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let's consider an example of transistor amplifier:

enter image description here

Let's consider the case in which power supply is not connected (so between VCC and GND there is an open circuit). What will I read at the output terminal? 0V (i.e. Collector and Emitter are short circuited when the transistor is off) or Undefined Voltage (i.e. Collector and Emitter are separated by an open circuit when the transistor is on)?

I have also the same doubt about MOSFET transistors.

I'd say that, since the material between collector and emitter is a semiconductor, it acts like a "mean" resistor (not open circuit but also not short circuit). So I'd say that without power supply, the output of the amplifier is 0. Is this reasoning correct?

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    \$\begingroup\$ Did you try simulating the circuit to answer the question for yourself? \$\endgroup\$
    – The Photon
    Jun 22 '20 at 13:31
  • \$\begingroup\$ I did not. Probably I should have done it before asking, but I'm interested not only at the result, but always at the physical reason of that behavoiur \$\endgroup\$
    – Kinka-Byo
    Jun 22 '20 at 13:34
  • \$\begingroup\$ In MOSFET I can tell that you can change the gate voltage and turn it on/off but in case of BJT I believe it must be off, but the best way to make sure is to simulate it. \$\endgroup\$ Jun 22 '20 at 16:31
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The resistors RL, R1, and R2 connect the output terminals, so if the input signal amplitude is small you'll read 0 V DC between them. If the input is strong you may see a DC offset due to rectification of the input by the b-c junction.

You'll also see some pass-through of the input AC signal. If the input amplitude is high enough to turn on the BJT's b-e and b-c junctions then you'll also see harmonics of the input at the output.

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  • \$\begingroup\$ About RL, R1 and R2, why do they connect output terminals? In case power supply is absent, I'd say that VCC and GND are not connected \$\endgroup\$
    – Kinka-Byo
    Jun 22 '20 at 13:45
  • \$\begingroup\$ Look more closely. RL connects the output node to the Vcc node. R1 connects Vcc to the base node, and R2 connects the base node to the output reference node. \$\endgroup\$
    – The Photon
    Jun 22 '20 at 13:46
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Are you applying Vin still while Vcc is open? You'll measure a high-passed version of Vin through the resistive connection from the base to Vout (R1 + RL).

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  • \$\begingroup\$ Why will I measure that quantity, if VCC is isolated from GND? \$\endgroup\$
    – Kinka-Byo
    Jun 22 '20 at 13:45
  • \$\begingroup\$ This reads more like a brief question back to the OP, and is perhaps better put in the comments. Then, if you come up with a good final answer after some discussion, feel free to summarize your comments as an answer. \$\endgroup\$
    – P2000
    Jun 22 '20 at 19:03
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It will be 0V, due to the leakage current of the (N)PN junction (and, as The Photon remarks, the current through RL/R1/R1, although it would still be 0 without these)

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The best answer will be to simulate with VCC=0V and the likely range of input voltages and any potential signals from other sources on the output.

You have to recognise that, absent power, you have two diodes present on the input side, and one on the output side.

Potential effects from these include:

Rectification, non-linearity, distortion of input signals which may be shared with other amplifiers. I had to take care with an audio PPM (peak program meter) that when unpowered, it didn't distort signals on a broadcast circuit any more than 0.1%. In an RF environment, strong interfering signals may mix to produce spurious noises on other frequencies.

Rectification, non-linearity, distortion of externally applied signals on the output side (if any).

Non-linear (distorted) crosstalk between input and output.

If the input is at the 1 mV level these won't happen. But if it peaks anywhere close to 0.6V the circuit may behave in unexpected ways when unpowered.

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With VCC floating, there is a resistive path between input and output.

And the Cob further obliges the collector to track the base movement.

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