmin= 3.20  Vpp= 0.026
Rs=1.00  Vmax= 2.91  Vmin= 2.89  Vpp= 0.021

C=0.050000
Rs=0.00  Vmax=27.69  Vmin=26.11  Vpp= 1.586
Rs=0.10  Vmax=14.73  Vmin=14.39  Vpp= 0.342
Rs=0.20  Vmax= 8.35  Vmin= 8.24  Vpp= 0.107
Rs=0.30  Vmax= 5.72  Vmin= 5.67  Vpp= 0.050
Rs=0.40  Vmax= 4.34  Vmin= 4.31  Vpp= 0.029
Rs=0.50  Vmax= 3.49  Vmin= 3.47  Vpp= 0.019
Rs=0.60  Vmax= 2.91  Vmin= 2.90  Vpp= 0.013
Rs=0.70  Vmax= 2.50  Vmin= 2.49  Vpp= 0.010
Rs=0.80  Vmax= 2.19  Vmin= 2.19  Vpp= 0.007
Rs=0.90  Vmax= 1.95  Vmin= 1.94  Vpp= 0.006
Rs=1.00  Vmax= 1.76  Vmin= 1.75  Vpp= 0.005

C=0.075000
Rs=0.00  Vmax=27.70  Vmin=26.60  Vpp= 1.093
Rs=0.10  Vmax=10.77  Vmin=10.65  Vpp= 0.120
Rs=0.20  Vmax= 5.73  Vmin= 5.70  Vpp= 0.034
Rs=0.30  Vmax= 3.87  Vmin= 3.86  Vpp= 0.015
Rs=0.40  Vmax= 2.92  Vmin= 2.91  Vpp= 0.009
Rs=0.50  Vmax= 2.34  Vmin= 2.33  Vpp= 0.006
Rs=0.60  Vmax= 1.95  Vmin= 1.95  Vpp= 0.004
Rs=0.70  Vmax= 1.67  Vmin= 1.67  Vpp= 0.003
Rs=0.80  Vmax= 1.47  Vmin= 1.46  Vpp= 0.002
Rs=0.90  Vmax= 1.30  Vmin= 1.30  Vpp= 0.002
Rs=1.00  Vmax= 1.17  Vmin= 1.17  Vpp= 0.001

C=0.100000
Rs=0.00  Vmax=27.70  Vmin=26.86  Vpp= 0.835
Rs=0.10  Vmax= 8.38  Vmin= 8.32  Vpp= 0.054
Rs=0.20  Vmax= 4.35  Vmin= 4.33  Vpp= 0.014
Rs=0.30  Vmax= 2.92  Vmin= 2.91  Vpp= 0.007
Rs=0.40  Vmax= 2.20  Vmin= 2.19  Vpp= 0.004
Rs=0.50  Vmax= 1.76  Vmin= 1.76  Vpp= 0.002
Rs=0.60  Vmax= 1.47  Vmin= 1.46  Vpp= 0.002
Rs=0.70  Vmax= 1.26  Vmin= 1.26  Vpp= 0.001
Rs=0.80  Vmax= 1.10  Vmin= 1.10  Vpp= 0.001
Rs=0.90  Vmax= 0.98  Vmin= 0.98  Vpp= 0.001
Rs=1.00  Vmax= 0.88  Vmin= 0.88  Vpp= 0.001

[ommsiva]

Hi all,

I found this circuit in google, Where they have used BD139 for current limiting function.

For example: now the LM317 produces 12V output, It has to pass through BD139(we assume that it is working in saturation), then drop between collector will be 0.2V. That means Vce=0.2V, Then a potentiometer is present between Collector to emitter terminal which takes this 0.2V.

1)How this 0.2V will forward bias base t0 emitter junction which is 0.7V? how current limiting is happening?

2) now the voltage has been reduced t0 11.8V and that will be given TIP3055, that takes minimum of 1V between base and emitter.
now output will be 10.8V, and input should be (12V+3V for regulator) 15 V for obtaining 10.8V. is it correct?

thank you all.