# Negative NF for LNA in Hspice simulation and 0dB of input and output matching

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#### Ngu Kek Siang

##### Newbie level 3 Currently I am trying to simulate the noise figure, S11 and S22 parameters for LNA at 2.4GHz in HSPICE. Below is the code I used:
Calculate noise figure:
.ac dec 50 400MEG 3g
.NOISE v(vout) vin 1

Calculate S11 and S22:
.net v(vout) vin rin=50 rout=50
.print s11(db) s11(m) s11(p)
.print s22(db) s22(m) s22(p)

However, I get negative NF after calculate from the following:
**** resistor squared noise voltages (sq v/hz)

element 0:r3
rs 9.1186f
1/f 0.
total 9.1186f
rx 5.2622k

...

**** total output noise voltage = 433.9253f sq v/hz

NF = 10log(9.1186f/433.9253f) = -16.77dB

and get 0 value for S11 and S22:
freq s11 s11 s11
db mag phase
400.00000x 0. 1.0000 180.0000
418.85142x 0. 1.0000 180.0000
...
2.41024g 0. 1.0000 180.0000

freq s22 s22 s22
db mag phase
400.00000x 0. 1.0000 180.0000
...
2.41024g 0. 1.0000 180.0000

Can I know did I do wrongly? Or why these values appear?

Thank you.

#### SunnySkyguy If using ratio of in/out noise power, then you must add signal gain (dB), otherwise use ratio of SNR's in/ out.

#### Ngu Kek Siang

##### Newbie level 3 Currently I am trying to use ratio of SNR's in/out, where:
For input noise, it is equal to 9.1186f sq v/hz:
**** resistor squared noise voltages (sq v/hz)

element 0:r3
rs 9.1186f

For output noise, it is equal to 433.9253f sq v/hz:
**** total output noise voltage = 433.9253f sq v/hz

Hence, the NF which I get at the moment from the testing circuit is:
= 10log(SNR in/SNR out)
= 10log(9.1186f/433.9253f)
= -16.77dB

I am not sure why I get negative value. Or how can I add signal gain to obtain the NF?

#### volker@muehlhaus .print s11(db) s11(m) s11(p)
.print s22(db) s22(m) s22(p)
and get 0 value for S11 and S22:
db mag phase
2.41024g 0. 1.0000 180.0000

2.41024g 0. 1.0000 180.0000
Your S-parameter tell us that you have a perfect short (S11, S22 mag=1.00 ang=180°) at input and output !?

- - - Updated - - -

Hence, the NF which I get at the moment from the testing circuit is:
= 10log(SNR in/SNR out)
= 10log(9.1186f/433.9253f)
The equation is correct, noise figure is calculated from signal to noise ratios.
The calculation is wrong. You have used noise voltages for calculation, not SNR values.

#### Ngu Kek Siang

##### Newbie level 3 Your S-parameter tell us that you have a perfect short (S11, S22 mag=1.00 ang=180°) at input and output !?
Can I know why I have a perfect short? The netlist I write, there are input, output, voltage source, and resistors. I am currently providing Vdd of 0.5V, from the output waveform, it is oscillating at about 0.49995V with an amplitude of 0.0028V.

The equation is correct, noise figure is calculated from signal to noise ratios.
The calculation is wrong. You have used noise voltages for calculation, not SNR values.
Can I know how can I get SNR values using HSPICE? I try to search online, but the examples which I found all are similar as below, which gives me noise voltages:
.NOISE vout vin 1

#### volker@muehlhaus Can I know why I have a perfect short?
I don't know, but this is what we get from your S-parameter data.

Can I know how can I get SNR values using HSPICE?
SNR is signal to noise ratio, so I would expect that you need the signal level also (not just noise voltage). Or if you compare the noise ratios, then you would also need to include the amplifier's signal gain, as SunnySkyGuy has mentioned earlier.

#### vfone Electronic circuits ALWAYS add noise in a system, so SNR at the output will ALWAYS be smaller than the SNR at the input.

Therefore, Noise Factor and Noise Figure will be always greater than zero.

Noise Factor(linear) = SNR_input(linear) / SNR_output(linear)
Noise Figure(dB) = 10*LOG (Noise Factor(linear))

#### SunnySkyguy ...
Hence, the NF which I get at the moment from the testing circuit is:
= 10log(SNR in/SNR out)
= 10log(9.1186f/433.9253f)
= -16.77dB

I am not sure why I get negative value. Or how can I add signal gain to obtain the NF?
If your signal gain is 20dB add this to -16.77 to get NF=3.23dB.
What is your gain assuming there is no BW reduction?

#### Ngu Kek Siang

##### Newbie level 3 If your signal gain is 20dB add this to -16.77 to get NF=3.23dB.
What is your gain assuming there is no BW reduction?
Thanks, SunnySkyguy, in this way, I am able to get low NF. My gain is around 17~19dB. By the way, I would like to ask another question, the formula which you use is as below:
NF = Gain + 10log(Input Noise Voltage/Output Noise Voltage)

Can I know why we can calculate using this method?

Besides, my S11 and S22 seems became shorted as the S11, S22 mag=1.00 ang=180° at input and output. What could be the possible reason it is shorted?

Thank you.

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