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I have to design an analog circuit to measure currents in the nano ampere range as part of a course, and from what I found on Wikipedia and some documentation available online from Tektronix, it seems that I need to make what is called a current sense amplifier. Is that right? like is a current sense amplifier a system which one can use to measure very small currents? because in all the designs I looked at, the output is a voltage proportional to the sensed current, so how do I actually display the output as a current? Do I just put a resistor at the output and measure current across it/calculate the current by hand by dividing the voltage by the resistance?

Am I thinking in the completely wrong direction? I don't know much about this, because so far all the circuits I have learned about have been some sort of voltage amplifier i.e. we tend to focus on the voltage gain of the system, except for power amplifier where you need to amplify current.

UPDATE #1: As a part of the project, I am not allowed to use any off the shelf operational amplifiers.Is there any way to design the circuit with simpler components i.e. transistors,resistors,capacitors etc. ?

Any help/reference/link to material where I can learn about how to design such circuitry would be appreciated!

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    \$\begingroup\$ eevblog.com/projects/ucurrent The EEBlog guy designed and sells one, but all the design info is published. \$\endgroup\$ Feb 26 at 14:32
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    \$\begingroup\$ @ScottSeidman Just want to point out that the course instructor may know about that one so just copying it could be a problem. \$\endgroup\$
    – GodJihyo
    Feb 26 at 15:09
  • \$\begingroup\$ Does the course instructor not have office hours so you could go ask them? \$\endgroup\$ Mar 1 at 1:08
  • \$\begingroup\$ New update: I can't use pre-built op-amps i.e. any off the shelf op-amps, is there a way to design one without a pre-built op-amp? I can only use transistors, caps, resistors, diodes and simple components. \$\endgroup\$ Mar 1 at 6:26

4 Answers 4

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Do I just put a resistor at the output and measure current across it/calculate the current by hand by dividing the voltage by the resistance?

Almost. I think you have the right concept but are using imprecise terms.

  1. The resistor is not "at the output" but "in series with the circuit", so that the current flows through it.

  2. You measure the voltage across it (not the current).

Other than that, yes, that's how you do it.

Some practical nodes:

  • The measuring device (the meter) must have an input resistance that is at least 100 times higher than the resistor.
  • If you were designing a practical instrument, you would use an op-amp with nanoAmp input bias current and place the resistor in the feedback.
  • Instead of using a single feedback resistor, you would use a T-network which allows you to use lower value resistors.
  • However, this method also amplifies the input offset error, compared to using a single resistor. Therefore, use a "zero-offset" chpper stabilized op-amp. For example, the LTC1052 has an input offset of 5 uV and a bias current of 30 pA.
  • Regardless of the feedback topology, great care must be taken to shield the '-' input through measures such as a guard ring around the '-' input of the op-amp and even mounting the '-' input on a Teflon terminal.

schematic

simulate this circuit – Schematic created using CircuitLab

Here my best advice: Rather that attempting to build your own, you're much better off checking out a "picoammeter" from your university's electronic shop.

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    \$\begingroup\$ If the assignment is "I have to design an analog circuit to measure currents in the nano ampere range" I don't think just picking a commercial digital picoammeter out of the university's electronic shop (assuming they have an electronics shop and it has such a meter) would go over too well with the instructor. Or they might see it as a stroke of genius, could go either way. :D \$\endgroup\$
    – GodJihyo
    Feb 26 at 14:58
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A current sense amplifier can mean many things but, what springs to mind is a transimpedance amplifier circuit. Basically you get volts out for a current in i.e. it has a gain of volts/amps = resistance (impedance).

This is the basic idea from Wikipedia - Transimpedance amplifier: -

enter image description here

Due to the virtual ground arrangement of the circuit, the input impedance will be very low for steady state input currents (what you need with any ammeter). The output voltage is negative for an input current in the direction as shown so, don't forget to use a spit supply like +/- 5 volts.

You use this for small currents (below around 10 mA maximum) and, you choose the resistor value in line with what the voltage output has to be for the current input. So, if you want 1 volt per nA, the resistor value would be 1 GΩ. But, you can't use any old op-amp for this because it will introduce bad errors due to: -

  • Input bias current being too high
  • You should aim for an \$I_B\$ of 10 pA maximum for 1 nA input (1% error)

You can certainly get op-amps with bias currents less than a pico-amp so, it shouldn't be a big problem finding one.

But also be aware of the promotion of the use of a potential divider in the op-amp feedback circuit in order to use lower values of resistor: -

enter image description here

You should also take note of thermocouple effects when using an amplifier configuration with significant gain (as per the three resistor feedback configuration used above). The thermoelectric effect of an op-amp lead and copper track is 35 μV/°C. At a constant temperature and with a circuit voltage gain of 1000, this manifests as a 35 mV DC error at the output. Of course, if the temperature rises by 10 °C, this will produce a 350 mV error.

This is why I recommend not using the three resistor feedback circuit and stick to a single feedback resistor (if you want decent and reliable accuracy).

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  • \$\begingroup\$ I suggest you replace the feedback resistor with a "T" network to let you use readily available resistors instead of a 1 GΩ resistor. (See my answer) \$\endgroup\$ Feb 26 at 14:34
  • \$\begingroup\$ That's what I'd recommend, but I just realized that if the current of interest doesn't flow "out" of a source, but within some system (which we don't want to influence by measuring), I'd need a precision current mirror? Or how would one approach "in-circuit" measurements? \$\endgroup\$ Feb 26 at 14:35
  • \$\begingroup\$ ", I'd need a precision current mirror?" - No. You would use a resistor in series followed by a differential amplifier. \$\endgroup\$ Feb 26 at 14:37
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    \$\begingroup\$ @DavideAndrea 1 Gohm resistors are readily available and I'll thank you for not blatantly promoting your answer in a comment below my answer (as if there was something really difficult about getting 1 Gohm resistors). \$\endgroup\$
    – Andy aka
    Feb 26 at 14:37
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    \$\begingroup\$ I agree with Andy on this one. The T-network in feedback also amplifies noise at the input by the same amount as the offset voltage. A single feedback resistor is best in a transimpedance amplifier. \$\endgroup\$
    – RussellH
    Feb 26 at 19:52
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You only need a current sense amplifier if you are required to have a very low† "burden" on the current input. Otherwise you can just use a resistor and an appropriate op-amp voltage amplifier (or buffer), which is exactly what ammeters generally do.

For example:

schematic

simulate this circuit – Schematic created using CircuitLab

Your job then would be to pick a combination of op-amp model, Rs and ratio of Rf/R1 to meet the design requirements. For example, you could have a resistor of 10MΩ that would drop 100mV at 10nA and then with a ratio of 9 you'd have 1V output for 10nA. Errors come from input offset voltage, input bias currents and resistor tolerances (and other sources).

In fact if you have a DMM with an accurate 10MΩ input resistance on the 200mV range you could use it directly as a nanoampere current meter (20 nA FS).

† very low meaning of the order of op-amp offset voltage, which is another design requirement that has to be met

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For measuring a leakage current: Charge the capacitor with this current for constant period of time and then perform sample-and-hold of capacitor voltage. This voltage is proportional to measured current according to $$i = C \cdot \frac{dV}{dt}.$$

(dV is cap voltage rise, dt is charging time).

After every period discharge the cap to 0V of course.

enter image description here

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