opto-coupler pole is BC capacitance?

[eem2am]

Hello,

If we use an opto-coupler (eg 4N35) to provide isolated feedback in an SMPS, then we must take into account the opto-coupler’s pole………(for feedback stability analysis)

….in fact, this pole is said to exist as a capacitor , -known as “C(pole)”.

In SMPS’s where the primary side opto-transistor has a pull-up resistor, the “C(pole)” capacitor is said to exist as being connected from the opto-transistor’s collectror to the opto-transistor’s emitter…

Regarding this “C(pole)” capacitor, is it really just the actual collector-base capacitance of the opto-transistor?

 

[FvM]

A capacitor alone is no pole, so what's the respective resistor? Of course it's not the load resistor Rc. I suggest β*Rc as a rough estimation. Or consider the transistor as a capacitance multiplier, in either case β*Ccb*Rc is the time constant.

P.S.: As a conclusion, if you want to achieve higher bandwidths in analog optocoupler applications, you should reduce the load resistance and/or operate the coupler in diode mode. If no reduction of voltage swing is acceptable, an active load is required, either an OP I/V converter or a BJT cascode stage.

 

[rfsystem]

The optocouplers BJT differs from a standard BJT that the area is much bigger to receive more infrared light and the base depth is much bigger to absorb a big part of the infrared. That has two consequences.

1. The internal Ccb is high.

2. The base transit time is high

If the collector current is not driven into low ohmic virtual ground the miller cap effect adds to the above two effects.

If you record the corner frequency of the coupler over current level like for BJT the Ft over Ic you can extract two time constants.

Tcoupler=Tc+Tb

1. Tc=Ccb*Vt/Ic

2. Tb=TransitTime

You can make a linear regression if you plot 1/Ft over 1/Ic

One can argue that base resistance is high but the photo current is generated if the minorities are recombined at the collector/base depletion zone. So there is no natural base resistance.