I am working on a project. The goal is to differentiate between short circuit and load. I am trying to use differential amplifier. At load there are 2 to 3 millivolts. I need to compare them with zero volt and get a reasonable volt at output of this opamp.

How do I differentiate between a load and short circuit on the output of a differential amplifier?

  • 2
    \$\begingroup\$ A differential amp is probably the correct way to do this, but you'll have to be careful about offset, layout and noise to accurately distinguish a couple of mV from a short. If the load is resistive you could try using a current sense amplifier. It's also important to know how quickly you need to react to a short. \$\endgroup\$ – John D Apr 22 at 14:56
  • \$\begingroup\$ What are you protecting? and how much current can the voltage-divider use to set the threshold? how fast MUST the short-circuit (200% of nominal?) be detected? Using 100Kohmresistors in a 1MHz bandwidth circuit, you will have 40 microvolt RMS, or 250 microvolt PeakPeak 6 sigma (1 ppm) noise. \$\endgroup\$ – analogsystemsrf Apr 22 at 17:13
  • \$\begingroup\$ @john D .for the problem i mentioned,how to use current sense amplifier you mentioned? \$\endgroup\$ – Muhammad May 2 at 1:38
  • \$\begingroup\$ @Muhammad take a look at products like these: ti.com/amplifier-circuit/current-sense/analog-output/… . There is lots of application information in the datasheets. \$\endgroup\$ – John D May 2 at 14:39

For this application I would recommend a comparator (since all you care about is whether or not your millivolt threshold is exceeded and not the information in the signal itself).

Inverting comparator configuration

Use the R1/R2 divider to set your reference voltage to a few millivolts. You will want to use high-grade components depending on the level of accuracy you are after. Alternatively you can change the divider out for a potentiometer which would allow you to set arbitrary thresholds.

In the inverting configuration, once the Vin is below Vref the output will go high indicating a "short".

  • \$\begingroup\$ It will be very difficult to get a single stage comparator to work when the threshold voltage is so small. On a 3.3V regulated supply you're looking for a threshold that is 500 ppm of the supply, or 0.05% of reference. You probably need a differential amplifier stage first. \$\endgroup\$ – scorpdaddy Apr 22 at 17:06
  • \$\begingroup\$ @ROB, Thankyou very much. input signal is AC.Secondly there are 2.5 millivolts across load.So if voltage across load are below 2.5millivolts,means there is short circuit. How can i proceed? Can i drive relay from ouptu of above meniode topolgy advised by you? \$\endgroup\$ – user219130 Apr 22 at 22:53
  • \$\begingroup\$ @scorpdaddy A pre-amp may be unecessary depending on the exact threshold desired and accuracy. The AD790 for example has an offset of 0.25mV and hysteresis of 0.5mV. \$\endgroup\$ – Rob Apr 23 at 3:12
  • \$\begingroup\$ @user219130 If it's an AC signal then you can't use the above circuit as it is because the output will oscillate according to the frequency. A simple solution would be to flip the inputs to form a non-inverting configuration but this would mean that the output during normal operation is a square wave and the output during a "short" is zero. It really depends on what kind of circuitry you are using to monitor the output. \$\endgroup\$ – Rob Apr 23 at 3:21
  • \$\begingroup\$ @Rob..can you please send me circuit for understanding ? \$\endgroup\$ – Muhammad Apr 23 at 7:47

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