741 Difference circuit with single supply

[fuper]

Hey -

I know how to setup an opamp like the 741 to operate as a differential amplifier. My question is would it be possible to do so without a split supply?

I know in general one simply uses a voltage divider to bias one of the inputs and then you can hook the opamp to V+ and ground. However looking at a vanilla differential circuit for an opamp the resistors already form a voltage divider, so this has me a bit confused.

I'm trying to measure [not worried about accuracy to much] the voltage drop on a shunt. I didn't want to use single supply opamp because as i have a box of 741 laying around and it always seems wasteful to use a dual or quad when you only require a single opamp.

 

[MikeMl]

As long as the voltage at the ends of the shunt are about 3V more positive than the VSS pin; and about 5V more negative than the VCC pin, and the voltage between the VSS pin and the VCC pin is 10V or more, then a crappy old 741 might work.

Do your self a favor and buy some modern CMOS rail-to-rail INPUT, rail-to-rail OUTPUT opamps and throw the 741s as far as you can throw them. I have some CK722 transistors in my junk box, but I will never use them in a project....

 

[fuper]

As long as the voltage at the ends of the shunt are about 3V more positive than the VSS pin; and about 5V more negative than the VCC pin, and the voltage between the VSS pin and the VCC pin is 10V or more, then a crappy old 741 might work.

Do your self a favor and buy some modern CMOS rail-to-rail INPUT, rail-to-rail OUTPUT opamps and throw the 741s as far as you can throw them. I have some CK722 transistors in my junk box, but I will never use them in a project....

Ah Ok. You might of saved me a lot of time. I could see myself trying to debug a simple circuit for hours not knowing the subtleties.

Do you have any recommendations for general purpose opamps? I bought a few of tl082. Could i use those for my application? I.E just wire it up as a differential across the shunt making sure the divider going into the reference are a few volts away from ground as you suggested?

 

[Hero999]

Look at spec's such as common mode range (the voltage range the inputs can work within) and the output saturation voltage.

My guess is that you at least need a single supply op-amp such as the LM358. The TL082 is fine for an audio amplifier (providing it's correctoy designed) but not for your application because it is not a single supply op-amp.

 

[MikeMl]

What are you calling "ground"?

What is the voltage at the ends of the shunt with respect to "ground"?

What is the differential voltage across the shunt with max current flowing?

What voltage do you have to run your opamp on measured with respect to "ground"?

 

[fuper]

Hi MikeMI

I'm calling ground 0 volts. I'm using a 5 volt supply on a breadboard to experiment so 5 volts used to power the op amp and everything else.

I am going to use a 1 ohm resistor for experimenting as that is the lowest I have. I was going to use a simple LED to represent the load. From memory a red led drops a couple of volts approximately. I usually use a 1k resistor to limit current into the led. Lets see if I can even calculate this out:

If we call the voltage drop of the led at 2 volts, and the combined resistance is 1000 + 1 or 1001 then the current into the led will be about (5-2)/1001 ~ .0029 amps or 3ma.

Now to tell what voltage drop across the shunt is, I'm a little unsure here:

.0029 amps * 1000 = 2.99 volt drop across 1k resistor
.0029 * 1 = .002 volt drop across the little shunt resistor.

What are you calling "ground"?

What is the voltage at the ends of the shunt with respect to "ground"?

What is the differential voltage across the shunt with max current flowing?

What voltage do you have to run your opamp on measured with respect to "ground"?

 

[audioguru]

The input common-mode voltage range of the TL08x and TL07x opamps includes the positive supply voltage so they can measure the current in a current-sensing resistor in series with the positive supply.

The lousy old LM358 dual and LM324 quad opamps have an input common-mode voltage range that includes ground, but a low power MC3317x or a normal power MC3407x is much better and come in single, dual and quad versions.

 

[Hero999]

The output of the TL08x doesn't include the positive rail though.

The output of the LM358 and MC3407x includes the negative rail but not the positive.

You need an op-amp with an input and output voltage of 0-3V?

 

[fuper]

Useful information guys, thanks! I see that the 358 has a more convenient common mode range than the 741. I did not know to look for that before. So to perform my experiment/circuit setup a 358 to measure the difference in volts across the shunt and to amplify that to whatever value is useful.

Hopefully I can get this to work.

 

[Roff]

Useful information guys, thanks! I see that the 358 has a more convenient common mode range than the 741. I did not know to look for that before. So to perform my experiment/circuit setup a 358 to measure the difference in volts across the shunt and to amplify that to whatever value is useful.

Hopefully I can get this to work.

That will work if you have a low side shunt. if you have a high side shunt, and you try to power the op amp from the same supply, you will exceed the input CMR.

 

[fuper]

Thanks Roff.

I know given the parameters of my circuit the power will be low but just for completeness sake. To calculate the power disipated by the shunt [1 ohm] I would go for V^2 / R.

I stumbled a bit figuring out what V would be. Would it be 5Volts or the tiny voltage drop of 1 ohm resistor at max current dissipation. I calculated that using 5 volts for V, you would expect the watts of power used in the shunt to be 25/1 or 25 watts which I realized was silly, so I calculated it with my tiny voltage drop i calculated earlier, .002 volts.

(.002)^2/1 = .000004 <- so I could use a 1/8 watt resistor if I wanted to.

In summary of using a shunt, it strikes me that you need to know a few operating parameters like Mike asked about earlier:

Max operating current of the Load you are measuring.
Voltage of the circuit.

From there you can calculate:

Voltage drop of the shunt.
Power consumed by the shunt etc.

Sound about right?

 

[Roff]

I think you missed my point. Shunts always need to be very low in vaalue compared to the load resistance. Put another way, the maximum voltage drop needs to be negligible compared to the total supply voltage. A typical shunt would drop maybe 100mV max.
What I was referring to is the location of the shunt in the ccircuit. A low side shunt is between ground and the load. A high side shunt is between +V and the load.