# Opamps Negative feedback Voltage Divider doubt or typo?

Discussion in 'Analog & Mixed-Signal Design' started by quique123, Jul 6, 2016.

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1. ### quique123 Thread Starter Member

May 15, 2015
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In this part of Ch8

"If R1 and R2 are both equal and Vin is 6 volts, the op-amp will output whatever voltage is needed to drop 6 volts across R1"

Shouldn't that read "...op-amp will output whatever voltage is needed to drop 6 volts across R2"?

2. ### AlbertHall Well-Known Member

Jun 4, 2014
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The first statement is true. The second is only true if R1 = R2.

3. ### dl324 Distinguished Member

Mar 30, 2015
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The correct way to look at it is: 6V at the non-inverting input causes R1 to have a current of 6mA. Since, in an ideal opamp, no current enters the inverting terminal; the output voltage will be
$\small 6V+I_{R1}*R2$

4. ### quique123 Thread Starter Member

May 15, 2015
34
2
dl324

Thanks. You seem to understand my confusion The reason that sounded weird to me is because it sounded like the Vout was making go through R1 from right to left.

5. ### crutschow Expert

Mar 14, 2008
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That's exactly what's happening.
Vout is the controlled voltage which is forced to go to 12V by the op amp so that the voltage at the (-) op amp input is 6V to match the 6V at the (+) input.
(The diagram left-to-right arrows show electron flow from negative to positive. Conventional current flow would be right to left. Is that what you find confusing?)

6. ### dannyf Well-Known Member

Sep 13, 2015
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That happens to be true here, only because R1 is the same as R2

The statement given in the book is true universally regardless of R1 or R2s values.

7. ### quique123 Thread Starter Member

May 15, 2015
34
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I've visited a few sites in YouTube, electronics-tutorials.ws where they always use conventional current flow.

8. ### crutschow Expert

Mar 14, 2008
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Yes. I always use conventional current flow (which I find easier to understand), as do many references, but whoever did your reference tutorial was familiar with electron flow (I think it's used in all military electronics courses, for example) and thus wrote it with electron flow designations.
I think the useage of electron flow comes from the days when electron tubes (valves) were the only active electronic component, and you need to use electron flow to explain tube operation.

9. ### EM Fields Member

Jun 8, 2016
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If we start at the very beginning, do you understand why an opamp is considered to have infinite gain ?

10. ### #12 Expert

Nov 30, 2010
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Do you understand that "electron flow" and "conventional flow" are models which people use to predict behavior? Do you understand that it doesn't matter which model you use, as long as you use it consistently? Both models work. You can use the one that is most convenient to you. What you can not do is insist that everybody does it your way.

11. ### quique123 Thread Starter Member

May 15, 2015
34
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EMFields
Well no, I don't know why other than they are modelled to have infinite gain in the ideal model.

12. ### Papabravo Expert

Feb 24, 2006
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Infinite gain is required in an ideal opamp so that the output will be exactly 0V, if and only if, the differential input voltage is identically 0V. AFAIK a yoctovolt is the smallest unit of voltage with a named prefix. It is 1 x 10^-24 volts. In a ideal opamp, a differential input of 1 yoctovolt would send the output to the rail. In a typical opamp a differential input of 1 yoctovolt and a voltage gain of 120 dB, the output would be 1 attovolt (1 x 10^-18). Can you now see the difference?

Jun 26, 2012
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