I want to make an adjustable power supply with -18 V and 18 V separately adjustable outputs. I found this circuit might be the best and simplest one to do the job.
How can I limit the output current to around 500-750 mA? Or should I just use an efuse?
For the transformer I was planning to use a 24V center tapped, but if it isn't enough to get an 18v output I can replace the transformer no problem.
You can use e-fuses (or any fuses) prior to the regulators, or you can do it the classic "active" way:
simulate this circuit – Schematic created using CircuitLab
Choose R1 and R3 according to maximum current permitted \$I_{MAX}\$:
$$ R_1 = R_3 = \frac{0.7}{I_{MAX}} $$
The value of 1.2Ω shown will limit current to about 600mA.
You should also consider the power that R1 and R3 will dissipate under full load. Here, 600mA through R1 would cause it to dissipate power \$P_{R1}\$:
$$ P_{R1} = I^2R_1 = (600mA)^2 \times 1.2\Omega = 0.43W $$
R1 should be rated 0.5W or more.
The most power will be dissipated in transistors M1 and M2, when the outputs are short-circuited. In that state M1 and M2 will dissipate about:
$$ P_{M1} = IV \approx 20V \times 600mA \approx 12W $$
The regulators will also be dissipating similar power when their outputs are near ground.
This is the biggest obstacle when building current limiting systems, and linear voltage regulators. The elements responsible for regulating can get very very hot. In this case, you really need to but the regulators and M1 and M2 on heat sinks, and maybe even put a fan in there.
Also, I haven't drawn in the usual capacitors that should accompany the regulators; don't omit them!
Every datasheet I've seen for the LM317 and 337 includes an application circuit for a current limiter. Here is a link to the current version of the 317 datasheet. See figure 14:
Here is an image grab from that document.
If you put this circuits (1-317 and 1-337) in front of the ones in your schematic, there will be two 317/337's in series for each output voltage, one as a current limiter followed by one as the voltage regulator. You will need a minimum of +/-24 V across the filter capacitors.
This is fewer parts than the classic 2-transistor current limit circuit. AND, the LM317 is an excellent current regulator. Regulation is based on the value of its internal voltage reference, which is more accurate and far more temperature-stable that the forward voltage of a transistor's base-emitter junction that the classic circuit relies on.
Compared to the traditional circuit, this approach requires a higher voltage across the current regulator. But the performance is a huge improvement.
Update:
Power dissipation in the components can be calculated with Watts Law:
P = E x I
and Joule's Law:
P = I^2 x R
The image shows a variable resistor, but many normal small pots cannot handle 750 mA. Better to start with a fixed resistor for a fixed current limit value. Use the given equation to select a standard resistor value. For example, a 1.6 ohm resistor yields a current limit value of 750 mA.
Using Joule's Law, the power dissipation at max current is 0.9 W. For better long-term reliability, use a resistor rated for a minimum of 2 W.
Power dissipation in the ICs depends on the voltage across the filter capacitors. Please update your question with your transformer output voltage under load and the value of the filter capacitors. As a cross-check, also show the voltage across the filter caps.
Below the current limit value, the voltage across the current 317 is fairly constant, and the power dissipation is directly proportional to the output current (Watt's Law). The voltage across the voltage 317 varies inversely with the output voltage. Its power dissipation depends on both the output voltage and current.
When the limiter kicks in, things change. The minimum power dissipation in the limit and voltage 317's will be at least 2 W each, demanding heatsinks.
More later, after more information.
One of simplest is to make followers with two Darlingtons, references from zeners, and current limiters from 1ohms+bjts (I_lim= 0.6V/1ohm=0.6A).
The input voltage should in +-24V to +-35V range.
... @Michal Podmanický and @AnalogKid have tickled my imagination, which is freewheeling roughly in this direction:
I.e. a long-tailed diff pair to do voltage comparison and negative feedback regulation, plus a current mirror whose emitter ends are used as another differential input, adjustable by a trim-pot, sensing the voltage drop across a shunt resistor. A double diff amp using just four extra BJT's :-) Hans Camenzind is my hero. I haven't simulated the circuit and haven't calculated the resistors, apologies...
