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schematic

simulate this circuit – Schematic created using CircuitLab

In this configuration of 2 NPNs, how is V_OUT following V_IN? Ignoring recombination current:

  • Assuming conventional current, current flows from Node_A to Q1 collector and Q2 base, but is somehow replenished by a voltage source that we can't see.
  • Assuming electron flow, current from Q1 collector and Q2 base are drained in Node_A which is a constant voltage source or ground that we can't see.

Now if we were to assume recombination current, we have an exception. But even after reading through some Recombination Current articles, I still can't see why this would be behaving like this.

Can somebody give an explanation?

I'm sure this isn't Reverse Mode, as V_IN causes Q1 to be forward biased and the emitter of Q2 is pulled to ground.

EDIT:

Dave answered that Q1 is acting like a dual diode. So I thought of testing if it was with this simplification:

schematic

simulate this circuit

I've actually had to ramp up the resistor to ground to 100k from the previous 10k to obtain a full swing. The collector still follows the emitter. Perhaps there's resistance at the emitter, so there's feedback in the collector? I checked the Gummel-Poon model and it seems to agree with this assertion. Just adding this little bit.

EDIT:

Something that I want to correct is that the the base-emitter of Q1 forms a closed loop of voltage source V2. The base of Q2 (on the first figure) and 100k R2 (of the second) only taps onto this. So it's following.

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  • \$\begingroup\$ The reason you had to use a much larger value resistor for R2 is that in your 1st circuit, the effective value of R2 'seen' at the base 'looking through' Q2 is 'amplified' by Q2's gain. \$\endgroup\$
    – brhans
    Apr 18, 2016 at 16:35

1 Answer 1

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In this configuration, Q1 is functioning more as a dual diode than as an active transistor. As such, current flows from the base to both the emitter and the collector. Since both diode junctions have about the same forward voltage drop, the voltage at the collector follows the voltage at the emitter fairly closely.

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  • \$\begingroup\$ Thanks for the reply. Never considered that because I thought the CB Breakdown Voltage needs to be reached first. \$\endgroup\$
    – Majin_Boo
    Apr 18, 2016 at 13:53
  • \$\begingroup\$ So... No recombination current going on at all, huh? I don't know why I came up with that explanation when recombination only strictly happens inside the component. \$\endgroup\$
    – Majin_Boo
    Apr 18, 2016 at 13:56

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