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I'm trying to design a current source / sink that will be a part of a step generator that will drive a BJT base. I only intend to drive small BJT transistors so a max base current of 10mA was decided upon (somewhat arbitrary).

The circuit I'm using is the following one: current source

The circuit can be re-configured by changing the polarity of the current source to sink. Using an op-amp with ultra-high input impedance / low input bias this circuit can go to nano amps and I intend to try and get close to that.

The op-amp of choice is TI (BB) OPA128 which has a typical current out rating of +/- 10mA. This seems about right, but I am a little worried that I will be stressing the device.

I have a couple of these devices, can I simply connect two of them in parallel or some other configuration so they will be able to drive a constant current of 10mA without too much "effort"?

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  • \$\begingroup\$ Given the 10ma is arbitary and limited by your choice of opamp, why not choose a different opamp? Or a transistor with a bigger beta? What's the transistor driving? \$\endgroup\$ – pjc50 May 25 '16 at 12:38
  • \$\begingroup\$ @pjc50 I could not find an op-amp with more suitable input characteristics really. The transistor will be the "device under test" in a curve tracer of sorts. \$\endgroup\$ – user34920 May 25 '16 at 12:49
  • \$\begingroup\$ If you're within its ratings you're within its ratings. Why make the design more complicated? (I'd be more worried at 10A, but at 10mA I'd be surprised if you could damage the opamp this way) \$\endgroup\$ – Brian Drummond May 25 '16 at 13:14
  • \$\begingroup\$ +/- 10 ma is the typical maximum output but it could be as low as +/- 5. Why not drive an N-channel MOSFET gate with the op amp output? You could use the voltage in the source-to-ground resistor as your current sense. Put a resistor between the drain and the base of your BJT so that the gate voltage is in the middle of your power supply. \$\endgroup\$ – John Birckhead May 25 '16 at 13:44
  • \$\begingroup\$ @JohnBirckhead Can you please include a schematic? \$\endgroup\$ – user34920 May 25 '16 at 13:50
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You could try an output buffer something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

As well as not exceeding the op-amp output capability you probably want to minimize the self-heating to improve the accuracy. The degradation with increasing die temperature is profound, and there's no sense throwing away performance of a stupidly expensive component (of course maybe you don't need to worry about low bias current and high current simultaneously in this application).

enter image description here


Edit: Below is a simulation you can play with. It's 100mA. You have to keep the output from saturating- so depending on op-amp and current the output might be able get within a few volts or less of each supply bus. The op-amp is supplying 1.4mA to get 200mA of load current (the output carries both the input and output current).

schematic

simulate this circuit

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  • \$\begingroup\$ What is R1 for? In addition, I understand what you say about the offset current degradation, however I can live with 1pA as the lowest I'm shooting for is in the nA range. Won't the transistors suffer from the same problem with even larger variations eventually hurting the output accuracy? Thanks! \$\endgroup\$ – user34920 May 25 '16 at 17:35
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    \$\begingroup\$ I think R1 will improve the stability near zero output current - otherwise it has to slew 2 Vbe near 0. The transistors don't enter into the accuracy nor does their leakage (or temperature) matter much- and they won't heat the sensitive front end. \$\endgroup\$ – Spehro Pefhany May 25 '16 at 17:44
  • \$\begingroup\$ I've simulated the circuit and when the original circuit sinks 100uA (both resistors = 10k), the transistors junction is sourcing 4.436mA with a load of 1k resistor. \$\endgroup\$ – user34920 May 25 '16 at 23:18
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    \$\begingroup\$ Above is a simulation with 100mA, the op-amp current is only 1.4mA. Of course you have to keep the output from saturating. \$\endgroup\$ – Spehro Pefhany May 26 '16 at 3:44
  • \$\begingroup\$ So instead of a stable current source, I'll need a bipolar voltage source like a DAC or something, correct? \$\endgroup\$ – user34920 May 26 '16 at 15:06
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enter image description here

See the picture. The source and drain current are the same. I hope this helps.

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