Author Topic: LM324 Power Supply with variable voltage and current (Read 178490 times)

mike_mike · « **Reply #450 on:** May 17, 2019, 04:32:36 pm »

Do you have any suggestions regarding the value for C1 ?
What would be the best value for this capacitor (C1) in order to eliminate any possible oscillation ?

PS: I also tried with 470pF, and the same image appears on the screen.

xavier60 · « **Reply #451 on:** May 17, 2019, 06:51:42 pm »

The CV op-amp is supposed to be saturated high and have no control while CC op-amp is in control. You can confirm this.
Why is the waveform not inverted? Where is the probe's ground clip connected?
What does the output look like?

mike_mike · « **Reply #452 on:** May 17, 2019, 07:26:09 pm »

Quote from: xavier60 on May 17, 2019, 06:51:42 pm

The CV op-amp is supposed to be saturated high and have no control while CC op-amp is in control. You can confirm this.
Why is the waveform not inverted? Where is the probe's ground clip connected?
What does the output look like?

1. Yes, the CV op-amp is 10.86V output while CC is in control.
2. The probe ground is connected to the marked point in the attached image. The other end of the oscilloscope probe is connected to the other end of the shunt resistor.
3. The output looks like in the attached screenshot (0015.jpg) whith Vout=26V, SlowCC=0.22A and Rload=10R.

iMo · « **Reply #453 on:** May 17, 2019, 07:45:16 pm »

@MM: how much time does it take to

1. charge 2x47uF with 220mA from 0V to 26V?
2. discharge 2x47uF (26V) via 10ohm to 0V with max 220mA?

mike_mike · « **Reply #454 on:** May 17, 2019, 08:09:22 pm »

1. Charging of the capacitor:
C*V=I*T
V=(I/C)*T
V=K*T
26V=(0.22A/0.000094F)*T
26V=2340*T
T=26/2340=0.011 [seconds]

2. T=R*C=10 $\$\Omega\$$ *0.00094F = 0.00094 [sec]
5*T=5*0.00094=0.0047 [sec]
the capacitor will be fully discharged after 5 time constants (T=RC)

iMo · « **Reply #455 on:** May 17, 2019, 08:52:56 pm »

Now look at the edges in your above picture. For example I can see an 11-12ms long rising edge there.

xavier60 · « **Reply #456 on:** May 17, 2019, 10:24:55 pm »

All looks well to me. It's best to always connect the probe ground to OUT-

xavier60 · « **Reply #457 on:** May 17, 2019, 10:54:27 pm »

The CV op-amp's compensation should only be altered to optimize load transient response. Keep this in mind when modifications are done that affect the gm of the output path.

mike_mike · « **Reply #458 on:** May 27, 2019, 07:02:39 pm »

Yes, the power supply is working well.
I do not see any issues for now.

xavier60 · « **Reply #459 on:** May 27, 2019, 07:14:17 pm »

Quote from: mike_mike on May 27, 2019, 07:02:39 pm

Yes, the power supply is working well.
I do not see any issues for now.

Good to hear. Is that with the low side fast current limiting working also?

mike_mike · « **Reply #460 on:** May 27, 2019, 07:39:46 pm »

The low side fast current limiting was not tested and was not included into the final schematic. The used FastCC is the one from @imo, the one with BD139.

ggchab · « **Reply #461 on:** May 28, 2019, 09:10:50 am »

I am following this thread with some interest and I am wondering why there is C1 and R14 in the feedback loop of U4 (CV) but only C5 in the feedback loop of U3 (CC).
I made some LTSPICE simulations with a similar circuit and the resistor seems to really improve stability, even if I don't clearly understand why.
Thank you for any answer to my beginner's question.

xavier60 · « **Reply #462 on:** May 28, 2019, 09:31:35 am »

Quote from: ggchab on May 28, 2019, 09:10:50 am

I am following this thread with some interest and I am wondering why there is C1 and R14 in the feedback loop of U4 (CV) but only C5 in the feedback loop of U3 (CC).
I made some LTSPICE simulations with a similar circuit and the resistor seems to really improve stability, even if I don't clearly understand why.
Thank you for any answer to my beginner's question.

I don't fully understand what's going on either. With the experience from the last two of my designs, I repeatedly observed that the CC loop becomes unstable with increased Proportional response and perfectly stable with mainly Integral response. In this design, the CC op-amp has an unavoidable Proportional gain of one because it is configured as non-inverting, but isn't enough to cause problems.
I see a lot of designs where the CC op-amp is configured purely as an inverting Miller Integrator.

xavier60 · « **Reply #463 on:** May 29, 2019, 01:35:59 am »

Designs can look similar but important operating principles can be quite different.
Mainly with the output stages.
There are the voltage follower type where the output stage follows the output voltage of an op-amp that is powered directly by the unregulated rail, yuck! This type has a characteristically low output impedance. As load resistance decreases, the output stage itself automatically sources more current.
The output stage of the design in this thread has a high impedance constant current characteristic. Voltage changes at the Base of Q1 control the output current regardless of what the output voltage is. It is termed as "Transconductance" amplifier. Input voltage controls output current.
These two output stages types usually require different compensation in their CV and CC control loops.
My preference has been for the high impedance type output stage.
The highly regarded Harrison topology is the high impedance type.

mike_mike · « **Reply #464 on:** May 30, 2019, 12:52:19 pm »

In the simulation, the diodes D1 and D7 were 1N4007.
If I change them to 1N5408, could the power supply starts to oscillate, if there were no oscillations with 1N4007 ?

iMo · « **Reply #465 on:** May 30, 2019, 01:23:30 pm »

99.999991% not..

mike_mike · « **Reply #466 on:** May 30, 2019, 01:29:42 pm »

... and if I change them to P600K ?

iMo · « **Reply #467 on:** May 30, 2019, 01:42:42 pm »

99.9~~9999~~1% not..
The only difference which may introduce an instability is in their capacitance, P600K is 150pF at 4V.
1N5408 is 30pF. 1N4007 is 15pF.
Add that cap diff to the D1/D7 (the output D1 diode's capacitance does not matter as there is the output capacitance large) and do a simulation...

iMo · « **Reply #468 on:** May 30, 2019, 02:32:54 pm »

When driving motors I would use a TVS diode, like 1.5KE30A or something like that..

mike_mike · « **Reply #469 on:** May 30, 2019, 05:32:58 pm »

If it is recommended to use a TVS diode, then why in the original schematic was used a 1N4007 ? What is the purpose of the 1N4007 in the original schematic (V17) ?

Kleinstein · « **Reply #470 on:** May 30, 2019, 07:40:08 pm »

D7 in the schematic I found is parallel to the output transistors, preventing the input voltage from being much smaller than the output. This diode can be important if there is a very large capacitor or other voltage source at the output and the supply is turned off. Here the 1N4007 is likely large enough to worst case charge the filter capacitor, when an external source is connected with the supply turned off. If in doubt (very large capacitor) one could use a larger diode, or 2 in parallel (works here because it would be a high current transient that could be critical).
Charing the cap is about as bad as discharging it through the diode (+ shunt + possible fuse).

mike_mike · « **Reply #471 on:** May 30, 2019, 08:20:16 pm »

@Kleinstein And what about the diode on the output of the power supply ?
Can I safely use there a larger current capacity diode ? For example P600K...

xavier60 · « **Reply #472 on:** May 30, 2019, 08:34:17 pm »

I sometimes use my bench supply to charge batteries.
Accidentally connecting a battery with reverse polarity would cause a large fault current through a diode.
I don't have a diode across the output. The design tolerates briefly applied reverse polarity.

mike_mike · « **Reply #473 on:** May 31, 2019, 05:13:38 am »

If powering a motor, for example a 3A motor, then the 1N4007 diode is fine for back EMF ?

xavier60 · « **Reply #474 on:** May 31, 2019, 05:56:41 am »

As far as I know, DC motors produce an inductive voltage spike only when the current path is broken. I can't see how this can be an issue for the power supply.
Just use 3A diodes. If you foresee ever charging batteries, consider omitting the output diode or mount the diode somewhere it wont ruin the PCB if it overheats.


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Author Topic: LM324 Power Supply with variable voltage and current (Read 178490 times)

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