comparing CS and Cascode frequency response

• posted

hi all, I calculated the frequency response of the CS and Cascode amplifiers using the time constants method. Here they are:

CS

f = ( 2*PI*( Rs*Cgs + Rs*Cgs*(1+Av) + Rd*(Cdb+Cl)) )-1

Cascode

f = ( 2*PI*(Rs*Cgs1 + Cds1*2*Rs + Rd*(Cgd2+Cdb2*Cl) + 1/gm2*(Cdb1+Cgs2) )-1

where:

Rs = resistance of the imput signal Rd = resistance under Vdd Cl = load capacity

Vdd | Rd |----------o Vout --- | Vcost--|| = --> | | ground | --- ----|| | -->

Rs | | | Vin ground | ground

I calculated f using

Rs = 50ohm Rd = 600ohm W = 592 micron

and I obtained

CS => f = 1.98EE+08 Cascode => 2.08EE+08

and I think that's a really little difference! where did I make the mistake?

thank you very much!

• posted

You should show your work as you did in your earlier posts.

• posted

Maybe it's better ;) here is my work about the cascode configuration

The cascode scheme and the small signal scheme

Here I calculate the time constants

Polarisation and results

Thank you very much!

• posted

When you say "frequency response", I would have thought you meant the transfer function, but apparently you mean the corner frequency, or the 3 dB down frequency. Could you clarify this point?

• posted

Sorry, f is the 3 dB cut-off frequency

• posted

Matteo, I will try to derive the transfer function of the cascode, and compute the

3 dB frequency from that. The transfer function of the CS was derived by both you and me in another posting. I will use that to compute the 3 dB frequency of the CS configuration.

If you want actual numbers, you will have to give me values of the following:

gm1 Cgs1 Cdg1 Cds1 gm2 Cgs2 Cdg2 Cds2

and anything else I might need.

• posted

but do you think that the time constants method brings to a wrong result? shouldn't it be correct?

the current in M1 and M2 is the same

using W = 592 micron Cgs1 = Cgs2 = 635 fF Cgd1 = Cgd2 = 124fF Cdb1 = Cdb2 = 770fF

Id=3mA and Vov = 0.15 V gm1 = gm2 = 0.04 S

thanks a lot! your help is very appreciated

• posted

I don't know this method, and rather than learn it, I will compute the 3 dB down frequency by another method, which will provide a check.

• posted

Thank you phantom, I'll wait for your results to check my work!

• posted

Matteo,

1. In this pdf:

you have Cds2 connected at the output. Shouldn't it be connected in parallel with Ro1; that is, across the gm2*Vgs2 source?

1. Also, I need to know what values you used for Ro1 and Ro2.

1. In the earlier schematic of the CS amplifier, you have an Rd value of

4.7 ohms. The cascode has an Rd of 600 ohms. Are you using 600 ohms in both amplifiers for this most recent calculation?

1. Are you expecting a large difference in the 3dB frequencies? I derived the transfer function for the cascode, and when I plot it versus frequency up to 1 GHz, I find the roll-off is dominated by the 600 ohm Rd and the 250 pF output capacitance. The amplifier parasitics don't matter much in either amplifier. I don't have Ro1 and Ro2 in the derivation, so when you give me those values, I'll put them in and see if they make much difference.

I'm getting values of around 200 MHz for the frequency at which the voltage output is down by half for both amplifiers. Perhaps the missing Ro1 and Ro2 may make a difference, but I don't think it will change by very much.

• posted

The Phantom ha scritto:

Cds2=Cdb2 goes from the drain terminal to the grounding, therefore I think it's correct. I also checked it on a book..

Ro1 = Ro2 = 2500 ohm because Id=2.5mA and LAMBDA=0.16

yes, I'm using 600 ohms in both amplifiers for this calculation

I already put Ro1 in my derivation but it doesn't appear in the 3db frequency; I think that Ro2 shouldn't make the difference because it's in parallel with Rd that is very little!

I found the same results! but I read everywhere that the cascode configuration is really better thant the CS's one! where can I find this "better"? I really don't understand..

thanks phantom

• posted

Isn't Cds2 the capacitance from the drain to the source? That's why it is designated Cds, the ds meaning drain to source, not drain to ground. It would go from the drain to ground only if the source were grounded. There is of course a capacitance due to the package which is from drain to ground, but Cds is internal to the chip, and goes from drain to source.

Since in the cascode, the source of the second FET isn't grounded, I think you need to connect it across the second current source which represents the effect of the second FET.

It doesn't have much effect connected the way you have it because it is completely dominated by the 250 pF output capacitance. But, if it is connected across Ro2, then it will have an effect, expecially if the 250 pF is removed. I'll run some more simulations and report on the results.

I will add Ro1 and Ro2 to the circuit and run the simulation again. Then I will remove the 250 pF output load and compare the performance of the two circuits. Perhaps then, without the 250 pF load, the cascode is better than the CS.

Are you in Italy? I guess it's Sunday there. I am a night person, and I'll be up for a few more hours. I'll probably be able to do this in the next hour or so.

• posted

Ok sorry, I didn't tell you that the *only* capacitance I'm going to put in my work are Cgd, Cgs and Cdb and NOT Cds.

take a look here at page 15:

( When I wrote Cdb2=Cds2 I meant (sorry) that I'll not consider Cds2 but only Cdb2. However I think (if I had used it) that Cdb2 should have gone in parallel with Cds2 because the source of the second MOS is connected to the ground and the bulk too )

you

I think that the source of the second mosfet (the upper one) is grounded because the source is connected to a costant voltage generator => in the small signal model is grounded. I checked it in some diapos I found in the web.

This'll be very interesting ..I would have done it if I had had a simulator!

the

yes I'm in Italy! it 11 in the morning here. yuor help is very appreciated! ..I'm only trying to understand why my cascode doesn't go better than the CS ;)

• posted

you

effect

Look carefully at the PDF you gave above. The source of the second FET is not at AC ground; there is a current source connected between the source and the constant voltage generator labeled Vss. Current sources have high impedance. The gate of the second FET is connected to a voltage source, not a current source. Therefore, the gate of the second FET is at RF (AC) ground. If the source of the second FET were also at RF ground, then the second FET wouldn't have any gain because both the gate and source would be grounded.

connected

In the earlier post where I said that I go corner frequencies of about 200 MHz, I had used a value of Rd of 4.7 ohms for both amplifiers.

Now that I have used 600 ohms for Rd in both amplifiers, and with 250 pF connected at the output, I get a corner frequency of 2.25 MHz for both amplifiers. The frequency response is totally dominated by the 600 ohm and 250 pF combination. I notice that your book doesn't show a large capacitor like 250 pF on the output of the cascode.

In order to get a large bandwidth, it is necessary to remove the 250 pF capacitor on the output. When I do this, I get a corner frequency of about 490 MHz for the CS amplifier and about 504 MHz for the cascode.

The purpose of a cascode is to eliminate the Miller effect reduction of bandwidth. If I reduce Cgd to zero, I see only a small effect in the simulation. I think that your FET has such a low Cgd (124 fF) that the Miller effect is almost negligible even in the CS amplifier, so that the cascode doesn't show much improvement.

I'll do some more simulations tomorrow.

the

be

• posted

would

Cds

you

effect

not

connected

I think I could not put Rd = 4.7 ohm because I want Vout = 1.8 volt in order to have a large swing. Vdd = RdId + Vout Id = 1.5/Rd choosing Rd=4.7 ohm means Id=319mA ..too much! 600ohm is better ;)

250

I tried to remove my 250fF capacitance but the frequency is the same.. in the cascode I used Rd=600 Id=3.1mA W=592micron and Vov=0.15 why do you get a different value?

yes, that's the central point ..I think you are right ..the miller effect is completely negligible because of Rd and 250fF at Vout

• posted

250
250
490

I think I have misread your schematic. I read Cl as *250 pF*, but it appears that it is actually 250 fF. Is this correct? I thought your script character was a "p" when it's actually a "f". It might be a good idea to print. :-)

With this change, I now get a corner frequency of 315 MHz for the CS amplifier, and a corner frequency of 984 MHz for the cascode.

By the way, I decided to double check your calculations from your original post:

hi all, I calculated the frequency response of the CS and Cascode amplifiers using the time constants method. Here they are:

CS

f = ( 2*PI*( Rs*Cgs + Rs*Cgs*(1+Av) + Rd*(Cdb+Cl)) )-1

Cascode

f = ( 2*PI*(Rs*Cgs1 + Cds1*2*Rs + Rd*(Cgd2+Cdb2*Cl) + 1/gm2*(Cdb1+Cgs2) )-1

and I obtained

CS => f = 1.98EE+08

• posted

ehmm ..yes, sorry for my script character :P

• posted

appears

I'm just using these values that you gave me:

"the current in M1 and M2 is the same

using W = 592 micron Cgs1 = Cgs2 = 635 fF Cgd1 = Cgd2 = 124fF Cdb1 = Cdb2 = 770fF Cl = 250 fF

Id=3mA and Vov = 0.15 V gm1 = gm2 = 0.04 S"

and:

"Ro1 = Ro2 = 2500 ohm because Id=2.5mA and LAMBDA=0.16

yes, I'm using 600 ohms in both amplifiers for this calculation"

W doesn't appear in the expressions you gave for f, I use the various capacitances you gave me, and Rs = 50, Rd = 600, Av = 24.

• posted

Rodger Rosenbaum ha scritto:

it's driving me mad.. I checked my calculations hundreds time and I always get the same values I posted days ago..

for the cascode I put in my calculator:

f = 1 / 2 / 3.14 / (50*635EE-15 + 124EE-15*2*50 + 600*(124EE-15 +

770EE-15 + 250EE-15) + 1/0.04*(635EE-15 + 770EE-15)) = 207804859.6 = 2.08EE+8 instead of your 7.2EE+8

am I stupid? where am I making the mistake? #|

• posted

I see two differences compared to the expression in your original post.

should be 770EE-15

*

About the first difference: in your posted expression you have Cds1; I used the value for Cdb1 for that.

The second difference is * instead of +. Maybe it should really be +, but in the expression you posted it's *.

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