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Old 1st March 2010, 08:36 PM   #1
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Default Source Followers and gain - ignorance requests instruction!

Hi there,

In trying to learn about amps and topologies over the years, I continually run into the same question and have been unable to determine the answer in my own studies.

For the sake of an example, just about any of the Pass designs I've seen around here operate with a source follower output stage... that's why I chose this sub-forum... your kindness and help is greatly appreciated in advance!

So, typically we have a differential input stageto allow for some error correction through negative feedback, followed by some variation of a voltage amp gain stage, followed by some well-biased array of source followers to deliver the necessary current to the speaker. Seeing BJT designs like this (or more complicated using extra stages and lots more feedback) never phased me. One day when reviewing one of Mr. Pass' designs, a question entered my mind...

The confusion I have is between the voltage gain stage output and the source follower input. If a bias voltage of 4-5V is at the gates for turn-on, and a Vgs of 20V will destroy the Mosfet, how can you amplify beyond that 20V input limit in a follower configuration? And often times a 9V Zener is used to provide such protection, further limiting input voltage.

My mind only sees a max 20V input, less Vgs drop of say 4V, leaves a Source voltage of 16V. 16V into 8 ohms is only 32Wpeak, 22Wrms.
Which leads me to two questions:
1 - How is it that we obtain more voltage at the output than at the input of a follower configuration?
OR
2 - How is it possible to deliver more than 20V to the gate without causing breakdown?

For some time I have been trying to figure this out in reading whatever App Notes and other circuit descriptions I can get hold of. I am at whits end...

Warm Regards,

Many, many, many thanks!
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Old 1st March 2010, 09:12 PM   #2
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Remember, the signal appears at the source. So as the gate voltage rises, the source voltage, ummm, follows. The gate-to-source differential is darn close to constant, so it won't go from 4 to 20V, more like 4 to 4.5V.
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Old 1st March 2010, 09:17 PM   #3
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The source is following the gate.

w
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Old 1st March 2010, 09:38 PM   #4
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Originally Posted by SY View Post
Remember, the signal appears at the source. So as the gate voltage rises, the source voltage, ummm, follows. The gate-to-source differential is darn close to constant, so it won't go from 4 to 20V, more like 4 to 4.5V.
Yup, understood. Probably poorly phrased questions above (I'm a bit wordy)...

Since it "follows" I don't see how it could ever be more than say 15.5V at the Source (assuming max 20V gate input, with a 4.5V drop, that leaves 15.5V at the Source) ... What am I missing?
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Old 1st March 2010, 09:46 PM   #5
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Ok.

Say you put 30 volts at the gate.
The Source pin in a follower will be 26v
the difference will be 4 volts.
No ode of Alan Bradly.
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Old 1st March 2010, 09:46 PM   #6
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By golly, a 4million Watt light bulb just went off. The breakdown voltage is relative to SOURCE, not Ground!

I kept relating a 50V input to the gate as though source is connected to ground... But it has the load (and source resistor) isolating it from ground potential... DUH!

I feel so stupid! Thank you!
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Old 1st March 2010, 09:47 PM   #7
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Remember, the max gate input is measured gate to source, not gate to ground.
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Old 2nd March 2010, 10:03 PM   #8
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By golly, a 4million Watt light bulb just went off.
Don't you love it when that happens?

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Old 3rd March 2010, 08:05 PM   #9
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Don't you love it when that happens?

For me I kind of dislike it. That amount of brightness blinds me and I can do much after that
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