I'm debugging a failed board for a humidity sensing extractor fan. RP1 is the humidity sensor. Q1 and Q2 (SOT-23) look like some sort of dual current mirror, but I don't understand this configuration or how it works. (The other side of R4 is connected to one of the output pins of a MC14049 Hex Buffer).

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Here's a picture of the part of the board. Ground plane at bottom. Q1 and Q2 bottom centre (marked KZG).

There are no through-holes on the board that I can see. Resistance between bases and ground appears to be infinite.

enter image description here

  • \$\begingroup\$ Are you sure this schematic is correct? The bases are left floating... is there supposed to be a ground in the net between Q1 and Q2? \$\endgroup\$
    – Lundin
    Feb 17, 2022 at 11:59
  • \$\begingroup\$ This does not make any sense, humidity or otherwise. Check the connections carefully and show us some photos of the board. Btw, humidity sensing failures are the sensor itself usually, so you should eliminate that possibility first before mucking with the rest of the circuit and possibly introducing new faults. \$\endgroup\$ Feb 17, 2022 at 12:02
  • \$\begingroup\$ Did you remove the component to the right of the IC (in the photo), or did it fall off? \$\endgroup\$ Feb 17, 2022 at 12:25
  • \$\begingroup\$ I haven't removed anything :-) There are two pin holes above that for CN4 - a connector that isn't implemented on this iteration/version of the board. I don't know if those should be bridged by something if CN4 is not connected - I've not got that far yet. \$\endgroup\$
    – Vince
    Feb 17, 2022 at 12:35
  • \$\begingroup\$ Is it back-to-back zener diodes in a sot-23 package by any chance? \$\endgroup\$
    – Buck8pe
    Feb 17, 2022 at 12:46

1 Answer 1


I looked up the case code and package ("KZG" and "SOT-23") and found this. It looks like it's a single 3.9V zener diode. Here's the pinout:

+--'--+    1 Anode
|_____|    2 NC
 '   '     3 Cathode
 1   2

The circuit has a pair of them cathode-to-cathode. This can be used to clamp/clip both the positive and negative sides of an AC waveform.

  • \$\begingroup\$ Thanks, Buck8pe also realised this, so I'm marking this answer correct, and handing him a credit as well :-) \$\endgroup\$
    – Vince
    Feb 17, 2022 at 13:01

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