Electronics > Projects, Designs, and Technical Stuff
Constant current mode for DIY power supply
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T3sl4co1l:
He's referring to the voltage drop. If you want voltage regulation as priority, you can't afford a load dependent 0-1.25V drop. :)

An electronic fuse is perfectly possible, by the way -- but such circuits are quite a bit more complicated than a pair of LM317s.  The OP may not be quite adventurous enough to do that.

Incidentally, electronic fuse behavior can be very annoying -- you're connecting a power supply to circuits with bypass capacitors very often, and those capacitors need a lot of charge.  Simply shorting an active supply to an electrolytic cap can draw 100A inrush easily, or take milliseconds to charge from around an ampere.  Milliseconds is in the range of blowing transistors from high dissipation.

An electronic fuse that opens in microseconds is easy enough to make, and adequate to protect power transistors, but also rather inconvenient to use.

An electronic fuse that opens in milliseconds is hard to make, for currents over a few amperes and voltages above 10 or 20V.  The problem is power dissipation in the pass transistor.

Best is both, a limited current of about double what you need for the target, with a modest active period (preferably dependent on voltage drop, so the on-time corresponds to peak temperature reached by the limiter -- maximizing the startup time you get).

The limiter can also be made switching, so that the pass transistor dissipates a tiny fraction of full power, and the dropped power is passed on to something more robust (a resistor or TVS, say), or even returned to the source.

Tim
aiq25:

--- Quote from: German_EE on February 24, 2019, 11:51:01 am ---Why should this be the case? In its simplest form a fuse is a current limiter and if you exceed the rated current the supply will 'disconnect'. It should be possible to build some form of electronic circuit that will disconnect the output of a regulated supply at a specific current draw and put this after the voltage regulation.

--- End quote ---

I have done this before for a work project actually, overcurrent and overvoltage protection. Not exactly a fuse but protection circuitry. But that's not exactly what I'm looking for here though, I want a constant current mode. I think T3sl4co1l explains it well above.
not1xor1:

--- Quote from: aiq25 on February 24, 2019, 07:46:08 pm ---
--- Quote from: German_EE on February 24, 2019, 11:51:01 am ---Why should this be the case? In its simplest form a fuse is a current limiter and if you exceed the rated current the supply will 'disconnect'. It should be possible to build some form of electronic circuit that will disconnect the output of a regulated supply at a specific current draw and put this after the voltage regulation.

--- End quote ---

I have done this before for a work project actually, overcurrent and overvoltage protection. Not exactly a fuse but protection circuitry. But that's not exactly what I'm looking for here though, I want a constant current mode. I think T3sl4co1l explains it well above.

--- End quote ---

You would have more chances of success by using a single opamp, but you have to take into account the working input voltage range and output limits and use a high side reference.
You might use a rail-to-rail opamp or higher/lower supply rails to keep the values within the working limits or some other tricks like in the circuit below where a couple of LT1004 (replaceable with LM385) are used to keep the maximum input voltage within the bounds of the LT1013 specifications and another one, plus a potentiometer, to provide a variable reference for the current control.

aiq25:

--- Quote from: not1xor1 on February 25, 2019, 03:05:37 pm ---You would have more chances of success by using a single opamp, but you have to take into account the working input voltage range and output limits and use a high side reference.
You might use a rail-to-rail opamp or higher/lower supply rails to keep the values within the working limits or some other tricks like in the circuit below where a couple of LT1004 (replaceable with LM385) are used to keep the maximum input voltage within the bounds of the LT1013 specifications and another one, plus a potentiometer, to provide a variable reference for the current control.

--- End quote ---
I understand the basics of this circuit but one thing I can't figure out is the actual equation to figure out what the current limit is... If you can explain it, I would appreciate it. This seems like an interesting circuit and will definitely consider it.
not1xor1:

--- Quote from: aiq25 on February 26, 2019, 05:00:24 am ---
--- Quote from: not1xor1 on February 25, 2019, 03:05:37 pm ---You would have more chances of success by using a single opamp, but you have to take into account the working input voltage range and output limits and use a high side reference.
You might use a rail-to-rail opamp or higher/lower supply rails to keep the values within the working limits or some other tricks like in the circuit below where a couple of LT1004 (replaceable with LM385) are used to keep the maximum input voltage within the bounds of the LT1013 specifications and another one, plus a potentiometer, to provide a variable reference for the current control.

--- End quote ---
I understand the basics of this circuit but one thing I can't figure out is the actual equation to figure out what the current limit is... If you can explain it, I would appreciate it. This seems like an interesting circuit and will definitely consider it.

--- End quote ---

Just remember that thanks to the feedback opamp inputs are kept at the same voltage.
Dismiss the 2.5V of U3 and U4 as they are not relevant for the purpose.
The circuit just tries to keep the voltage through R1 (0.33*I in this test circuit) equal to the value between the cathode-like terminal of U2 and the potentiometer cursor.
When the voltage drop trhough R1 is less than that value the darlington (Q1+Q2) is in saturation and there is no current regulation.

The maximum current is when the pot cursor is connected to the anode of U2.
So let's call:
w = wiper position ranging from 0 (min current - close to the cathode) to 1 (max current - close to the anode)
I = current
R = R1 value
V = U2 voltage
Now we have
R*I = V*w
i.e. the regulated current I is I= V*w/R

The potentiometer value is not critical. High values are more affected by opamp bias, but in any case the intrinsic non-linearity of the potentiometer would likely be greater.
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