High Side Driver

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Where do you get one with dissimilar Vz?

Tim

You don't, but a dual 20 volt zener would work.

He won't because he can't.

Due to Miller, the actual value is Vdd-dependent (in addition to device-dependent). It's easier to overfill L1 and clamp off whatever you don't need. Fewer headaches.

Cheers, James Arthur

Question: Does total Qg include Miller effect?

Yes, and that's why it's Vdd-dependent.

Best regards, Spehro Pefhany

Spehro -

The graph I looked at for an example showed two Vdd lines so close together that I would have trouble distinguishing the difference on the graph. Never-the-less, I have been warned.

Many thanks.

Cheers, John S

Ain't no Miller in a follower (high-side) configuration. ...Jim Thompson

Huh! VDD-dependency is due to depletion layer spread. Ain't no "Miller" in a follower (high-side driver) configuration. ...Jim Thompson

Sorry, I thought we were talking about common source switching.. mixing two different conversations, I'm afraid.

Best regards, Spehro Pefhany

P.S. Here's a decent application note on MOSFET gate charge:-

Best regards, Spehro Pefhany

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Thanks! That's a good app-note! ...Jim Thompson

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"Miller" is just a word. There's certainly charge from Cdg in a follower. Lots of charge if Vd is high.

The fet doesn't know it's in a follower circuit. All it sees is the voltages on its three leads.

There's no gate current caused by the voltage changing across Cdg?

Sure there is, Cdg, the Miller capacitance, right? That's what I call it, anyhow.

As the FET turns on, dv/dt across Cdg saps off the charge you're trying to force into the gate, by an amount that varies with Vdg(t0).

Depending on the FET (and since you are depending on the FET's particular characteristics), it could matter.

Cheers, James Arthur