# PSRR and CMRR in a diffferential amplifier

Please, can anyone help me how to find the PSRR and CMRR in this simple differential amplifier circuit as shown below in figure 1?

I have searched on the internet and I decided to follow this methodology to find it:

First of all, I calculate the differential gain of the amplifier (the figure 1) which is for a single-ended output 4.510 (so for differential output is 9.021) as shown in figure 2.

After this step, I apply an ac voltage source in series with the DC supply voltage (1.2V) of the circuit, to insert the supply noise.Also in the transistor, I apply only dc voltage values.Again I measure the gain which is very small for single ended output:7.58*10^-6 (I have named this gain as Asupply.Also a differential output does not make sense because Asupply is zero) as shown in figure 4.

So the PSRR = Adm / Asupply = 9.020 / 7.58*10^-6 = 1.189*10^6

PSRRdB =20log(PSRR) = 20log (1.189*10^6 ) = 121dB

Because I am little trouble can someone advise me if the methodology that I follow is correct? My result for PSRR is correct? Am I doing something wrong?

Furthermore, I insert a mismatch between W (W=1um right hand and W=2um left hand)of the input transistor to become more realistic the circuit and my schematic become as shown in figure 7 and results is in figure 8.

Also, I try to calculate the CMRR of the amplifier.We know from figure 1,2 that the differential gain is 9.021. To find the CMRR I follow these steps. First of all as shown in figure 5 I apply the same input in the input transistors and calculate the gain as shown in figure 6.So

CMRR = Adm / Acm = (9.021/3.11*10^-6) = 2.9 *10^6 CMRRdB = 20log(2.9 *10^6) = 129dB

Is this result correct for CMRR?

Figure 1

Figure 2

Figure 3

Figure 4

Figure 5

Figure 6

Figure 7

Figure 8

• Does SRC163 have a conducatance/shunt resistance value? Your schematic would be more clear for others to understand if you showed this explicitly. Otherwise, I don't see how your power supply noise source can be affecting the output at all. – The Photon Sep 16 '17 at 21:04
• OK actually I do see...it looks like your power supply noise is coupled through the FET backgates. Your design would be more realistic if you explicitly included the the current source design using FETs. And also if you included a realistic amount of mismatch between the FETs and between the resistors. – The Photon Sep 16 '17 at 21:07

1. The PSRR should probably be measured in terms of the differential output, $v({\rm out2})-v({\rm out1})$, rather than a single-ended output.

2. In a real circuit, the PSRR depends mainly on mismatch between the components. To estimate what this design would produce, you should include some mismatch between R43 and R44, and between M327 and M328. In ADS is fairly easy to set up parameters with tolerances and do monte carlo simulations to see the range of effects with process variation.

3. You might rather do an AC simulation instead of transient, so you can see how the PSRR varies with frequency.

4. Voltage sources add when placed in series. It would be slightly more obvious to others to just put your AC voltage source in series with the DC source than to make a new source with both AC and DC components. But what you did won't give wrong results.

• Thank you so much The Photon for your response but I can't understand your thoughts if my results for PSRR are correct or may need some changes. Also, I refer you that the DC voltage sources in my simulation don't have any resistance. This voltage source on ADS doesn't give the opportunity in the designer to add resistance so the only way to do this is to put a resistor in series with the voltage source. – gr1 Sep 17 '17 at 10:07
• @gr1, my points 1 and 2 are the most important ones. If you brought me these results at work and said this simulation was supposed to predict the PSRR of a real circuit, I'd ask you to fix these things before looking at your results. – The Photon Sep 17 '17 at 15:08
• Thank you again @ The Photon. But how can I measure the v(out2)−v(out1)? This result if we haven't got any mismatch gives zero because of the supply noise is common for the two outputs! I rewrite my question above measured the gain with a mismatch.So now the result for PSRR = (9.021)/(1.044*10^-4) =8.64 *10^4 ? What your opinion about my methodology to find the CMRR? Thank you again for your answers.I ask you maybe silly questions but I am beginner in electronics and I haven't got previous experience – gr1 Sep 17 '17 at 15:41
• 1. You can simply write "out2 - out1" instead of just "out2" on the y axis of your graph (or in the "expression" field after double clicking the trace on the graph). 2. The fact this gives 0 is exactly why my 2nd point is so important. 3. If you'd take my advice in point 3, you wouldn't have to do any calculation to get the PSRR (you can even plot "(vout2 - vout1)/1 mV") 4. My opinion is you haven't measured PSRR in any useful way at all. – The Photon Sep 17 '17 at 15:44
• Thank you so much @The Photon for your time. I have 3 last questions if you can help me.My first question is: I have rewrite my question above and the result which we discuss appears in the figure 8. So now PSRR = (9.021)/(0.160) = 56 => PSRRdB = 20log(56) = 35dB? Correct? My second question is :Is the asymmetry I add is appropriate? If I had, for example, W=10um for the transistor in the left hand and W=1um for the transistor in the right hand the result may be very different for PSRR because now the asymmetry in widths is very big. – gr1 Sep 17 '17 at 16:12