Differential Amplifier Common-mode Output Resistance

In summary: You can also do the problem with a high value of VA, and then use the shortcut of saying that the transistor with the larger emitter current takes over the current of the other. That's what you get with the infinite VA assumption. So, the answer to this problem is that Roc is a somewhat ill defined concept.In summary, the conversation is about determining the common-mode output resistance in a differential amplifier. The lack of clear definition for this parameter is causing confusion and the individual is seeking clarification on its derivation. The conversation also touches on the use of infinite VA assumption and the resulting effect on the calculation of Roc. The solution to the problem is that Roc is a somewhat ill defined concept.
  • #1
hisotaso
27
0

Homework Statement



I am learning about differential amplifiers, and am having trouble determining the common-mode output resistance. In the text, we have the two sections "Differential-mode gain and input and output resistances" and "Common-mode gain and input resistance." Notice the lack of common mode output resistance.

The only place I can find reference to common-mode output resistance is in the section "Two-port model for differential pairs", and it is given by Roc =~ 2*μf*Ree (in this section, no derivations are shown). At the end of this section is an "Exercise", and it asks for Rod and Roc. Rod = 3.2MΩ and Roc = 4.8GΩ. In other words, Roc is much bigger.

Now, in the homework problem I am working on, VA is given as infinite, which means ro =~ VA/Ic is infinite, which means μf = gm*ro is infinite. However, in the solutions (authors solution manual) given my prof., Roc is given as Rc/2 which is much less than Rod (in this problem Rod =~ 2*Rc = 94kΩ, and and the given solution to Roc = Rc/2 = 13.5kΩ).

So my question is: what is Roc? I have googled around a bit and have not been able to find any reference to it. I would be happy with a link to a derivation (preferable actually) as opposed to just an expression, though at this point any help would be appreciated.

Homework Equations





The Attempt at a Solution

 
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  • #2
Ok well I have a line or reasoning that I think works out, please correct me if this is invalid.

Roc=[itex]\frac{voc}{ic1+ic2}[/itex]

ic1=ic2=ic

=[itex]\frac{\frac{vc1+vc2}{2}}{2ic}[/itex]

=[itex]\frac{\frac{ic*Rc1+ic*Rc2}{2}}{2ic}[/itex]

Rc1=Rc2=Rc

=[itex]\frac{ic*Rc+ic*Rc}{2*2ic}[/itex]

=[itex]\frac{2Rc}{4}[/itex]

=[itex]\frac{Rc}{2}[/itex]
 
Last edited:
  • #3
Don't know why itex isn't working, at least on my end...
 
  • #4
hisotaso said:
Ok well I have a line or reasoning that I think works out, please correct me if this is invalid.

Roc=[itex]\frac{voc}{ic1+ic2}[/itex]

ic1=ic2=ic

=[itex]\frac{\frac{vc1+vc2}{2}}{2ic}[/itex]

=[itex]\frac{\frac{ic*Rc1+ic*Rc2}{2}}{2ic}[/itex]

Rc1=Rc2=Rc

=[itex]\frac{ic*Rc+ic*Rc}{2*2ic}[/itex]

=[itex]\frac{2Rc}{4}[/itex]

=[itex]\frac{Rc}{2}[/itex]

Don't use itex; it's just tex. And, don't use the SUB /SUB construct within tex (but you can use it in the non-tex ordinary part of your post); use the underline prefix instead. For example your first tex expression should be like this:

Roc=[tex]\frac{v_{oc}}{i_{c1}+i_{c2}}[/tex]

That didn't work out as intended. It looks like you get an extra carriage return when following the non-tex part with some tex. Maybe you should put everything inside tex:

[tex]R_{oc}=\frac{v_{oc}}{i_{c1}+i_{c2}}[/tex]
 
  • #5
hisotaso said:
Now, in the homework problem I am working on, VA is given as infinite, which means ro =~ VA/Ic is infinite, which means μf = gm*ro is infinite. However, in the solutions (authors solution manual) given my prof., Roc is given as Rc/2 which is much less than Rod (in this problem Rod =~ 2*Rc = 94kΩ, and and the given solution to Roc = Rc/2 = 13.5kΩ).

So my question is: what is Roc? I have googled around a bit and have not been able to find any reference to it. I would be happy with a link to a derivation (preferable actually) as opposed to just an expression, though at this point any help would be appreciated.


Maybe you should ask your instructor what the definition of common mode output impedance is.

If you put an AC short (a big capacitor) between the differential outputs and calculate the impedance from that point to ground, I think you get the result you quote for Roc.
 

1. What is a differential amplifier common-mode output resistance?

A differential amplifier common-mode output resistance is the resistance at the output of a differential amplifier in its common-mode state. It is the measure of the amplifier's ability to reject common-mode signals and amplifying only the differential signal.

2. Why is the differential amplifier common-mode output resistance important?

The common-mode output resistance of a differential amplifier is important because it determines the amplifier's ability to reject unwanted signals and accurately amplify the desired signal. A higher common-mode output resistance indicates a better ability to reject common-mode signals.

3. How is the differential amplifier common-mode output resistance calculated?

The differential amplifier common-mode output resistance can be calculated by dividing the input offset voltage by the input offset current. It can also be measured experimentally by applying a known common-mode voltage and measuring the resulting output voltage.

4. What factors affect the differential amplifier common-mode output resistance?

The differential amplifier common-mode output resistance can be affected by the precision and matching of the amplifier's components, such as the transistors and resistors. Additionally, parasitic capacitance and inductance in the circuit can also affect the common-mode output resistance.

5. How can the common-mode output resistance be improved in a differential amplifier?

The common-mode output resistance in a differential amplifier can be improved by using precision and matched components, minimizing parasitic effects, and incorporating techniques such as common-mode feedback or active load circuits. Additionally, using a differential amplifier with a higher gain can also improve the common-mode rejection ratio.

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