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I'm trying to build a 0 to 2A variable current source following the Linear app note.

Here's the schematic:

Variable current source schematic

The circuit is a difference amplifier (LT1995) providing gain to the voltage developed across a sense resistor. The amplifier output voltage is fed into a inverting integrator (LTI1880) with a RC network (R = 1k, C = 10 nF). The output from the integrator is fed into a low pass RC filter (R = 100, C = 10 nF) and that voltage is placed on the gate of a P Power MOSFET (IRF9530).

The source of the IRF9530 is tied to the "low side" of the sense resistor (terminal going into the input P1 on the LTI1995 in figure). The drain of the FET I have tied to ground for testing but I plan to have an inductive coil load in the future.

The output current in my measurements does not scale as \$\dfrac{V_{in}}{5 \times R_s}\$. This is my goal for this circuit.

With no load attached (drain left floating), I provide voltage on the input (Vin) pin the amplifier output will develop a positive output as it should but the integrator will quickly integrate down to the negative rail (-15V). The IRF9530 is operating with a very large (-15V) gate to source voltage in this case.

When the drain was tied to ground, and the voltage at the input was tied to ground the the FET gate voltage was -2.2V.

Can someone provide a better picture of the operation of the circuit and places where perhaps I'm assembling this wrong. I have prototyped the circuit on the breadboard.

Datasheets LTI1995:

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Can you provide a schematic of your circuit? – Ricardo May 17 '14 at 20:07
Hi Ricardo, Tanks for response. I have attached a schematic of the entire circuit on imgur: – shraken May 17 '14 at 23:44
If you're trying to build the circuit exactly as in the app note, then there was no need to get the schematic out. Sorry, my bad. In any case, I've inlined the schematic into the question for you. – Ricardo May 18 '14 at 0:55

2 Answers 2

Your circuit is apparently working fine, you just don't realize it.

With the drain left floating, the MOSFET cannot pull any current, and therefor the integrator keeps providing a lower and lower gate drive, trying to draw current.

With the drain grounded and the reference at 0 volts, the integrator goes just low enough to start pulling a tiny amount of current and satisfies the loop.

What you now need to do is get a set of power resistors, such as 10 ohm / 25W, 20 ohm / 25 watt, etc. and use them as loads. From the voltage across your load resistors you can determine the current through them.

However, before you do that, you need desperately to provide a heatsink for your MOSFET. Its' peak power dissipation at a current of 2A will be nearly 30 watts, worst case (for very low load resistances).

You also need to make sure that your shunt resistor is of high enough power. If you're following the app note religiously and using a .2 ohm resistor, it needs to be a 1-watt or better unit, since at 2 amps it will dissipate (2 x 2 x .2) = .8 watts.

You would also do well to reconsider your power wiring. The contacts on a solderless breadboard may not handle 2 amps well.

Finally, you need to make damn sure you can't provide a reference voltage greater than 2 volts.

And finally finally, I suggest you do a good deal of testing with resistors, in order to convince yourself that the circuit really is working. I suspect that you may find it doesn't work nearly as well when you give it a highly inductive load like a coil. It may work, and it may not, since the effects of the inductance will tend to counteract the effects of the loop integrating capacitor, and the result may be instability.

Otherwise, congratulations.

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Hi WhatRoughBeast, Thanks for response and considerations. I have a programmable resistor instrument that can sink 150W so i'll experiment with sweeping the load resistance up to the 500 ohm limit. I have a good heatsink for the MOSFET that I am using on a prototype PCB. Yea that's one of my primary concerns in regards to driving a large coil. First step is to verify that it can drive a programmable 0 to 2 A current into a fixed 50-ohm load. – shraken May 18 '14 at 1:46
Keep in mind that, for a 50 ohm load, you will not be able to drive more than 0.3 amps into it with your 15 volt power supply. (15 / 50 = .3). If you want to run the whole 2A range, set the load to ~7.5 ohms. – WhatRoughBeast May 18 '14 at 4:00
Could someone tell me what the integrator is for? I'd think that direct connection of in amp and mosfet gate would do. Or is it preventing oscillation? – Nicolas D May 19 '14 at 17:09

The work-around I ended up using for this circuit was getting rid off the LPF formed by the 100-ohm and 10 nF capacitor. There must be instability in the loop for when the LPF is included the gate is pulled down to the negative (-15 Volt rail).

With the LPF removed the expected functionality is achievable. The current delivered to the load doesn't follow the equation exactly but it's linear and good enough.

Thanks for the feedback and support. I ran into another problem where output of a DAC is unable to drive the voltage reference (REF). I ended up putting a buffer between and this fixed it.

An alternative topology I found in an Analog Device app note is:

Anyways I appreciate the responses. Cheers!

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