There is often a 100 ohm resistor between the +sense and +ve and between the -sense and -ve outputs of bench power supplies. This protects the output from high voltage if the sense wires are accidentally disconnected. So why are these resistors low value like 100 ohms? What if it's 1K?

I'm making a 6V power supply and can't seem to see much difference in the output whether it is 100 ohms or 1k. The output voltage always falls if the sense wire gets disconnected, whether the resistor is 100 ohms or 1K.

Also, it seems 1K is better, if the actual power line gets disconnected. The 1k will survive, while the 100 ohms will burn up, possibly taking the sense wires with it.

So, what am I missing?


I think the question was not clear. There are no feedback resistors involved here. These are the resistors I am talking about: Rs and Rs shown below. They will burn up if they are low value and the output cable is disconnected (they are usually 100 ohm 0.5 watt; sometimes even 5 ohm or 10 ohm). But they will not burn up if they are 1K 0.25W. Image from here and here.

enter image description here

I cannot seriously think the resistor is low value to decrease noise. There must be dozens of much higher value resistors in the power supply itself. And this is not a feedback resistor, it is a protection resistor (protecting against inadvertent high voltage at the output if the sense wires are accidentally disconnected). Almost no current should be flowing through them during normal operation.

PS: my power supply has only the +ve sense wire, so will have only the one resistor. Ground return is directly connected, no sense wire or protection resistor.

update: to make it more clear, this is the output stage that I currently have (after putting in the back to back diodes). The power supply is rated 6V 1A, but with sensing failure may go upto 9V 1A. The 40m ohm resistance comes from a dual mosfet protection system. The 1N5408 is to protect against accidental high input voltages from output (and prevent the crowbar from nuisance tripping). It can be bypassed. The 20m ohm is output cable resistance. Please ignore the 10k resistor, it is temporary. Local sensing can be done before the mosfet switch or at the load (indicated by the unconnected bent wires going from the op amp). Circuit works fine with all resistor values. If I were to put in parasitic capacitance to simulate, where should it be?

enter image description here

  • \$\begingroup\$ The higher the value, the noisier the result. \$\endgroup\$ – Ignacio Vazquez-Abrams May 8 '18 at 23:30
  • \$\begingroup\$ Power supply is usually supplying DC \$\endgroup\$ – Indraneel May 8 '18 at 23:35
  • \$\begingroup\$ The resistors don't care. \$\endgroup\$ – Ignacio Vazquez-Abrams May 8 '18 at 23:43
  • 1
    \$\begingroup\$ Given the schematic you show, how much current is likely to be passing through the resistors? \$\endgroup\$ – Ignacio Vazquez-Abrams May 9 '18 at 9:39
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    \$\begingroup\$ Then you haven't looked closely enough. There is an error amp in between the resistors. \$\endgroup\$ – Ignacio Vazquez-Abrams May 9 '18 at 11:23

The error amp generally has an input impedance in the few k region, as it is often really an opamp with feedback, so you want the sense resistors to be small (But large enough to be swamped by the cable conductance) to minimise error.

Secondly, an unconnected set of sense leads look like a capacitor, so you want the sense resistors small so that the resulting pole is way above the loop frequency to avoid impacting the loop stability.

Thirdly resistors add voltage noise, more for higher values, it's a physics thing.

One canny approach is to parallel the resistors with a pair of back to back diodes of sufficient robustness to handle the output current, means that you get at most a 1.4V error.

  • \$\begingroup\$ Thanks. Does this mean that if my error amp is not a differential amplifier, but just an op amp with no feedback from the output pin (because I just have one sense wire on the +ve line, no ground sensing), the value of the resistor will not matter as it is not influencing any op amp feedback resistors? Also, with one sense wire there will not be any capacitance between it and ground. Thanks for the diode tip, that is working perfectly. \$\endgroup\$ – Indraneel May 9 '18 at 12:41
  • \$\begingroup\$ Of course there will be capacitance, there is always capacitance, in this case, likely to the common return wire. I would suggest that almost any opamp likely has sufficient GBP that you would need at least some capacitance in feedback to get the thing to be stable. There is a reason almost everyone uses resistors of less then 1k here. \$\endgroup\$ – Dan Mills May 9 '18 at 12:49
  • \$\begingroup\$ I updated with the output section of my supply. I'm using slow op amps (LM358), can oscillation still be an issue? \$\endgroup\$ – Indraneel May 9 '18 at 12:55
  • \$\begingroup\$ Of course oscillation can still be an issue, even the LM358 has a GBP of ~1MHz, and what power device are you using as the pass transistor? Do a small signal Bode plot (Easy here you can just place a signal generator between the output and the sense pin and plot the output gain and phase). \$\endgroup\$ – Dan Mills May 9 '18 at 15:07
  • \$\begingroup\$ I'm using a TIP122. I've never done a bode plot before, will have to look it up. There is also a differential amp for high side current sensing, not shown in my partial schematic above. So far everything is simulating without any oscillations (not even with the 1mF output cap visible above). \$\endgroup\$ – Indraneel May 9 '18 at 15:27

Well, like you said, these resistors are clearly there for the sense lines to follow the output voltage. The only reason I can think of that they are low value is because with low value the sense line „react“ faster. Think of some small capacitance between the sense lines. If the output changes, then this capacitance is charged faster with low Rs.


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