I'm designing a circuit that will allow me to do transistor testing (similar to a curve tracer). One part of the circuit is a power supply that will apply different voltages to the collector / emitter.

** I have posted a question regarding the same circuit, however this is a completely different question. **

I need to monitor the current following into the transistor so I'm using high side sensing because:

  1. I want to detect possible shorts at the power supply (I>250mA)
  2. I have a few AD8418 current sense amplifiers that fit the job.

The AD8418 has a gain of 20 V/V so I can get the range I'm interested in using a stable 1 Ohm resistor. Some online reading and searching brought to my attention that "high" values such as 1 Ohm are rarely used for current sensing. However, using a much lower values resistor will result in a much lower output voltage which will have to go additional amplification before being sampled by an ADC. I'm afraid that another stage will add noise, offset error and gain error.

Should I go ahead and use a "stable" 1 Ohm resistor or try another, more complex, solution?


2 Answers 2


This is mostly a question of:

  1. How much voltage are you willing to drop?
  2. How much voltage do you need to drop?
  3. How much power can you afford to lose?

If you're measuring 10 A, you're going to dissipate 100 watts through your resistor, but for you, 1 Ω may be perfectly fine. Let's run the numbers:

  1. At 250 mA, the voltage drop is 250 mV. From what I understand, this is the highest current you are interested in. Since you seem to have control over your input voltage, this doesn't sound like a problem.
  2. You have done this calculation already. At 250 mV and a gain of 20, you're looking at 0 to 5 volts full range.
  3. The power through the resistor is I²R, 62.5 mW. As long as you use a large enough resistor (physically), this is not a problem. Even a tiny 0805 is usually rated at 125 mW, but you may want a larger package to combat self-heating.

I'm afraid that another stage will add noise, offset error and gain error.

Sure, it might. But the AD8418 isn't perfect either. The key to a good design is to calculate what kind of accuracy you need. The offset and gain error can be ignored: That's just a multiplication and addition in the microcontroller. The extra noise, well. That can be filtered out, if it's a problem.

Should I go ahead and use a "stable" 1 Ohm resistor or try another, more complex, solution?

It seems to me that using 1 Ω is perfectly reasonable.

  • \$\begingroup\$ Thanks for the quick answer! I'm a bit worried about the offset, gain error and other "distortion" sources because everything will be required calibration to some degree. The ADC gain error and offset errors are easy to null. The non linearity resulting from the resistor + current sense amp will keep me busy until I can get rid of them, as much as possible. I am hoping I can lump them together and fix the calculations using a second order formula, nothing higher. I'm an optimistic. \$\endgroup\$
    – user34920
    Commented Sep 25, 2016 at 11:46
  • \$\begingroup\$ @user34920 I guess I would be most worried about resistor self-heating in combination with its temperature coefficient. That's pretty much what you can't remove by calibration. Of course, if you're going to make a lot of these you may not want to calibrate, but it sounds like this is just a one-off. \$\endgroup\$
    – pipe
    Commented Sep 25, 2016 at 11:57
  • \$\begingroup\$ yes this is a single unit. I will get the resistor with the lowest temp. coef. I can find. \$\endgroup\$
    – user34920
    Commented Sep 25, 2016 at 12:00

If you use a non-inverting op-amp configured with a 10R resistor fed back to the inverting input, the voltage at the inverting input is precisely what is demanded by the non-inverting input whereas the voltage across the 10 ohm resistor will tell you what current flows. This gives you a high signal level but of course it's only good for a few tens of mA into the load so, buffer the output of the op-amp with a source follower to give much more current drive capabilities: -

enter image description here

This will simultaneously provide an accurate DC voltage onto the device under test and give you 1 V per 100 mA across the 10 ohm resistor. You do need a power rail that is several volts above the voltage to be fed to the device under test so be aware of this. Also, at 250 mA the resistor power dissipation will be 0.625 watts so choose this resistor carefully.


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