Differential amplifier gain

[Sambhav_1]

Hi,

i am lookin for designing 100MHz bandwith diff. amp but I am getting this kind of gain plot. I want to know why is this peak coming and how can i eliminate?

 

[martinnl]

You apply the same signal on both inputs. What you see is your common mode gain.

 

[Sambhav_1]

You apply the same signal on both inputs. What you see is your common mode gain.

Do you mean this is not the correct way to determine bandwidth? apologies if i am understanding wrong.

 

[martinnl]

The differential input signal is the difference between the two inputs which in your case is 0. You only change the common mode level of the 0 amplitude input signal.

Set two voltage sources to, for example, 0.5 V AC amplitude and whatever DC level you have. One should have 0 degrees phase and the other 180 degrees.

This means that your input signal will be 0.5-(-0.5)=1 V in amplitude.

 

[FvM]

You don't show which output signal is used to plot the gain, we can just guess that it's a single ended output, resulting in relative large common mode gain. You should use output voltage difference instead.

 

[Sambhav_1]

The differential input signal is the difference between the two inputs which in your case is 0. You only change the common mode level of the 0 amplitude input signal.

Set two voltage sources to, for example, 0.5 V AC amplitude and whatever DC level you have. One should have 0 degrees phase and the other 180 degrees.

This means that your input signal will be 0.5-(-0.5)=1 V in amplitude.

with this, how can i check my differential gain? i need 15dB differential gain. so should i check like this, amplitude at output/amplitude at input and converting it into dB.

 

[KlausST]

HI,

differential_gain = differentail_output / differential_input

where:
differential_input = (IN+) - (IN-)
differential_output = (OUT+) - (OUT-)

Klaus

 

[martinnl]

And in dB 20log10(differential_gain)

 

[Sambhav_1]

Thanks Everyone,it helped.
I have one more question related to differential amplifier design,

I am trying to Find SNR in ltspice but there is only noise analysis.
so i am thinking to calculate signal and noise power seperately. I have calculated the nosie (uV/sqrt Hz) over the frequency range (nearly60dB)
I am confused with the signal power. SHould i do transisent sim with sine wave and calculate the power?

 

[martinnl]

Best way is probably to refer the noise back to the input and compare with the expected input signal.

Since you have a certain gain your output signal depends on the input signal amplitude and your SNR will only be valid for a certain input/output amplitude.

No need to run a transient. Run an AC to see your gain.

Note that your units for the noise is a bit strange. When you "calculate" noise over a frequency band your unit won't be V/sqrtHz. Also you mention that it is 60 dB, dB is unit less and that result doesn't make sense.

 

[Sambhav_1]

Best way is probably to refer the noise back to the input and compare with the expected input signal.

Since you have a certain gain your output signal depends on the input signal amplitude and your SNR will only be valid for a certain input/output amplitude.

No need to run a transient. Run an AC to see your gain.

Note that your units for the noise is a bit strange. When you "calculate" noise over a frequency band your unit won't be V/sqrtHz. Also you mention that it is 60 dB, dB is unit less and that result doesn't make sense.

I am getting this noise plot, 60 dB was coming from 20log(Anoise)
. [IMG alt="
\mathrm{SNR_{dB}} = 10 \log_{10} \left [ \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right )^2 \right ] = 20 \log_{10} \left ( \frac{A_\mathrm{signal}}{A_\mathrm{noise}} \right ) = \left ( {A_\mathrm{signal,dB} - A_\mathrm{noise,dB}}\right ).
"]https://wikimedia.org/api/rest_v1/m...f2fbec646247d05d7d35efc260be9d2aa1bed3d[/IMG]
i am little confused now. Am i doing till the abeve step correctly?

Thanks

 

[martinnl]

Okay then I understand where you got it from. The unit in this case is dBV, when you take 20log10(Volt) your unit is not dB but dBV.

But I would stop earlier in the equation, you have A_noise, just find A_signal and you're done.

 

[Sambhav_1]

Okay then I understand where you got it from. The unit in this case is dBV, when you take 20log10(Volt) your unit is not dB but dBV.

But I would stop earlier in the equation, you have A_noise, just find A_signal and you're done.

Thanks martinl,

In the schematic shown, i am planning to use cs stage with source degeneration to increase linearity.But as expected my gain is reducing. ANy suggestion on this?