Help with the output (load) current of this differ. amp circuit?

[ChrisP58]

You have all three of your shunt references in parallel, so you will only see the lowest voltage in all positions of your switch. You need to give each of them their own dropping resistor. Or, tie the switch common to the resistor and op-amp pin, and the three switch points to the individual references.

But, you can probably use just one reference, and a three resistor divider to make the other two voltages. Just make sure that the values in the divider are significanly lower than the resistance values in the op-amp.

 

[SandiegoSD]

You have all three of your shunt references in parallel, so you will only see the lowest voltage in all positions of your switch. You need to give each of them their own dropping resistor. Or, tie the switch common to the resistor and op-amp pin, and the three switch points to the individual references.

But, you can probably use just one reference, and a three resistor divider to make the other two voltages. Just make sure that the values in the divider are significanly lower than the resistance values in the op-amp.

Yes..good heads up. Switch should come before the zeners, not like what i sketched.Also the references are cheap and resistors are hard to match, so I think i'll go with 3 references.

 

[SandiegoSD]

(almost) Fried the protoboard today..

Hi folks, I received the parts for the circuit I've been talking about (I ordered mostly thru hole parts) plugged them onto a breadboard to test it. Oh well.. about some 5 seconds after I turned on the power supply I smelled the frying plastic/something else's odor and decided to touch every part to find out which was overheating. I instantly regretted it when I put my index finger on the back of the LM2940. That thing was HOT!!

I then realized that with (24-10=14)V across this linear regulator, and the 1A current which I'm trying so hard to get, it's taking some 14W on that little piece. The power rating for LM2940 is a measly 2~3W w/o heatsink.
**broken link removed**

I guess i'll use another voltage supply(probably 12V instead of 24V) to power the LM2940 regulator so the voltage drop is ~2V and total power will be closer to 2W and it shouldn't overheat too much.

Hopeful by then the load current will turn out to be 1A (I'm secretly worried because for the first few seconds the load current was only .74A before the LM2940 goes into thermal shutdown )

p.s. the circuit I'm talking about:
**broken link removed**
There's a load resistor in series with that amp-meter not shown in this graph.