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Hi, i need help understanding how this NPN,PNP transistor work because it make me really confuse from what i'm understand. The circuit is from laptop motherboard schematics. The transistor is IMD2AT108 . Datasheet is here : http://rohmfs.rohm.com/en/products/databook/datasheet/discrete/transistor/digital/emd2t2r-e.pdf

  1. Voltage on pin 1&2 is 19.36V
  2. Voltage on pin 3 is 19.46V
  3. Pin 5 & 6 is connected to ground.
  4. Pin 4 is 3.25V

My question is :

  1. Please explain why voltage on pin 1&2 is 19.36V ?
  2. Suppose for PNP transistor in the circuit is in situration mode because Vbe is -ve but from measurement the PNP transistor is in cut-off mode. Please explain.

Thanks all for your help. Ryan

  • \$\begingroup\$ As you say, the PNP transistor is in cut-off mode, in other words, pin 1&2 is floating somewhere above \$19.46-V_{be}\$. It can be \$19.46 V\$, it can be \$18.86 V\$, but with all the pico and femto amps flowing through it settles at \$19.36 V\$. That's what happens when there's no pull-up or pull-down resistor. \$\endgroup\$ Jul 20, 2017 at 13:53
  • \$\begingroup\$ Look at the schematic OP provides. Each of the 2 transistors has resistor across EB to handle leakage, and a resistor in series from control-logic-input to the base to limit current. \$\endgroup\$ Jul 20, 2017 at 13:56
  • 2
    \$\begingroup\$ Looks like pin 5 isn't connected to ground, but to some logic level signal. When that goes high, the transistors will turn ON, putting somewhere over 19V on pin 4. \$\endgroup\$
    – user16324
    Jul 20, 2017 at 14:00

1 Answer 1


Both transistors are OFF.

The 19.36 volts (down from 19.46 volts) is the effect of your meter needing a small current from pin 1/2.


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