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Explanation of the Right Half Plane zero concept

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alex.ang1983

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Hi All,

I would like to understand a bit more in details and clearly the concept of right half plane zero expecially how can I detect it (kind of) from a circuit and a bit of maths more (for example in a simple common source device). I read Razavi and Allen book but they also don't explain so much.

Everyone who can help me on this will make me happy. If you have papers to suggest me (or something like this..) this is good as well.

Many thanks
 

sutapanaki

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rhp zero explain

I'm not going to go into math here. The math you can find in the books. The zero is caused by a direct path from input to output of a common source stage for high frequencies. This direct path (imagine a capacitor between gate and drain that is going to a short for high frequencies) is transferring the signal without inversion - it is just a short. On the other hand, the proper common source stage is inverting the signal. Because the capacitor is trying to change the polarity of the stage, the zero it introduces is in the right-half plain. At the output of the stage there are 2 effects that fight against each other in terms of current. One is the controlled current source of the transistor gm*vgs. The other is the current through the capacitor between gate and drain. If the gate voltage is increasing wrt the source, the controlled current source is sinking current from the drain. On the other hand the current in the capacitor is being pushed into the drain and this current increases with frequency. For a given frequency the capacitor current becomes equal to the current of gm*vgs and is completely absorbed by the controlled current source. This is then the frequency of the zero because no current is left to go into the load and the output voltage is 0 (in Laplace domain, not on the jw axis ). If the output voltage is 0, then both the current in the capacitor and the current in the controlled current source are defined by vgs. It follows then that:

sCgd=gm
s=gm/Cgd

So, the RHP zero is w=gm/Cgd
 

sapphire

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left hand plane zero vs rhp zero

Hi sutapanaki

But the current from Cgd is 90 degree out of phase when compared to gm current. No way these two current can cancel each other. Could you please explain that?

Thanks

sapphire
 

sutapanaki

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significance of rhp zeros

That is the reason why I said that they are equal in Laplace domain, not on the jw axis. We use phase difference when describing things in the frequency domain i.e. on the jw axis of the complex plain. In the frequency domain you can not say that at the frequency of the zero the transfer function is 0 i.e. the output of the stage is not 0v. There is just a break point there, accompanied by a change in slope. However, in the complex plain it is perfectly ok to have H(s)=0 for certain s. Remember that H(s) is a 3-dimensional surface in the complex plain and at the zero locations it really reaches zero, and at the pole locations the surface goes to infinity. But this doesn't mean that it does the same on the frequency i.e. jw axis. It can reach zero or infinity on the jw axis only in the very specific case when the zero or the pole are placed on the jw axis. But this is not usually the case with the RHP zero in gain stages - in fact in this case it is a real zero meaning placed on the horizontal axis.
 

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rhp zero transfer function

laplace domain and complex frequency are so confusing. Why not think everything in frequency domain? What's the purpose to use laplace domain? Thanks
 

sutapanaki

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Re: RHP zero help

It could be confusing at first, but once you grasp the idea everything falls in place. In frequency domain we usually work for steady-state processes. In this case transients are already gone. Laplace transform in general is used when we have a transient process going to a steady state, or even not reaching steady-state at all if there's instability. In general the natural complex frequencies "s" of a system are in the complex plain, not on the jw axis. H(s) is a characteristic function of the system and is defined over the complex plain and as I said before is a 3-dimensional surface. If you dissect this surface with a vertical plain going through the jw axis you get the frequency response, which in fact is equivalent to substituting s with jw.
After all, Laplace transform is just a means to solve easily integro-differential equations. The system really is described physically/mathematically by differential equations which we map into the Laplace domain.
 

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