Hi all,

can someone tell me where i've gone wrong here. I have a amplifier whose load is an attenuator as such

VDD
  |
  |
 <
 <50 ohm
 <         50 ohm
  ------vvvvv------GND
  |
  |
 <
 <975 ohm
 <
  |
  |
 gm.Vin

In other words rather than having a 1 kohm resistive load I have a attenuator which is used for output matching for measurement purposes. So the attenuator has a voltage gain of 1/40. I can measure the S21 and the NF, and I know the source impedance is 50 ohm. Therefore I believe I can write,

NFamplifier=NFtotal - Vattn^2/V50^2*S21

which should approximately equal to NF - 2/S21 if we assume 25<<975. Vattn^2 is the attenuator output noise voltage squared which I calculated as 4kT(1/25+1/975)*25*25 based on this diagram

     __4kT/975__
    |                  |
    |      975       |
-------vvvvvvv--------------
                           |        |
                           <       |
                           <       4kT/25
                           <       |
                            |____|
                              GND

The results are completely off so i've definately made an error in theory here. Any help is greatly appreciated.

thanks,
Aaron

Sorry about that...let me see. The entire circuit consists of an amplifier followed by an attenuator. The attenuator provides 50 ohm matching.

Let the entire circuit's output noise power in volts squared be Pa
Let the attenuator's output noise power in volts squared be Pb
Let the source (50 ohm) noise power at the input be Pc


For the entire circuit including the attenuator,

Noise Factor, F1 = (Pa + Pc*S21^2)/(Pc*S21^2) = (Pa/S21^2 + Pc)/Pc

This is the noise figure which i can measure. I can also measure the S21. Take the attenuator output noise Pd and refer it to the input of the circuit:

Pb/S21^2

Now we should be able to say that the NF of the amplifier minus the attenuator is

F2 = (Pa/S21^2 + Pc - Pb/S21^2)/Pc = F1 - Pb/Pc*S21^2

In other words refer all noise back to the input, and take the measured noise factor and subtract the (input refered attentuator noise)/(source noise). Is this correct?

Also, the attenuator noise can be approximated as 4kT*25 while the source noise is kT*50 so the noise figure is F1-2/S21

I must have made a mistake somewhere...any help would be great

thanks,
Aaron

hmm...i don't think so...Pb is measured at the output of the attenuator not the input. Maybe i'll put this attenuator into simulation and check the output noise to double check.

Hi aw,

thanks for your help. The matching is very good. S11 is better than -30 dB at the operating frequency. Also I have had no luck in working back with simulation results. Well I guess I'll keep trying. Thanks again for the help

Aaron

what happens if you do this is in simulation:

VDD
 |
<
< 1313 ohm port 2
<
 |
LNA=gm.vin

with the LNAmatched to 50 ohm port 1, what S21 (LnaAv should equal this S21+10lg(1313/50) and NF do you get ? Confirming these numbers should help towards understanding your initial result. Also i guess you are using ideal VDD (ie GND and VDD are the same for RF purposes with no impedance?)

cheers
aw

The noise figure of an x dB attenuator is x dB...
so using Friis nois factor power cascaded lineup

NFtotal=NFLNA+[(NFattn-1)/S21LNA]

power gain S21LNA = 17dB =50
NFLNA=4.4dB=2.75
NFtotal=11.3dB=13.5

This then means the NFattn must be 27dB, so it power loss is 27dB.
This attnuator voltage gain is 29/(29+1284)= -33dB.  The 6dB dicrpenacy is to mismatch between 1284 and 29.

what do you think to this train of thought ?
cheers
aw

hmmm...i think i got it...i think the LNA noise gets lost in the noise floor after the attenuator. Therefore it cannot be de-embedded...someone please tell me i'm wrong...haha

Aaron