Can a single supply op amp be used for a summer?

The question is unclear; a summing node with resistors can do this without any amplifiers. To add a 'single supply' op amp would presumably either be required because the output range (to +5V) is at the power supply positive power rail, OR because the input range (to GND) is at the negative power rail? Or, both?

And, do you need the op amp to keep the input signal unloaded, or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this to a 1 to 5V source. The linear mapping of this would be: Vout =3D 0.8*Vin +1. Given typical dual supply op amps, this can be done with a summer and an inverter. However, because a summing op amp has an inverting gain, the single supply op amp wouldn't be able to do this. So, I was wondering if there was a way of making that conversion, as stated earlier, using single supply op amps.

thanks

You can 'sorta' do this if you don't mind a bit of error. This brute force approach would be to hang a small resistor to ground from the non-inverting input. (Let's use 1 k ohm.) Then 'sum' through two bigger resistors. From a one volt supply put 'say' 10k to the noninverting input and from the varibale voltage put 8k ohm to the non- inverting input. Then have a gain of ten in the opamp. The result is 'sorta' what you want. You can make it more 'accurate' by throwing away more signal and adding more gain... but it looks like a losing proposition to me.

If you allow for more opamps you might be able to make the resistor current sources look more ideal. Does a Howland current source (Gack!) work from a single supply?

George H.

because it's a homework question.

You can do this with a pair of resistors and no op amps, as long as you don't care about loading the supply and having an output with a non-zero source impedance.

So, a 1k resistor from input to output, and a 4k pullup resistor from output to +5V, does it. Add a voltage-follower at the input end, or a voltage-follower at the output end, or both, if they're needed. It isn't clear whether they ARE needed.

Excellent, Thanks

George H.

I'm trying to draw this out. Please clarify. I assume we are talking about an op amp.

"1k resistor from input to output" --- is this the inverting or non- inverting input, the feedback loop?

" and a 4k pullup resistor from output to +5V, does it"

So the circuit input goes to the non-inverting input?

thanks

--- What he meant was: (View in Courier:)

. +5V . | . [4k]R1 . | . +---Vout . | . [1k]R2 . | . Vin

R1 doesn't have to be 4K and R2 doesn't have to be 1k, but their ratio has to be 4:1.

If you needed a lower output impedance you could also:

. +5V . | . [4R1] . | . +-------|+\\ . | | >--+--->Vout . [R1] +--|-/ | . | | | . Vin +--------+

or, if you needed a higher input impedance:

. +5V . | . [4R1] . | . +-->Vout . | . [R1] .Vin---------|+\\ | . | >--+ . +--|-/ | . | | . +--------+

or, if you needed both:

. +5V . | . [4R1] . | . +-------|+\\ . | | >--+--->Vout . [R1] +--|-/ | .Vin---------|+\\ | | | . | >--+ +--------+ . +--|-/ | . | | . +--------+

JF

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I get it now. Nice and simple. This will work well.

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=A0 =A0 =A0

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Given that resistor combination, Vout =3D (5-Vin)/5

Vin Vx

This doesn't work.

--- E1 .                 +5V Non-inverting input .                  | input .                [4R1] / . Non-inverting    | E2 / . input       +-------|+\\   . \\        |       |  >--+--->Vout .         \\      [R1]  +--|-/   | .Vin---------|+\\   |    |        | .            |  >--+    +--------+ .         +--|-/   | .         |        | .         +--------+

Version 4 SHEET 1 880 680 WIRE 144 16 -208 16 WIRE 288 16 144 16 WIRE 144 64 144 16 WIRE 368 96 224 96 WIRE 288 128 288 16 WIRE 224 144 224 96 WIRE 256 144 224 144 WIRE 368 160 368 96 WIRE 368 160 320 160 WIRE 144 176 144 144 WIRE 256 176 144 176 WIRE 144 208 144 176 WIRE -208 288 -208 16 WIRE 64 288 -208 288 WIRE 144 320 144 288 WIRE 144 320 0 320 WIRE 64 352 64 288 WIRE 0 368 0 320 WIRE 32 368 0 368 WIRE 144 384 144 320 WIRE 144 384 96 384 WIRE 32 400 -80 400 WIRE -208 432 -208 288 WIRE -80 432 -80 400 WIRE -208 544 -208 512 WIRE -80 544 -80 512 WIRE -80 544 -208 544 WIRE 64 544 64 416 WIRE 64 544 -80 544 WIRE 288 544 288 192 WIRE 288 544 64 544 WIRE -208 608 -208 544 FLAG -208 608 0 SYMBOL Opamps\\\\LT1218 288 96 R0 SYMATTR InstName U1 SYMBOL res 128 48 R0 SYMATTR InstName R1 SYMATTR Value 4k SYMBOL res 128 192 R0 SYMATTR InstName R2 SYMATTR Value 1k SYMBOL Opamps\\\\LT1218 64 320 R0 SYMATTR InstName U2 SYMBOL voltage -80 416 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 5 0 1) SYMBOL voltage -208 416 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 5 TEXT -192 576 Left 0 !.tran 2 uic

I appreciate your taking the time to write this up, but I'm going to have to simply believe you that this circuit is workable with a single ended op amp. I cannot make heads or tails out of what you sent. But thank you for your effort.

You need LTSPICE .. It's free.. you import that and it'll show you and simulate for you.

But

I've got PSPICE. But I can't imagine such a simple circuit would require simulation. It's probably two or three lines of algebra. But I'll get LTSPICE if it will get me to an answer.

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I've got PSPICE. But I can't imagine such a simple circuit would require simulation. It's probably two or three lines of algebra. But I'll get LTSPICE if it will get me to an answer.

snip

You might want to check your equation.

JF, I found a good document from TI explaining single ended op amp design, so I will continue from there. I really do appreciate you effort, but unfortunately there is some data being lost in translation here. This is apparent by your equation defining a voltage divider. A divider consisting of two resistors will create a center voltage of: Vout =3D (E1-Vin)*R2/(R1+R2). Your equation shown above as Vout =3D (E1- Vin)*R2/(R1+R2) + Vin is the description of another circuit. This is the circuit which I cannot see, but it clearly consists of more than two resistors in series. Anyway, if you are interested, google "single supply op amp design Mancini" and you will find the paper from TI. It shows diagrams of several typical op amp circuits along with the equations for Vout =3D f(Vin).