Input resistance of a circuit

The Electrician

Joined Oct 9, 2007
2,970
Sorry, I still don't quite understand.
You said that the gate of M2 must be grounded at high frequency and we can do that by placing a capacitor between gate of M2 and ground.
However, could you explain why we need to ground gate of M2 at high frequency while it is not necessary at low frequency?
It IS necessary at low frequency if you want to have the highest gain possible. Jony said in post #14 "Also I don't see any capacitor between M2 gate and GND?"

You responded:

"yes, you are right. I used capacitor in place of R2 in the topology. However,
last week, I go to a university and asked a friend of mine for help. He said that I can use resistance instead of a capacitor provided that they have the same impedance. Therefore, I draw a resistor here."

For a cascode, you should have ZERO impedance to ground. Since the resistors R1 and R2 are providing the bias for M2, they can't be zero ohms; only a capacitor can provide an impedance approaching zero ohms without shorting out the DC bias voltage.

Even at low frequencies it's a good idea to have a capacitor to ground at that node. Furthermore, neither R1 nor R2 goes to ground in your circuit because Rs is not zero ohms.
 

WBahn

Joined Mar 31, 2012
30,060
Denominator units look like Ω + Ω = Ω to me. Did your eyes fool you into thinking that the parenthetical (RS+RD+R1+R2) was (RS*RD*R1*R2)?
Yep, you're right. I don't know if the parens fooled me or if my vision was just enough blurry this morning that I saw '+' as '*' in there.

But that still leaves a units descrepancy in the numerator as the voltage gain will be dimensionless but the current gain will have units of 1/Ω. Unless the "current gain" is actually a transconductance. I haven't been following the thread closely enough to know.

Thanks for pointing out my oversight.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
It IS necessary at low frequency if you want to have the highest gain possible. Jony said in post #14 "Also I don't see any capacitor between M2 gate and GND?"

You responded:

"yes, you are right. I used capacitor in place of R2 in the topology. However,
last week, I go to a university and asked a friend of mine for help. He said that I can use resistance instead of a capacitor provided that they have the same impedance. Therefore, I draw a resistor here."
Sorry, I meant this circuit in which C is used in place of R2.
(from the book, page 47: http://books.google.com.vn/books?id...BA#v=onepage&q=" SELF-BIASED CASCODE"&f=false)



For a cascode, you should have ZERO impedance to ground. Since the resistors R1 and R2 are providing the bias for M2, they can't be zero ohms; only a capacitor can provide an impedance approaching zero ohms without shorting out the DC bias voltage.
In the circuit above, there is no R2 but a capacitor. I wonder if the capacitor is used for two purposes:

1. bias M2
2. ground gate of M2 at operating frequency

But from the component values from the book:

\(V_{G} = 0.8V, \qquad V_{in} = 0.8sin( \omega t),\qquad R_{b} = 0.75k \Omega, \qquad C_{b} = 2.4pf, \qquad \omega = 5.25GHz\)

And therefore:

\( X_{c} = \frac{1}{\omega C_{b} } = \frac{1}{5.25GHz* 2.4pf } = 79.36 \Omega \)

It is not small at operating frequency. I wonder how is it possible to provide ground here?

Even at low frequencies it's a good idea to have a capacitor to ground at that node. Furthermore, neither R1 nor R2 goes to ground in your circuit because Rs is not zero ohms.
Yes, my mistake. Rs should be removed.
 

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The Electrician

Joined Oct 9, 2007
2,970
Sorry, I meant this circuit in which C is used in place of R2.
(from the book, page 47: http://books.google.com.vn/books?id...BA#v=onepage&q=" SELF-BIASED CASCODE"&f=false)





In the circuit above, there is no R2 but a capacitor. I wonder if the capacitor is used for two purposes:

1. bias M2
2. ground gate of M2 at operating frequency

But from the component values from the book:

\(V_{G} = 0.8V, \qquad V_{in} = 0.8sin( \omega t),\qquad R_{b} = 0.75k \Omega, \qquad C_{b} = 2.4pf, \qquad \omega = 5.25GHz\)

And therefore:

\( X_{c} = \frac{1}{\omega C_{b} } = \frac{1}{5.25GHz* 2.4pf } = 79.36 \Omega \)

It is not small at operating frequency. I wonder how is it possible to provide ground here?


Yes, my mistake. Rs should be removed.
The capacitor doesn't provide bias.

The impedance of 79 ohms is a lot less than the thousands of ohms that will probably characterize Rb.

The circuit designer may in fact want Cb to be its given value for various reasons other than just providing a ground.

In high frequency RF design, one often wants impedances to be near the characteristic 50 ohms of a transmission line. The value of Cb may work together with the rest of the circuit components and parasitics to tune the network.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
I have read that Rb and Cb is used to ensure the voltage at G2 changes in according to D2 and therefore VG2D2 will not exceed breakdown threshold.
As Jony said, the small signal model is worthless at high frequency. Can you suggest a way to analyse this? I think we can use small signal model but Cgs, Cgd and Cds should be included. I intend to draw a full model for MOSFET at high frequency and then analyse it. However, it seems very complicated as there are many connections created by parasitic capacitances.
 

The Electrician

Joined Oct 9, 2007
2,970
I have read that Rb and Cb is used to ensure the voltage at G2 changes in according to D2 and therefore VG2D2 will not exceed breakdown threshold.
As Jony said, the small signal model is worthless at high frequency. Can you suggest a way to analyse this? I think we can use small signal model but Cgs, Cgd and Cds should be included. I intend to draw a full model for MOSFET at high frequency and then analyse it. However, it seems very complicated as there are many connections created by parasitic capacitances.
Nodal analysis will work. Jony gave you a link to my exposition of Shekel's method. If you want a symbolic analysis, the algebra will be so cumbersome that I won't be feasible to do without some mathematical software. If you already have Mathematica, which I think you do, that will do nicely.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
Yes, Mathematica is really great! I have just known it but with its help I think there is no problem!

I'd like to ask you if this model is good at 1.7GHz.



And the value of parasitic capacitor depends on bias voltage. Is there ready values for them according to bias voltage or I have to simulate to get exact these values?
 

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The Electrician

Joined Oct 9, 2007
2,970
Yes, Mathematica is really great! I have just known it but with its help I think there is no problem!

I'd like to ask you if this model is good at 1.7GHz.



And the value of parasitic capacitor depends on bias voltage. Is there ready values for them according to bias voltage or I have to simulate to get exact these values?
I can't comment on the goodness of the model. Where did you get it? What is it a model for? It looks like a model of a dual gate MOSFET, or some similar device. Usually, at high RF frequencies models like this aren't used; S parameters are used instead. You should get yourself a good text on microwave RF design.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
I can't comment on the goodness of the model. Where did you get it? What is it a model for? It looks like a model of a dual gate MOSFET, or some similar device. Usually, at high RF frequencies models like this aren't used; S parameters are used instead. You should get yourself a good text on microwave RF design.
I find it in the page:
http://reference.wolfram.com/applications/insydes/Tutorial/TransistorModels.html
It is a model for MOSFET at high frequency. I have just tried to find analyse the circuit using that model and the results are extremely complex! Even when all Cgd, Cbs, Csb are removed, the expressions also complicated..
 

LvW

Joined Jun 13, 2013
1,759
But that still leaves a units descrepancy in the numerator as the voltage gain will be dimensionless but the current gain will have units of 1/Ω. Unless the "current gain" is actually a transconductance. I haven't been following the thread closely enough to know.
It was simply a typo - and was clarified already in my post#66.
 

LvW

Joined Jun 13, 2013
1,759
Oh. Okay. Like I said, I haven't been following this thread too closely.
Yes, no problem - and I have made a similar error since The Electrician did point to my typing error already in his post#55 (and I didn`t notice that).
 
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