# can somebody here explain this KCL equation to me.

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#### the_falcon

##### Member level 4 HI,
The attached diagram has the small signal equivalent diagram of a common gate amplifier.I am trying to understand the noise figure expression of the same.

But I kind of got stucked in the right beginning not able to get the KCL equation.Could anyone teach me that.Also please tell me as how they can interpret source resistance(Rs) as (1/gm) in the problem.

Falcon #### analogTechie

##### Junior Member level 1 Hi,

KCL is written at the two nodes in the model.

The first equation is for the total current flowing into 'node Vs', and the second is for the total current flowing out of the output node 'Vo'. It looks like a pretty good model because it includes noise currents (and noise voltages converted to the equivalent currents) in addition to the signal currents.

I believe there are a couple of errors in the equations though: First of all, the first term of the first equation should be (vin +VnRs - vs)/Rs (looks like they left the vin term out by mistake). Secondly, I am not sure that you can add noise currents quite in this fashion since the direction of the noise current is unknown. So, if you are doing a worst case analysis, you must determine which direction would give you the worst case and use that one. Alternatively, you would need to analyze a whole set of equations. [Others might want to weigh in on this]

As for the Rs=1/gm, that is because looking into the source of a common-gate transistor, you see an impedance equal to 1/gm (actually 1/(gm+gmb))

rgds,

js

### the_falcon

points: 2

#### the_falcon

##### Member level 4 Hi dude
thanks for taking time to explain me.that was really appreciable.but can u also tell me as how they got the positive sign for ind in the first equation.

i kinda getting confused as they use different signs for currents flowing in the similar directions.

regards
falcon

#### snafflekid

##### Full Member level 4 The purpose of this circuit is to create an impedance match between the signal input and the amplifier. Rs in this case is the signal resistance and the amplifier is common-gate configuration. The transistor can be sized and drain current set so that the impedance looking into the transistor source will equal Rs. That is why the problem says Rs=1/gm, you assume that you make Rs=1/gm

VnRs is the noise voltage of this signal resistance.
When calculating a noise factor, the input is grounded and the sum of the noise powers is found at the output. Or noise factor can be found by using input referred noise, it is the same answer.

I show the KCL equations on page 1. It can be tricky and of course one mistake on a sign makes the whole problem wrong, so it is best to become very good at doing this. I hope it helps.

As for the way this problem starts to find noise factor, I would never do it that way. I don't know how your teachers want you to find F, but I would draw each circuit as having a single noise source, then use superposition to add the noise sources at the output, like I have shown. Actually, I would leave out RL because it makes the problem more complicated (this does not affect F very much in practice) and find F as a sum of noise currents, but your problem includes RL so I did too.

Good luck! I know how helpful it is to see one of these problems worked out in detail. F should be unitless, so if you end up with units somewhere, you made a mistake.

### the_falcon

points: 2

#### the_falcon

##### Member level 4 Hi Snafflekid,
Really really thanks a lot.Cant believe you took such a pain to explain me

CHeers dude.Good luck for you.. Falcon

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