# Single Ended Amplifier Question

Thread Starter

#### jegues

Joined Sep 13, 2010
733
See figure attached for problem statement as well as my attempt.

This is an old midterm for an course I'm going to be taking next term and I thought I'd give it a crack to see if I can get ahead.

I think I was able to accomplish part a of the question but part b I have some confusions about.

What is a "single-ended amplifier"? Also, how does one have a negative gain?

Hopefully after I figure those two things out I can work out a solution.

Thanks again!

#### guitarguy12387

Joined Apr 10, 2008
359
Single ended amplifier is the opposite of differential amplifier (to use laymans terms haha!) Basically, you have one signal referenced to ground rather than taking the difference of two signals, which would be a differential amplifier.

Negative gain = 180 degrees phase shift

#### guitarguy12387

Joined Apr 10, 2008
359
Also, your work is a little hard to follow (i.e. what is R2, R1). But i think you probably want to choose values for the resistors, not node voltages.

I'm guessing part b is asking you to find the circuit topology that implements the gains. (I.e. single ended inverting/non inverting amps)

Thread Starter

#### jegues

Joined Sep 13, 2010
733
Single ended amplifier is the opposite of differential amplifier (to use laymans terms haha!) Basically, you have one signal referenced to ground rather than taking the difference of two signals, which would be a differential amplifier.

Negative gain = 180 degrees phase shift
So what does this mean in terms of a transfer function?

In a differential amplifer we see,

$$\frac{R_{2}}{R_{1}}(v_{2}-v_{1})$$

You're saying it's the opposite, so is it

$$\frac{R_{2}}{R_{1}}(v_{2}+v_{1})$$

?

If one signal is referenced to ground (let's say V1) then it should be,

$$\frac{R_{2}}{R_{1}}(v_{2}-v_{1}) = \frac{R_{2}}{R_{1}}(v_{2}-0) = \frac{R_{2}}{R_{1}}v_{2}$$

Are any of these correct?

#### guitarguy12387

Joined Apr 10, 2008
359
Try grounding one of the inputs

Thread Starter

#### jegues

Joined Sep 13, 2010
733
Try grounding one of the inputs
Didn't I do that in the post above? V1 was set to 0v.

#### guitarguy12387

Joined Apr 10, 2008
359
In a differential amplifer we see,

You're saying it's the opposite, so is it

?
No i meant that differential = difference of two where single ended is not the difference. It has a single input wrt gnd, not difference btw two signals.

Didn't I do that in the post above? V1 was set to 0v.
Yeah, sorry i didn't check that before. Looks right to me though. Depending which input you ground gives inverting or non-inverting.

I don't konw what you are designating as R2 and R1 though. All Rs are the same in the schem you gave.

#### guitarguy12387

Joined Apr 10, 2008
359
HINT: changing the ratio of the series to shunt resistors is what gives you different gains. Since they are the same in the given plot, that is why it is unity gain.

#### guitarguy12387

Joined Apr 10, 2008
359
That is a pretty confusing way to put it. That only makes sense when dealing with sinusoids. DC doesn't have phase.

You want to think in terms of amplitude and phase. To say that you measure -5v from 1v input seems to suggest that the amplitude is somehow negative. In fact, this is not the case.

For clarity, if you have an amp with gain of -1, the amplitude of the input AND output will be 1 but the output will be 180 degrees out of phase from the input. Use the frequency domain to get the whole story.

#### shteii01

Joined Feb 19, 2010
4,644
That is a pretty confusing way to put it. That only makes sense when dealing with sinusoids. DC doesn't have phase.

You want to think in terms of amplitude and phase. To say that you measure -5v from 1v input seems to suggest that the amplitude is somehow negative. In fact, this is not the case.

For clarity, if you have an amp with gain of -1, the amplitude of the input AND output will be 1 but the output will be 180 degrees out of phase from the input. Use the frequency domain to get the whole story.
Yes, I see what you mean.