1
\$\begingroup\$

I'm a grad student in physics that hasn't looked at circuits since early in my undergrad. I was recently handed a voltage divider that my group used to bias a micro-channel plate detector stack as I'll need to build a new one for a new system.

What I'm confused about is the purpose of some surprise resistors (R1 and R4 on the included schematic) as I don't understand why they would need to be included. Any insight would be much appreciated as the person who built this graduated 8 years ago.

As for the bit with the capacitors in the bottom right along with the output labeled 'Sig', that was to read the bias off of the rear micro-channel plate capacitively to provide timing information on a detector event.

R1=1Mohm

R2=4Mohm

R3=470kohm

R4=150kohm

All capacitors are 4.7uF

Vin is a DC voltage that was biased at -2.5kV.

A microchannel plate detector is a radiation detector with many small channels typically constructed out of a lead glass that function as an electron multiplier. When an incident ion/electron strikes the wall of a channel, it kicks up secondary electrons. Those electrons then hit the wall further down the channel leading to a multiplicative avalanche of electrons that produces a readable voltage pulse on the anode. The resistance across the plate/plates is in the 10^9 ohm range, so as far as I understand it, the resistance of the load between each node (Vfront> Vback>Vanode) is approximately that much.

enter image description here

\$\endgroup\$
2
  • 2
    \$\begingroup\$ You'll need to include more of that circuit if you want an answer. For instance, you should add the circuits that are currently not shown that connect to the nodes on the right. You should also give details of Vin. I also see that your capacitors are unspecified; they need to have values and reference IDs. REASON: nobody will understand what this is: a micro-channel plate detector stack. \$\endgroup\$
    – Andy aka
    Mar 16, 2021 at 17:57
  • \$\begingroup\$ "Vin is a DC voltage that was biased at -2.5kV" - then your battery symbol is backwards. \$\endgroup\$ Mar 16, 2021 at 18:46

1 Answer 1

3
\$\begingroup\$

Looks similar to a PMT (photomultiplier tube) circuit with a series of dynodes attached to a voltage divider. Image from here

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.