I'm working on a simple (to most of you) constant current load. It's a spin on the one that you see everywhere that Dave did an EEVBlog on. (drawing attached) The OpAmps are from an old ta75324 quad package, and the fet is an irfz24. I realize there are better options, but I have a bunch of these.

The load works fine when I put a supply on the L+/L- and drive the first opamp, but I'm seeing 16.3V at the mosfet gate when no load is attached, or when no voltage is applied to the non-inverting input of the first opamp. It would be completely functional as-is as an analog load, but I plan to drive it with a mCU/DAC, and I would prefer not to have to mess with ensuring that the DAC is applying a voltage to the set pin before powering the load circuit. Instead, I would prefer if the fet was off when no power is applied to the set pin (I'm calling the non-inverting input of the first op-amp "set pin.")

I'm assuming that the problem is with the lack of feedback when there is no load voltage, but I can't think of a way around it without effecting the basic circuit.

Is my best option to drive the opamp supplies through a transistor with the base tied to the "set pin?"

Thanks

 

I should have mentioned, because I know some of you will call me on it, that I realize it needs some passives here and there to help with oscillation etc, and it will get them, but I'm still learning, so I'm starting as simple as possible and building up. More components == more points for troubleshooting.

 

You can't leave an op amp input with an open circuit, as the input bias current has no place to go, thus the op amp output will saturate at one of the rails.
Try connecting a 10kΩ resistor from the (+) input to ground of the left op amp.

But it's a constant-current current and will always try to generate a current, even with no load.
So you either have to connect a load to it or, if there's no load and you want it still to be operating in the linear region, replace the load with a short.
That way it will still work, generating a constant current through the short.

The basic rule is that no load for a voltage supply is an open circuit, but no load for a current supply is a short circuit.

Is my best option to drive the opamp supplies through a transistor with the base tied to the "set pin?"

I have no idea what you are asking.
Why do you want to "drive the op amp supplies"?
What transistor?
What "set pin"?

 

The resistor is perfect. I'm a dumb***. Thank you. I was over on the right-side of the circuit thinking way too complicated, and never even considered anchoring that input.

The basic rule is that no load for a voltage supply is an open circuit, but no load for a current supply is a short circuit.
I have no idea what you are asking.
Why do you want to "drive the op amp supplies"?
What transistor?
What "set pin"?

Again, this was overcomplicating it, but the idea was to use the input (the + pin of the left-most opamp) to turn off/on a transistor as a switch for the supply to the opamps. So that they have no power until they're being driven.

 

So are you satisfied with its operation now?

 

It isn't done by any stretch, but I'm sure that will take care of that problem. I say I'm sure because I killed the amp by mixing up the supply lines. Another reason I should have gone to bed a few hours before. When a pulldown resistor at the floating input on the opamp never occurs to me, and I can't keep +/- straight in my head, I should probably go to bed.

I still need to work out some things in the feedback loops (filtering and balancing ripple and response time etc), but I don't expect I can do any better than what you did in a recent post here for the peltier sink, so I'm going to start as close to there as I have the components for and actually build an analog version to have if I need it. Then I can get back to the digital version that is the end goal.

 

You can't leave an op amp input with an open circuit, as the input bias current has no place to go, thus the op amp output will saturate at one of the rails.
Try connecting a 10kΩ resistor from the (+) input to ground of the left op amp.

But it's a constant-current current and will always try to generate a current, even with no load.
So you either have to connect a load to it or, if there's no load and you want it still to be operating in the linear region, replace the load with a short.
That way it will still work, generating a constant current through the short.

The basic rule is that no load for a voltage supply is an open circuit, but no load for a current supply is a short circuit.
I have no idea what you are asking.
Why do you want to "drive the op amp supplies"?
What transistor?
What "set pin"?

Crutschow is right!