Improving Current limit

[jonnybgood]

Dear jonnybgood
Again Hi
As i told you , we have many many different way for current controlling . but this way is usual and very good , and when the out put voltage is short circuit , the voltage can not increase ( remember that this circuit called hiccup current limiter , and at one time the current limiter , do it's duty , and at the other time , the voltage feed back will do it's duty .
BTW : i built 25A power supply with this way about 2 years ago.
Best Wishes
Goldsmith

In fact I figured out that I was putting the current sense resistor to the emitter not to ground.... I will try with it connected to ground today
thanks for now.

---------- Post added at 11:31 ---------- Previous post was at 09:53 ----------

With current limit resistor, the voltage at the output terminals fall below 9V (0.1 Ohms * 2.7Amps = 0.27V drop). I know that the op amp is trying to keep a 9V across the load and current sense resistor in series not the load only. Since the feedback is from a voltage divider from emmitter to ground. Can I take feedback only from the output with respect to current sensing resistor instead of directly to ground?

 

[mtwieg]

In fact I figured out that I was putting the current sense resistor to the emitter not to ground.... I will try with it connected to ground today
thanks for now.

---------- Post added at 11:31 ---------- Previous post was at 09:53 ----------

With current limit resistor, the voltage at the output terminals fall below 9V (0.1 Ohms * 2.7Amps = 0.27V drop). I know that the op amp is trying to keep a 9V across the load and current sense resistor in series not the load only. Since the feedback is from a voltage divider from emmitter to ground. Can I take feedback only from the output with respect to current sensing resistor instead of directly to ground?

As you've noticed, sensing current in the ground path can cause some issues, but they are generally solvable.

For the issue described above, you can either use a differential error amplifier to sense only the voltage across the load and not the current sense resistor. Another possibility is to reference all the circuitry to the other side of the current sense resistor (the negative terminal of the load), essentially making it the actual "ground" of your circuit.

Another issue with ground current sensing is that if you want to connect other supplies common to its output, you must be careful to connect them at the negative load terminal, on the output side of the current sense resistor. Otherwise you are likely to cause severe issues with the current sensing, or get errors in the output voltage. Also when doing this you should make sure that the input supply voltages to each regulator are isolated from each other, so that they do not short out the current sense resistors.

 

[jonnybgood]

As you've noticed, sensing current in the ground path can cause some issues, but they are generally solvable.

For the issue described above, you can either use a differential error amplifier to sense only the voltage across the load and not the current sense resistor. Another possibility is to reference all the circuitry to the other side of the current sense resistor (the negative terminal of the load), essentially making it the actual "ground" of your circuit.

Another issue with ground current sensing is that if you want to connect other supplies common to its output, you must be careful to connect them at the negative load terminal, on the output side of the current sense resistor. Otherwise you are likely to cause severe issues with the current sensing, or get errors in the output voltage. Also when doing this you should make sure that the input supply voltages to each regulator are isolated from each other, so that they do not short out the current sense resistors.

I can say that the circuit work very well today thanks to every one who helped especially to goldsmith and mtwieg's trick worked well as I put all the grounds of the voltage feedback stage to the negative point of the load.

One thing is left for the last improvement as the output voltage of the regulators (current limit stage) is dropping by 30mV which is effecting the accuracy of the current limit. Are there any voltage regulators with better regulation or its just simply the ones with a lower drop out?

 

[goldsmith]

Hello my friend
there is the better way for voltage regulation available that called LDO Method ( low drop out) . the other advantage of LDO is that you can control the output voltage from zero up to input voltage ( suppose that your input is about 90 v and the supply of the control op amp is about 15 v , with LDO you can control it from 0 up to 90 volt simply and your op amp don't need 90 volt as supply voltage !!!!)

Here you can see simple LDO :
you can change the out put voltage with changing the v4.
Good luck
Goldsmith

 

[jonnybgood]

Hello my friend
there is the better way for voltage regulation available that called LDO Method ( low drop out) . the other advantage of LDO is that you can control the output voltage from zero up to input voltage ( suppose that your input is about 90 v and the supply of the control op amp is about 15 v , with LDO you can control it from 0 up to 90 volt simply and your op amp don't need 90 volt as supply voltage !!!!)

Here you can see simple LDO : View attachment 68169
you can change the out put voltage with changing the v4.
Good luck
Goldsmith

Just to make sure I was understood well, I was refering to the voltage regulator used for the reference voltage of the op amp.

I assume this is a schematic of an LDO regulator. I found it very interesting and I will look further into it.

I need some suggestions for some preferably TO-92 LDO's if they have better regulation. Otherwise I was going to try and put a 9V zener diode across my 78L05 (series with maybe 1K) to minimize the fluctuation in the input of the regulator what do you think? Here the regulator would not be loaded at all!

 

[goldsmith]

Dear Jonny
Hi
I can't understand that what is your problem clearly ! can you describe your new problem , please?
All The best
Goldsmith

 

[jonnybgood]

Dear Jonny
Hi
I can't understand that what is your problem clearly ! can you describe your new problem , please?
All The best
Goldsmith

My op amp that is used for the current limiting stage (driving 2n2222a) takes its reference from a 10k multiturn preset that is connected to the output of a 78L05. Now the input of the 7805 will vary from 17V to about 12.5V once the psu is loaded with max load. This change in input of the voltage regulator is causing the output from 5V to drift by 20-25mV. This is causing the noninverting input to drop also causing a nonlinear current limiting over different loads. I need a more precise voltage reference than the 78L05, this can be anywhere from 2V -5V. What are some common voltage precision reference that are cheep. If it drifts by a millivolt it is oki. I will power it by the same regulator.

 

[goldsmith]

if you want the fix voltage for your reference , why you don't use infinite feed back constant voltage ?
Do you know about it's circuit ? ( with 2 transistor and some other elements) ( very constant voltage)
Best Things
Goldsmith

 

[jonnybgood]

No but I would like to Shall I start a new thread?

 

[goldsmith]

The voltage across D2 is a 5.1 volt but very very constant

Respectfully
Goldsmith

 

[FvM]

Now the input of the 7805 will vary from 17V to about 12.5V once the psu is loaded with max load. This change in input of the voltage regulator is causing the output from 5V to drift by 20-25mV. This is causing the noninverting input to drop also causing a nonlinear current limiting over different loads. I need a more precise voltage reference than the 78L05, this can be anywhere from 2V -5V. What are some common voltage precision reference that are cheep. If it drifts by a millivolt it is oki. I will power it by the same regulator.

Available regulators and references form all manufacturers have specifications for line and load regulation, initial accuracy, tempearture and long term drift, whatever you need.

If better line regulation is your only concern, you may consider to simply power the reference regulator from the regulated OP supply.

Referring to the post #24 circuit, I fear the control loop is unstable without additional compensation.

 

[jonnybgood]

Cheers ! Are those zener diodes? I will build it tomorrow and let you know. I assume you tried lowering the voltage down to at leas 10V and saw that output remained constant.

 

[goldsmith]

Dear FvM
Hi
How are you ?
Is it possible that you guide us to add compensation network to that ? i think we should add a series RC network at feed back loop . am i right?
BTW : if we don't use the compensation network at this feed back loop , what will happen ? is it possible , that you guide us , please ?
Thanks in advance
Goldsmith

 

[FvM]

I fear, it's easier to diagnose an unstable feedback loop in a simulation or in a hand calculation than to suggest a good compensation that fulfills all requirements.

The reason for stability problems in the present case is the large voltage gain between OP output and input by the cascaded transistor stages. The external gain (feedback factor > 1) makes it more difficult to achieve a suitable loop gain roll-off.

In the previous post #9 circuit, you apparently could achieve stability by brute force methods, making the feedback loop very slow, which however brings up problems in dynamic behaviour, e.g. with load current steps.

A method, that works in many cases is to give the error amplifier a PI characteristic by placing a RC series circuit as local feedback accross the OP. On the other hand, you don't want separate integrators for I and V feedback because they cause overshoots in transition between I and V regulated mode.

 

[goldsmith]

Dear FvM
Is it possible that you tell us , that , at this application what value for that RC network is better , please?
Best Regards
Goldsmith

 

[FvM]

Honestly speaking, I would try it in a simulation. You need to provide a dynamic variable load, e.g. a pulsed current source or a load resistor with a time controlled switch.

 

[jonnybgood]

The voltage across D2 is a 5.1 volt but very very constant
View attachment 68212
Respectfully
Goldsmith

Can you give me a brief explanation of how the constant voltage circuit works please?

 

[goldsmith]

Again Hi
Yes , it is very simple. according to the volt- ampere curve of each zener diode , it's voltage can be very constant if it's current become very constant. i used two current source with infinite feed back loop : when the voltage of D1 is approximately constant , the voltage across R1 is approximately constant . and the thus IC1 is approx , constant , and the current of D2 will be constant ( approx) . thus the voltage across R is constant ( approx) and naturally the current of Q2 is constant , and the voltage of D1 will be more constant than past . and this behavior will be continuous and thus the voltage across each zener diode will be very very constant . and the temperature stability of 5.1 volt zener diodes is more than the other types of Zener diodes . thus it's stability is very incredible.
Good luck
Goldsmith

---------- Post added at 13:22 ---------- Previous post was at 13:18 ----------

BTW: R3 is used for transient time.( to inject a little current to drive Q1 for start time)

---------- Post added at 13:27 ---------- Previous post was at 13:22 ----------

Dear FvM
Again Hi
Thank you for your guide to us , but is it possible that you tell us your mean more clearly , please? and is it possible that you offer us , that what value can use for compensation network at this aim , please ? i think that RC network can shift the poles of this circuit to the left of S plane to improve it's transient response . but wee need your again guidance to select them.
Respectfully and with Appreciate.
Goldsmith

 

[jonnybgood]

Thanks, wonderful explanation Regarding the current limit stage, I was talking to an engineer and told him about my circuit. He suggested that I use a differential amplifier with inputs connected to ground and current sense resistor +ve terminal, than use its output to drive the 2n2222A. This differential amplifier should have closed loop gain adjusted finely until I get my required limiting. Do you think this is better or have some advantages?

 

[goldsmith]

Dear Jonny
The differential amp can use to obtain the difference between the voltage across the resistor and it can amplify you current error signal . but at that circuit because of that , the current sensor ( resistor ) is under the load to ground , i think it is not necessary to use diff amp , here .
All the best
Goldsmith