Op-amp/Potentiometer problem

[Deleted member 113489]

Hi,

First of all, I am an absolute novice with electronics. In fact, novice is probably being generous. I know nothing. I haven't done any kind of electronics since school 4 years ago, and that wasn't particularly advanced and was all theoretical anyway.

However, I'm doing a research degree and it's become necessary for me to build a circuit as part of my apparatus. My supervisor helps me where possible, but I've encountered a problem that I can't explain and would really like to understand and know how to fix before I go back tomorrow morning, as it'll bug me all evening otherwise.

The circuit is an op-amp with a potentiometer. The op-amp is a 324, with a 9V power supply. The problem I'm having is that for some reason, the output is always very close to 9V, regardless of how the potentiometer is set, even though the voltage from the potentiometer is varying from 0V to 9V when I turn it, as it should be. I tried replacing the IC, and had the same problem but with the output value always being very close to 0V instead. I swapped the two several times, and one was always around 9V, and the other always around 0V.

From your experience, is this likely to be a problem with both ICs, or with the potentiometer, or perhaps with something else?

Here's a drawing of the circuit. It's from memory, so excuse any errors - I'm pretty sure the real thing is as it should be, although if you spot something obvious, please point it out!

**broken link removed**

 

[duffy]

https://www.electro-tech-online.com/threads/basic-opamp-circuits.35258/

 

[Deleted member 113489]

Okay...but I'm not sure how that answers my question. Is my circuit wrong in some way?

 

[duffy]

Look at the one on the upper-right. See how that input has two resistors making a voltage splitter? In your circuit, you are relying on the input (IN) being midpoint biased. This could be the problem. If the input isn't perfectly balanced at half the supply voltage (say there's a cap on the other side) then that 1M resistor on the non-inverting input will pull your circuit to ground. If there's no input, it will DEFINITELY pull it to ground.

 

[Deleted member 113489]

Okay, I think I follow. So do I need to change the value of the 1M resistor, or is it more complex than that?

I should mention - the input has indeed always been 0 when I've been testing the circuit - I've been doing static tests. The input will ultimately be from a BNC connector, and the output will also be a BNC. The purpose of the potentiometer is to adjust the gain of the apparatus connected to the BNC input.

Thanks for all the help so far.

 

[duffy]

No, you need TWO resistors instead of one. You need another 1m resistor connected from the top of that one to 9v.

I wish you would draw that schematic right, you don't even show a potentiometer at all.

 

[Deleted member 113489]

No, you need TWO resistors instead of one. You need another 1m resistor connected from the top of that one to 9v.

I wish you would draw that schematic right, you don't even show a potentiometer at all.

Okay, thanks. The potentiometer is the thing before the 47K resistor going into the -ve part of the op-amp. Sorry if that's not the standard way of showing it - I'm just copying a 63 year old.

 

[duffy]

I figured that. Try it with the second resistor.

 

[Deleted member 113489]

I figured that. Try it with the second resistor.

Will do, thanks. By going to 9V...does that mean it needs to be connected back to the +ve power pin of the IC?

Sorry that this is all so basic - I didn't even know how to solder properly until a few days ago.

 

[duffy]

See where it says "+" on the amp? That's the non-inverting input. One end goes there. The other end goes to +9V, and yes, that's a power pin on the IC.

 

[Deleted member 113489]

I drove back and added a 1M resistor as shown in red, before I saw your last post. It didn't work - in fact, the inputs now have voltages of around 7.5V where previously they were 0V. But perhaps I got it wrong?

**broken link removed**

Based on the diagram you linked me to, do I perhaps need another capacitor between +9V and the bottom mini-circuit as shown in blue? Thanks again for all the help.

 

[duffy]

That's the right place to put it, and 7.5V is closer to the midpoint bias than 9V or 0V. What was the maximum and minimum output voltage after you added the resistor and turned the knob on the pot?

I'm assuming VO and OV are the same things in your sketch, and that they both really mean ground.

 

[Deleted member 113489]

Maximum and minimum were both around 7.5V - turning the knob didn't make much notable difference. Ideally I'd like the range to be the full 0-9V...

Yeah, 0V and V0 are both ground. I should really have more consistent symbols...

 

[duffy]

Starting to sound like a wiring problem. You say the wiper on the pot puts out 0-9V, right? What's the voltage on that inverting input when you do that? Look at the spec sheet on the LM324 and put pin #'s on that schematic, then double-check that everything's going where it's suppose to go. Pin4 is 9V, right? That's Vcc, not Ve, check to see you don't have that backwards. Look at the way the pins are numbered, they go up one side and down the other, not the same way on both sides.

 

[Deleted member 113489]

I'm pretty sure everything is wired up properly - and yes, Pin4 is 9V.
I'm still confused about the original problem of one IC having outputs of 0V and the other of around 9V - surely if there's a problem, it should present itself identically with identical chips?

 

[duffy]

Not if it's a floating input. Everyone is always sure everything is wired up properly, including me. Next comes the astonishment at discovering you mirrored something or counted backwards or have an isolated ground or something.