Constant Voltage Using LM308

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
I found an LM308 op-amp chip because i realize the educational set-top op-amp box has resistors built in for gain adjustment.

I am trying to connect up the attached image on a breadboard.

Rf=200
Rw=10
R1=1000
R2=100
V1=0-5V

I hooked it up with the neg connection of the adjustable 12 V DC power supply to one of the rails, and the positive 12V DC power supply to the other rail. I am assuming this will give me the desired +/- max 15 V power required for the op-amp.

Unfortunately, the + input of the op-amp is hooked up to ground. But I am reading a volt or so across the + to - op-amp inputs, so something is not right.

Any suggestions would be a help.
 

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Ron H

Joined Apr 14, 2005
7,014
The problem is, if the op amp were ideal, its output would be at +11.5V, and it would be supplying 105ma to R2. An LM308 with ±12V supplies cannot do either of these. What are you really trying to do?
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
I am a Mechanical Engineering grad student who will be using a constant voltage hot-wire device to do flow measurement research. In this device, the resistor, Rw, from that diagram, is actually a 10 ohm tungsten probe that varies in resistance with respect to flow speed.

My professor suggested that i build a simple circuit following the diagram to see for myself that the voltage across Rw will remain constant as Rw resistance is varied. For this simulation, Rw is just a 0-20 ohm potentiometer. I tried to pick good resistance values for the resistors that would not overload the LM308 op-amp. I also tried an LM358 but had no luck so i went to the LM308, since it is powered with both +/- 15V instead of +15 and ground.

All i intended to do was hook a couple resistors to a cheap op-amp for a very simple circuit and this has turned out much harder for me to do than I ever intended. I also have access to an LM317 voltage regulator but i do not think that is the right product as all i want is a suitable op-amp.
 

Ron H

Joined Apr 14, 2005
7,014
Whether or not you can use a common op amp depends on how much voltage you want to maintain across Rw. You can't get 1V (as your circuit is attempting to do) with a common op amp.
If you want design help, tell us the maximum and minimum values of Rw, and how much votage you want across it.
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
Thank you Ron,

Ideally, the resistance for Rw would be between 5-10 ohms. As this resistance is varied, the voltage across it would hopefully remain nearly constant. About 0.5 volts to measure across Rw would be nice, but i am not too picky. I have tried sorting through the LM308 and LM358 data sheets but they are a bit overwhelming.

2 equations i have been using to size resistances are:

V_s = V_1((R_f*R_2)/(R_1*R_w)+R_2/R_1+R_f/R_1)

and

V_w = (R_f/R_1)*V_1 (this is substituted into KCL at node W for the above equation)
 

Ron H

Joined Apr 14, 2005
7,014
Don't you have access to a lab power supply that is adjustable down to 1V? Could you borrow one from the EE department?
Or are you getting sucked in to doing some circuit design?
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
I have 2 DC power supplies that are pretty modern...will go down to 0.1 V i believe. I have access to a wide range of cheap resistors. My selection of precision type is much more limited.
 

Ron H

Joined Apr 14, 2005
7,014
I have 2 DC power supplies that are pretty modern...will go down to 0.1 V i believe. I have access to a wide range of cheap resistors. My selection of precision type is much more limited.
My point is, why can't you use a power supply to heat the probe?
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
Well then i guess you could say sucked into circuit design. The point is to use an op-amp to maintain a constant voltage across the pretend probe. This is just a "simple" little experiment for me to keep my hands busy until some equipment arrives.
 

Ron H

Joined Apr 14, 2005
7,014
Well then i guess you could say sucked into circuit design. The point is to use an op-amp to maintain a constant voltage across the pretend probe. This is just a "simple" little experiment for me to keep my hands busy until some equipment arrives.
A simple op amp can't provide the 200mA or even 100mA, that your probe requires. I can show you how to add a transistor inside the feedback loop to provide the current, but it won't have short circuit protection without some other parts, and the transistor will probably require a heat sink.
You could use a voltage regulator IC. Is that allowed?
How far do you want to pursue this?
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
Haha. Don't want to pursue too far.

I think the voltage regulator might be too simple, if its purpose is to regulate voltage.

If i am fine with 100 mV (even 50 mv) across Rw with a resistance of 10 ohms, that would only be 10 mA of current through Rw, plus a bit more to flow back into the inverting input. Could that work without the transistor?
 

Ron H

Joined Apr 14, 2005
7,014
Haha. Don't want to pursue too far.

I think the voltage regulator might be too simple, if its purpose is to regulate voltage.

If i am fine with 100 mV (even 50 mv) across Rw with a resistance of 10 ohms, that would only be 10 mA of current through Rw, plus a bit more to flow back into the inverting input. Could that work without the transistor?
Sure, 10mA is fine.
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
I tried a simple check without using R_w which is a pot as follows:

I tried hooking up a 500 ohm resistor before the negative op-amp input and after the 5 V source, then a 1000 ohm resistor in the negative feedback. This means only R_f in feedback and R_1 before the op-amp referencing this diagram. That should give me 10 Volts at the output, since V_s = (R_f/R_1)*V_1. I assume the output current would be V_s/R_f = 10/1000 = 10 mA. However, something is not right as there is slightly under a volt measured across the op-amp inputs.

Thanks for your patience.
 

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Ron H

Joined Apr 14, 2005
7,014
I tried a simple check without using R_w which is a pot as follows:

I tried hooking up a 500 ohm resistor before the negative op-amp input and after the 5 V source, then a 1000 ohm resistor in the negative feedback. This means only R_f in feedback and R_1 before the op-amp referencing this diagram. That should give me 10 Volts at the output, since V_s = (R_f/R_1)*V_1. I assume the output current would be V_s/R_f = 10/1000 = 10 mA. However, something is not right as there is slightly under a volt measured across the op-amp inputs.

Thanks for your patience.
What is the value of R2?
It would be much easier to read your schemaric if you would put the parts values on them.
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
I have attached a picture with labeled resistances and calculated output voltage, as well as current for the slightly more complicated circuit using the pretend hot-wire probe for R_w which is a potentiometer.

The more simple circuit i tried just for troubleshooting uses 1000 ohms for a single resistor in feedback and 500 ohms at the op-amp input.
 

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Ron H

Joined Apr 14, 2005
7,014
The analysis is correct. Will it work for you to power your probe, or is it just an exercise?
You may have concluded that R2 is theoretically redundant, as it has no effect on Vw. It does affect frequency response if Rw has shunt capacitance, and it may provide better short circuit protection for the op amp if it is large enough. i believe most modern op amps have built-in short circuit protection.
 

Thread Starter

lfgrdwill

Joined Dec 9, 2011
27
What do you exactly mean by power the probe? This is just an exercise but i would like the exercise to work. Both the LM308 and LM317 op-amps aren't working as i intended them to (or at all) with this circuit.
 
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