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aaron_do

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source degen
Feb 12^th, 2007, 7:47pm

If we use source degeneration for linearization, the degen resistor can vary by up to 50% if it is on chip. So how do we deal with the huge change in circuit performance?

If i'm using a differential structure where the tail current is biased by a bandgap, I don't need to worry much about DC current variation, but obviouly there will be huge changes in gm, IIP3, and NF with resistor size.

thanks,
Aaron

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Prabhu

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Re: source degen
Reply #1 - Feb 13^th, 2007, 12:54am

Hi Aaron,

Degeneration using resistors is usually done in BJT differential pairs.It is the only way to linearize since a BJT's IIP3 is fixed independent of its bias current.
But in the case of MOS differential pair there is no need to use degeneration resistor.Instead the effect of degeneration can be obtained by increasing the overdrive of transistor which depends on the device size and current.
This should perform better than resistive degeneration in terms of process variations.

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aaron_do Senior Fellow Offline Posts: 1398	Re: source degen Reply #2 - Feb 13^th, 2007, 6:06pm Hi Prabhu, thanks for the reply. Do you have any references or equations for your answer? I tried simulating what you said where I removed the degen res and adjusted the transistor size to get the same overall gm. Unfortunately the resulting IIP3 was about 6 dB worse than when using source degen. Aaron
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Prabhu

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Re: source degen
Reply #3 - Feb 14^th, 2007, 1:37am

Hi Aaron,

Consider 2 differential pairs both with the same tail current I and length.

pair 1: Width = W, Degeneration resistor = R, effective transconductance = gm / ( 1+ gmR )
pair 2: Width = W / x , no degeneration, transconductance = gm / √x

Equate the transconductances in both the cases and find the differential voltage at which the tail current is completely switched to one arm in both the cases.
You will find both of them to be equal.
This justifies that both the pairs will exhibit similar distortion characteristics.

I also simulated the two cases and found their IIP3 to be close within 1dB.

May be you were using very small overdrives for the devices in your simulation.

Raja Prabhu

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aaron_do

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Re: source degen
Reply #4 - Feb 14^th, 2007, 8:08am

Hi Prabhu,

I simulated using the active filter I was designing. I simply removed the source degen and modified W/L to get the same gm and VDS. I will resimulate using a simple circuit. Are you sure the resistor you are using is large enough? It should be large enough to cause substantial DC voltage drop.

If you are talking about 1dB compression point then I agree that both methods would yield similar results when a tail current is used. However, I believe source degen is used specifically for IIP3 improvement. I remember reading something about it and I think there's a related post here somewhere...

Aaron

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Ed1

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Re: source degen
Reply #5 - Feb 16^th, 2007, 5:58am

Prabhu wrote on Feb 13^th, 2007, 12:54am:

Hi Aaron,

Degeneration using resistors is usually done in BJT differential pairs.It is the only way to linearize �since a BJT's IIP3 �is fixed independent of its bias current.
But in the case of MOS differential pair there is no need to use degeneration resistor.Instead the effect of degeneration can be obtained by �increasing the overdrive of transistor which depends on the device size and current.
This should perform better than resistive degeneration in terms of process variations.

Can you please point me to some "Read" about why IIP3 of BJT is fixed ?

thanks,

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Prabhu Junior Member Offline Posts: 23	Re: source degen Reply #6 - Feb 18^th, 2007, 4:23am Hi Ed1 You can just write down the BJT current equation IC as a function of VBE and derive its IP3 . It will be a function of only the thermal voltage. You can also refer to, "Distortion in Elementary Transistor Circuits" paper by Sansen.
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aaron_do

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Re: source degen
Reply #7 - Feb 20^th, 2007, 6:47pm

Hi Prabhu,

thanks for posting that paper. BTW I believe there is a mistake on page 318 (the 4th page), section II. In the paragraph before equation (11), it should say a reduction of 1 dB is a reduction to 0.891 (not 0.122). It is simply 10^-0.05. Hence,

a1.Vin - 3/4a3.Vin^3 = 0.891.a1.Vin
Vin = sqrt(0.109).IP3
P1dB = IP3 - 9.63 dB

The answer is very close but the analysis is different.

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Ed1 New Member Offline Posts: 4	Re: source degen Reply #8 - Feb 20^th, 2007, 6:59pm aaron_do wrote on Feb 20^th, 2007, 6:47pm: thanks for posting that paper. Can you please give me a link to the paper? thanks,
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aaron_do

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Re: source degen
Reply #9 - Feb 20^th, 2007, 7:39pm

Hi Prabhu,

I resimulated using a single transistor with and without source degeneration. For the 75 ohm source resistor with DC current of 2.3 mA, I found there was an improvement of around 5 dB in the third order IM. I ran a swept PSS and PAC analysis at 1 MHz and chose a point where the gradient of the 3rd order IM was linear and equal to 3. I believe this is expected, and I remember reading a paper which explains this a while ago but i don't remember the title.

btw I used ideal components except for the transistor. I probed the output current with an ideal voltage source as the load to ensure VDS is constant and there is no compression of VDS.

Aaron

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