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Infineon IPA50R140CP N-channel mosfet
Infineon IPA50R140CP N-channel mosfet
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Old 26th January 2016, 05:37 AM   #1
exeric is offline exeric  United States
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Default Infineon IPA50R140CP N-channel mosfet

Here is an interesting n-channel mosfet for use in a tube grid powerdrive application. It has a ruler flat Crss curve at 2.5 pf from 40 to 500V. If I haven't missed an important parameter that is required for that usage then it may be the current best. Here is the spec sheet:
http://www.mouser.com/ds/2/196/IPA50...v2.1-80391.pdf
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Old 26th January 2016, 06:46 AM   #2
Wavebourn is offline Wavebourn  United States
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Why?
Input capacitance is about 2 nano Farad, it is more significant; however it is bootstrapped by about 15 A / Volt transconductance. Output capacitance goes up sharply below 100W, reverse capacitance goes up sharply below 50 Volt, up to one nano Farad.
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Old 26th January 2016, 07:18 AM   #3
Hearinspace is offline Hearinspace  Canada
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Infineon IPA50R140CP N-channel mosfet
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Originally Posted by Wavebourn View Post
. . . . . . . . . .however it is bootstrapped by about 15 A / Volt transconductance.
By that , do you mean it's made effectively greater?
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Old 26th January 2016, 07:29 AM   #4
Wavebourn is offline Wavebourn  United States
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Originally Posted by Hearinspace View Post
By that , do you mean it's made effectively greater?
No!

Effective capacitance is Cg-s * (1 - Click the image to open in full size.)

I.e. the higher is the transconductance, the lower is the capacitance. However, 15 A/V is a large current transconductance. On the grid drive current it will be lower.
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Old 26th January 2016, 07:34 AM   #5
exeric is offline exeric  United States
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Perhaps you're right. I was thinking that as long as you could keep the Crss above that 40v it could be operated pretty linearly. Someone I respect thinks it is important for that to be linear. But perhaps I was putting to much emphasis on that and ignored the fact that the linearity comes in at a pretty high level, 40 volts, and goes up abruptly below that 40volts.

Do you have an n-channel mosfet that you recommend for a follower in that application?
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Old 26th January 2016, 03:54 PM   #6
Hearinspace is offline Hearinspace  Canada
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Infineon IPA50R140CP N-channel mosfet
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Originally Posted by Wavebourn View Post
No!

Effective capacitance is Cg-s * (1 - Click the image to open in full size.)

I.e. the higher is the transconductance, the lower is the capacitance. However, 15 A/V is a large current transconductance. On the grid drive current it will be lower.
OK, that's something you taught me a few years ago but I couldn't hear the tone of voice in the first "however" and needed to check.

I still don't really understand the use of "Bootstrap". It seems to mean a few different things. Th word suggests pulling something up, though it often appears in ways that aren't that.
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Old 26th January 2016, 04:29 PM   #7
Shoog is offline Shoog
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Bootstrap usually means an effective component value (usually capacitance) is multiplied or reduced by the amplification of the active device doing the bootstrapping.

Shoog
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Old 26th January 2016, 05:10 PM   #8
smoking-amp is offline smoking-amp  United States
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Bootstrap is likely referring to source follower mode.

Take a look at this old standby:
IXYS IXTU 01N100
http://www.mouser.com/ds/2/205/98812-72889.pdf
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Old 26th January 2016, 06:19 PM   #9
Wavebourn is offline Wavebourn  United States
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Quote:
Originally Posted by Hearinspace View Post
OK, that's something you taught me a few years ago but I couldn't hear the tone of voice in the first "however" and needed to check.

I still don't really understand the use of "Bootstrap". It seems to mean a few different things. Th word suggests pulling something up, though it often appears in ways that aren't that.
Yes, source is pulling the "lower leg" of the "capacitor" up, so if the follower was ideal, the resulting capacitance would be zero.
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Old 27th January 2016, 02:08 AM   #10
Hearinspace is offline Hearinspace  Canada
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Infineon IPA50R140CP N-channel mosfet
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Originally Posted by Wavebourn View Post
Yes, source is pulling the "lower leg" of the "capacitor" up, so if the follower was ideal, the resulting capacitance would be zero.
Were talking about the capacitor that's below the resistor in series with the source in a Tubelab Powerdrive circuit? If so, how does it do that?
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