Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.

On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

John

"John Larkin" wrote in message
...
On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

John



Thanks.

Is there a cheat sheet anywhere on the web to estimate the % of Cin that
disappears with various amounts of source degeneration?

On Wed, 1 Sep 2010 16:17:08 +0100, "ian field"
wrote:


"John Larkin" wrote in message
.. .
On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

John



Thanks.

Is there a cheat sheet anywhere on the web to estimate the % of Cin that
disappears with various amounts of source degeneration?


The small signal gain (gate-to-source) of an (R-loaded) MOS follower
is ...

(gm*R)/(1+gm*R)

Take it from there ;-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

Democrats are best served up prepared as a hash
Otherwise the dogs won't eat them :-)

On Wed, 1 Sep 2010 16:17:08 +0100, "ian field"
wrote:


"John Larkin" wrote in message
.. .
On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

John



Thanks.

Is there a cheat sheet anywhere on the web to estimate the % of Cin that
disappears with various amounts of source degeneration?


Well, the effective source resistance is 1/Gm. That makes a voltage
divider with whatever impedance the follower drives, like the source
resistor. So the gain might be, say, 0.9. That will reduce the charge
pumped into Cgs to 1/10 of the original value, so Cgs drops by 10:1.

There's a little additional error due to the slope of the drain
curves, which makes the follower gain less than 1 for even an infinite
source load, like a current sink.

Again, you can also bootstrap the drain if every pF matters. Or use a
dual-gate mosfet or a phemt. Or an opamp.

John

"John Larkin" wrote in message
...
On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

Its proving difficult to get relevant hits on google - any hints on a
suitable search string?

On Wed, 1 Sep 2010 21:58:01 +0100, "ian field"
wrote:


"John Larkin" wrote in message
.. .
On Wed, 1 Sep 2010 15:00:35 +0100, "ian field"
wrote:

Does a source follower present lower apparent input capacitance?

I'm thinking in terms of 2N7000/BS170 sized devices.

Thanks.


Yes. If the follower voltage gain is close to 1, the gate-source
capacitance is bootstrapped and goes away. All that's left is
drain-gate capacitance. You can bootstrap that, too, with another
follower.

Its proving difficult to get relevant hits on google - any hints on a
suitable search string?


But I've just explained it all!

John