0
\$\begingroup\$

So I'm trying to make a power supply using the L200 regulator. I'm using the circuit in Figure 1. Programmable Voltage Regulator with Current Limiting. I made the circuit after that design. Except that I put a potentiometer in series with R3 to make the current variable. And it does work. The problem that I'm having is this.

I want to current to go up to 2 amps. R3 in my circuit is 0.22 ohms. so 0.45/0.22 = 2.04 amps. BUT the pots I have will only go down to 0.4 ohms, I've tried like 5 of them, is this normal or do I have crappy pots? I don't know how to make the supply go up to two amps.

Is it possible to make the pot go down to maybe 0.1 ohms at least?

\$\endgroup\$
4
\$\begingroup\$

The datasheet shows you how to do it properly.

enter image description here

Figure 23. Programmable Voltage and Current Regulator.

You've already run into the limits of the pot resistance at each end. The other problem to watch is current through the track. For example, a 100 Ω, 1 W pot can take $$ I = \sqrt {\frac {P}{R} } = \sqrt {\frac {1}{100}} = 100 mA~max $$ This is the maximum current that any part of the track can handle. Don't, for example, think that because you are at 50% setting that it can handle 200 mA.

The Figure 23 solution uses the pots as a control signal. The end of track resistance should be negligible on a 10k or 100k pot.

[OP's question:] Will this work with any opamp or do i need to use the 741?

The 741 was one of the early opamps and nearly anything made since then would be better. Read the datasheet note about connecting point 'A' to a negative supply if you want to get the output down to 0 V. Other than that, just check the operating voltage and how close to the supply rails the device is able to get.

\$\endgroup\$
2
  • \$\begingroup\$ Oh i didn't thank of that. Thanks for the tip!! Will this work with any opamp or do i need to use the 741? \$\endgroup\$ – Xane Jul 11 '16 at 6:38
  • 1
    \$\begingroup\$ See my edit regarding 741. \$\endgroup\$ – Transistor Jul 11 '16 at 8:26
3
\$\begingroup\$

That is completely normal, particularly with carbon pots. What you are measuring is a combination of the contact resistance of the wiper, and the fact that the wiper doesn't go exactly to the end of the element. A wirewound pot may be able to go lower, but even so, 0.1Ω is a tall order.

You may need to adjust other parameters of the design instead.

Also relevant is that your multimeter may not even measure resistances below 1Ω accurately. For instance, the $20 DMM on my desk measures 0.4Ω when the probes are shorted together, while the Agilent U1273A at work (~$400 DMM) measures something like 0.06Ω when shorted. If you really want an accurate measurement, use a 4-wire resistance measurement and (ideally) a quality bench DMM.

\$\endgroup\$
1
  • \$\begingroup\$ I see thanks for your answer! then i might add a switch to turn the current limit on/off. When it's on i might be able to go up to 1 amp with the limit, when it's off it will work with the full 2 amps. Thanks :D \$\endgroup\$ – Xane Jul 10 '16 at 16:17
0
\$\begingroup\$

It's impossible to get value near 0.1 ohm , considering that any conductive item like the pot present itself some resistance also in the external contact. Resistance very near to 0 ohm is only possible by using superconductors.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.