ldo capacitor problem

[devop]

The fig illustrate the factors which determine the stability of LDO, but I don’t know where is the Cb in a real chip ,since I read most spec of LDO,and there is only one capacitor connect to Vout.Or the Cb embedded in the chip itself??
And why don’t we concern the capacitor of load ??
and how can we determine the size of C0???
thanks

 

[peterwang]

Cb is bypass capacitor. This diagram is from Figure 2 of the book:

Study and Design of Low Drop-Out (LDO) Regulators
Authors: Gabriel Alfonso Rincon-Mora & Phillip Allen
Publication: GIT
Volume: 28 pages

The answer for your quesiton is at pp.10, you can get the book at:


**broken link removed**

 

[huojinsi]

In my viewpoint, the LDO structure should be the same as the picture. The output has only one large capacitor Cb, and this Cb capacitor equates the series of a capacitor and a ESR. The effect of Cb is to generate a zero and raise the phase margin of loop, so stabilize the whole loop.

 

[renwl]

The capacitor Co is used to reduce the low frequency ripple.

and the capacitor Cb is used to bypass the high frequency noise.

 

[safwatonline]

On-Chip decoupling cap

 

[devop]

On-Chip decoupling cap

I think so. but the paper said it will make the first nodominant pole 1/2*pi*Resr*Cb,how could it be ,the esr is so small?

 

[safwatonline]

if it is the first non dominant pole then it is called the bypass capacitor and it used as Off chip decoupling with very low ESR and value range of about 0.1uF (ceramic), so even if the ESR is usually small, the Cb is still large.
Note: this cap is used to supply the high freq. current

 

[devop]

thanks safwatonline,you helped me a lot,

 

[peterwang]

Cb is normally much smaller than Co. The pole of 1/2*pi*Cb*Resr refers to the situation of high freq, where Co is virtually shorted, so the Resr of Co becomes in paralell to Cb, and becomes the next pole after Co.

In fact, there are three pole/zeros in play:

po = 1/2*pi*ro*Co (where ro is the output resistance in paralell to Co).
zo = 1/2*pi*Resr*Co (note, this is a zero)
pb = 1/2*pi*Resr*Cb (this is pole, it happens when freq is high and Co becomes shorting path)

Normally in LDO, po < zo < pb, no matter your compensation is internal or external.

 

[lijianheng]

Cb is a uF offchip bypass cap.