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Driven switch body. (what about greater than one bootstrap)

Nice. Take no prisoners over-the-top everywhere.

Reply to
jlarkin
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I don't understand the drawing. Where does the Q9 BF862 JFET's source current go?

--
 Thanks, 
    - Win
Reply to
Winfield Hill

This is a version they sent me to check--Q6 is the tail current source for Q9, but it's drawn upside down. (My output is paper schematics plus BOMs, so their layout guy did the schematic capture.)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

I had to change K1 to a 9-V coil part, because it didn't pull in reliably with a 12 V coil. Otherwise it worked fine. I still have one someplace.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

Normally a TIA's output voltage is relative to the summing-junction return, FGND in this case, but here it's relative to FGND - Vgs of the JFET, where Vgs is a poorly-defined value. But you added a 2nd BF862 to match, trying to bias to Vgs=0 (for Idss = 10 to 25mA). Was this a design where you only needed to know relative output changes?

--
 Thanks, 
    - Win
Reply to
Winfield Hill

2nd

FGND is "floating ground". The probe tip has to have a bias voltage (AC and /or DC), and the simplest approach was to float the whole front end.

Cheers

Phil Hobbs

Reply to
pcdhobbs

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