0
\$\begingroup\$

I'm working on a circuit that can act as a current sink or source. The goal is to control it using some digital means so it can output the required current. Right now I am not sure about the implementation of the resistor control, perhaps someone has a suggestion? The circuit has to be very accurate and stable.

schematic

simulate this circuit – Schematic created using CircuitLab

The output is set by adjusting R2. If R1 = R2 the output will be 1mA (I1), if R1 = 2 x R2 the output will be 2mA (2 x I1). If the current is fed sourcing, this will act as a current source. If the current is fed sinking (reverse current source polarity) this will act as a current sink.

The design requires stable current source / sink. The design requires very low input bias current op-amp - something that costs $15+ and can be obtained easily The design calls for very accurate resistors as the relations between them sets the output. I would like to put 2 or 3 resistors with relays instead of R2 and switch between them.

What do you think about a digital potentiometer?

\$\endgroup\$
  • \$\begingroup\$ How accurate and how stable? \$\endgroup\$ – EM Fields Mar 27 '16 at 21:36
  • \$\begingroup\$ Can you fix "The design required for a very stage current source / sink - no problem" which isn't a sentence? What is the range of the required current? +/- how many mA? \$\endgroup\$ – Transistor Mar 27 '16 at 21:39
  • \$\begingroup\$ Your current level is apparently 1 mA. Op amps with offset currents under 1 nA are trivially easy to find. Are you really claiming that you need nA accuracy? That's an accuracy level of 10^-6. So what do you need, really? \$\endgroup\$ – WhatRoughBeast Mar 27 '16 at 22:10
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Dual-rail adjustable constant-current source.

Would the circuit of Figure 1 satisfy your requirements? The output current \$I_L\$ is set by \$V_{IN}\$. Relationship is given by \$ I_L = \frac {V_{IN}}{R3}\$.

For example: \$V_{IN} = +1~V\$ will result in +1 mA through the load.

Note that neither end of the variable load can be grounded with this circuit.

For micro control replace R1, etc., with a bipolar DAC output or add a negative offset to a regular 0 - 5 V ADC or filtered PWM output.

I don't see why you would want or need to use relays or digital potentiometer.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ This has the problem that neither end of the load can be grounded. That may or may not be a problem for the OP. \$\endgroup\$ – Transistor Mar 27 '16 at 23:57
0
\$\begingroup\$

A typical solution for a bipolar voltage-to-current converter is a Howland current source aka Howland current pump.

enter image description here

You need to ensure you have sufficient voltage across the sense resistor to get decent accuracy, and as the schematic shows, resistor matching will determine how high your output impedance will be (you'd prefer infinite for a current source), so a trim (as shown) may be desirable. The gain selection can be done by relays if necessary. You could try to use one resistor value, but the accuracy will tail off pretty quickly if you need to cover several orders of magnitude in current).

The schematic is from this TI document, which has an in-depth analysis of the various pitfalls with this circuit.

| improve this answer | |
\$\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.