# Total noise in an amplifier with multiple noise sources

I have some doubts about how to evaluate the total noise in a circuit. The noise evaluation theory is well explained on the web, there are not specific practical examples similar to that I need. I'll consider, for instance, the following common emitter amplifier:

The previous picture shows the ideal circuit. Now, suppose that:

• All resistors are noisy (Thermal Noise): so they can be represented as ideal resistors with parallel current sources or series voltage sources. For instance, I'll consider the voltage source model.

• There is a leakage noise current between collector and emitter, represented by a current source.

• All the previous noise are known in terms of RMS

So, the overall real circuit will be:

My purpose is to evaluate the total noise at the amplifier output. In this circuit there are many noise sources. It's proved that in this situation, the overall noise may be evaluated by summing the square of all noise voltages or all noise currents, and then taking the square root. We'll get the total equivalent rms current or voltage noise:

$$i_{tot} = \sqrt{\sum_k(i_{k}^{2})}$$

or

$$v_{tot} = \sqrt{\sum_k(v_{k}^{2})}$$

But, I have these doubts:

I) In this case there is no simple connection (series or parallel) between noise sources. Is it correct in any situation (so also in this case) to apply the previous equation to find the total noise?

II) In this case there are both current and voltage sources. Of course, I may convert resistor series voltage models to parallel current ones. But suppose I keep the situation like that in figure: how can we apply the previous equation?

III) Once I've evaluated the total noise current (rms), how can I add it to my signal? I'd do this procedure:

• Consider the ideal circuit, without noise sources.
• Find the Norton or Thevenin equivalent circuit, in order to represent it like a signal voltage source or a current voltage source
• Add to the signal ideal model, the noise contribution found before.

In this case, this is a voltage amplifier, so it can be represented by a voltage source with a series output resistance:

Now, after I have found the total voltage noise with the equation:

$$v_{tot} = \sqrt{\sum_k(v_{k}^{2})}$$

I may add this noise voltage source in series to my ideal equivalent circuit:

Is this analysis correct?

IV) If the previous analysis is correct, I'll have a signal voltage source and a noise voltage source. The last one is expressed by its rms value. How can I manage this situation? Should I treat it like a mean value constant in time?

I'll only answer point (1). Noise powers add; so you can use an RMS sum to add noise voltages. (square each voltage, add, and take the root of the sum.

BUT...

all sources must be normalised to the same gain. So you can:

1. RMS add V1,V2, then multiply by gain A of TR1, then RMS add that and v3

2. scale V1,V2 by A, then RMS sum ... sqrt((A.V1)^2 + (A.V2)^2 + V3^2)

3. scale V3 by 1/A then RMS sum with V1,V2 if you need the input referred noise.

• Thank for your excellent explanation. So: noises sources connected to the same node may be added with rms sum, while noises sources connected to different nodes may be summed in the same way but weighted by the gain between those two nodes, correct? – Kinka-Byo Dec 25 '20 at 6:38
• Is it correct also If I use current noises and thebamolifier is a voltage amplifier? – Kinka-Byo Dec 25 '20 at 6:40
• Assuming you scale current to voltage (or vice versa) using the correct impedance, yes. – user_1818839 Dec 25 '20 at 11:52