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comparing CS and Cascode frequency response

M

Matteo

Jan 1, 1970
0
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!
 
T

The Phantom

Jan 1, 1970
0
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!

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

Matteo

Jan 1, 1970
0
T

The Phantom

Jan 1, 1970
0
hi all,
I calculated the frequency response of the CS and Cascode amplifiers
using the time constants method. Here they are:

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?
 
M

Matteo

Jan 1, 1970
0
The Phantom ha scritto:
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?

Sorry, f is the 3 dB cut-off frequency
 
T

The Phantom

Jan 1, 1970
0
The Phantom ha scritto:

Sorry, f is the 3 dB cut-off frequency

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.
 
M

Matteo

Jan 1, 1970
0
The Phantom ha scritto:
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.

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

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
 
T

The Phantom

Jan 1, 1970
0
The Phantom ha scritto:

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

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.
 
M

Matteo

Jan 1, 1970
0
The Phantom ha scritto:
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.

Thank you phantom, I'll wait for your results to check my work!
 
T

The Phantom

Jan 1, 1970
0
Matteo,

1. In this pdf:
http://www.thebags.it/listing/uni/Cascode_Configuration_Small_Signal.PDF

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

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

3. 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?

4. 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.
 
M

Matteo

Jan 1, 1970
0
The Phantom ha scritto:
Matteo,

1. In this pdf:
http://www.thebags.it/listing/uni/Cascode_Configuration_Small_Signal.PDF

you have Cds2 connected at the output. Shouldn't it be connected in
parallel with Ro1; that is, across the gm2*Vgs2 source?
Cds2=Cdb2 goes from the drain terminal to the grounding, therefore I
think it's correct. I also checked it on a book..
2. Also, I need to know what values you used for Ro1 and Ro2.
Ro1 = Ro2 = 2500 ohm
because Id=2.5mA and LAMBDA=0.16
3. 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?
yes, I'm using 600 ohms in both amplifiers for this calculation
4. 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 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'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.

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
 
R

Rodger Rosenbaum

Jan 1, 1970
0
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..

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.
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..

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.
thanks phantom

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.
 
M

Matteo

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
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.
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:
http://www.thebags.it/listing/uni/lecture25annotat.pdf

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

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 ;)
 
R

Rodger Rosenbaum

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
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:
http://www.thebags.it/listing/uni/lecture25annotat.pdf

( 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 )

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.

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.
I checked it in some diapos I found in
the web.

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.
 
M

Matteo

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
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.


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.

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

Rodger Rosenbaum

Jan 1, 1970
0
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?

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

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 <<<For this one, I get 1.107E+08
Cascode => 2.08EE+08 <<<For this one, I get 7.2915E+08

You might want to check your calculations again.
 
M

Matteo

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
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.
ehmm ..yes, sorry for my script character :p

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 <<<For this one, I get 1.107E+08
Cascode => 2.08EE+08 <<<For this one, I get 7.2915E+08

You might want to check your calculations again.

Wow .. but ..Can you tell me which values of W, Vov=Vgs-Vth, Rd and Rs
did you used in this calculation please?
thank you again
 
R

Rodger Rosenbaum

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
ehmm ..yes, sorry for my script character :p



Wow .. but ..Can you tell me which values of W, Vov=Vgs-Vth, Rd and Rs
did you used in this calculation please?
thank you again

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
3. 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?
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.
 
M

Matteo

Jan 1, 1970
0
Rodger Rosenbaum ha scritto:
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

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.

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? #|
 
R

Rodger Rosenbaum

Jan 1, 1970
0
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:

I see two differences compared to the expression in your original post.
f = 1 / 2 / 3.14 / (50*635EE-15 + 124EE-15*2*50 + 600*(124EE-15 +
should be 770EE-15
770EE-15 + 250EE-15) + 1/0.04*(635EE-15 + 770EE-15)) = 207804859.6 = *
2.08EE+8 instead of your 7.2EE+8

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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