Electronics > Projects, Designs, and Technical Stuff
Is 550uF too big for a power supply that has CC limit?
Doctorandus_P:
I once tried setting the ouput of my power supply to 30V and the current limit to 20mA, and then connect a led.
It should just light up with 20mA right?
(Note: In those days a led really needed 20mA to be somewhat visible).
In my case the led exploded. Didn't expect that.
But over time you learn.
The power supply was ok
Kleinstein:
Ideally a LED would just get the set 20 mA. However with real life power supplies there is usually a much higher initial current. This extra current comes from the capacitor at the output, that can be quite large in some cases (e.g. up to 1000 µF for some 3 A supplies) and also from the time it takes for the current regulation to set in.
Here one kind of has the choice of 2 evil. There are mainly 2 types of regulator circuits: those with an output stage that sets a current and those where the output stage sets a voltage with a low impedance.
For the 1st. type a reasonable size capacitor in mandatory for stability, but the cross over to CC mode can be fast. If the regulator is fast and well tuned the capacitor can be relatively small - so no real need for 500 µF, but some 10 µF may be required.
For the 2 nd type one does not need a capacitor at the output or least can use a very small one (e.g. 1 µF). However the onset of CC mode can be slow, especially if the voltage needs to drop. So there is a kind of simulated capacitance, that can be about the size required in the other case. Often there is also some delay before the current limit sets in. This is sometimes good (e.g. when testing a SMPS or fan) and sometimes bad (e.g. the LED).
So it is really difficult to get a fast acting current limit. Normally one can live with the limitations. There is no need to connect a LED to the supply set to 25 V and hope for the current to be fast enough - set the supply to 0 V and only turn up the voltage after the load is connected.
bloguetronica:
I can only assess the capacitance that is needed when I have the board on my hands and when I manage to solve the issues. Everything else is a guess. Anyway, it wont be 10uF, as it is too low for this design.
Kind regards, Samuel Lourenço
Kleinstein:
The circuit shown here uses a low output impedance power stage and could in principle work essentially without capacitor at the output. So 10 µF is not such a bad target for this type of circuit. The difficulty in this circuit is more in getting the current control work well, especially if the differential amplifier is not perfect.
bloguetronica:
--- Quote from: Kleinstein on February 03, 2019, 10:24:08 am ---The circuit shown here uses a low output impedance power stage and could in principle work essentially without capacitor at the output. So 10 µF is not such a bad target for this type of circuit. The difficulty in this circuit is more in getting the current control work well, especially if the differential amplifier is not perfect.
--- End quote ---
Well, today I had an epiphany about that differential amplifier. I think I should try a 10K resistor in series with that capacitor in the feedback from the output to the inverting input. That, in conjunction with the 10K resistor that goes in series to the inverting input, will cause the AC gain to be -1. I think that the AC gain of that stage was set too low, to almost 0, if that makes sense. Perhaps then, it is possible to eliminate that extra capacitor named C12.
Given that all conditions are stable (all op-amps compensated), then it is a matter of testing at full load, in order to find the minimum capacitor.
Kind regards, Samuel Lourenço
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