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Is it possible to saturate a transistor by applying a voltage on the base that is comming from a different source than the source of the voltages of the collector and the emmiter?

schematic

simulate this circuit – Schematic created using CircuitLab

Edit:

I wanted to start a desktop computer with an arduino that is powered by an adaptor. I solved this problem by connecting the 2 ground pins with each other.

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    \$\begingroup\$ What matters is the voltage of V1 referred to the circuit's own ground i.e. V2- on the schematic. \$\endgroup\$ – user_1818839 May 8 '16 at 12:17
  • \$\begingroup\$ Yes if the grounds are connected ofcourse! \$\endgroup\$ – Jasser May 8 '16 at 12:17
  • \$\begingroup\$ Yes, if it's of an appropriate level. There isn't really such a thing as a 'foreign' voltage. \$\endgroup\$ – user207421 May 8 '16 at 12:17
  • \$\begingroup\$ grounds are not connected \$\endgroup\$ – AXANO May 8 '16 at 12:18
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    \$\begingroup\$ Then, no you cannot saturate the transistor since you require a closed loop for the current to flow which is not present as the negative terminal of V1 is not connected to -ve terminal of V2. \$\endgroup\$ – Jasser May 8 '16 at 12:25
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Voltages don't "come from" someplace, they exist between two nodes. The notion of a "foreign" voltage is non-sensical. In the context of electronics, you can't talk about a voltage in absolute terms, unless it is implied what other node the voltage is relative to. If the circuit has a node labeled ground, then it is understood that voltages talked about as being absolute are really relative to ground. That's actually the main purpose of labeling a node as ground in a schematic.

Your circuit doesn't show any node as being the implied ground. However, from context and the way the schematic is drawn, we would usually assume that the implied 0 V reference is the negative lead of the power supply and the cathode of the LED. Forcing people to make assumptions like this is a bad engineering, so ground should really be clearly marked if you're going to talk about voltages on single nodes without explicitly mentioning the reference node.

The transistor only reacts to voltages between its pins. To turn on a transistor, you have to put enough voltage on B with respect to E so that sufficient base current flows. Since B-E looks like a diode to the outside circuit, it is usually better to specify the B-E current instead of the B-E voltage. The voltage will usually be 600-750 mV over the useful operating range, but the current will vary much more widely over that range.

The circuit as you show does nothing because the negative end of V1 is not connected to anything. You can't say that it's putting 5 V on the base of the transistor, since that voltage isn't referenced to anything in the transistor's circuit. With the negative end of V1 open, current can't flow thru V1, so it becomes irrelevant.

If you were to do this with a real power supply, it would have some capacitive coupling back to the power line and the transistor circuit. Connecting the positive end of V1 to the transistor base would couple some capacitively picked up noise onto the base that is ultimately referenced back to the transistor circuit. If the transistor were driving a speaker, you'd probably hear some hum. Unless you have a very unusual situation, the part of the capacitively coupled signal referenced back to the transistor circuit won't be strong enough to light the LED.

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  • \$\begingroup\$ Right on! Transistors are current driven. \$\endgroup\$ – Bradman175 May 8 '16 at 13:20
  • \$\begingroup\$ @Jasser You don't have to say anything, but you could read the text in the input box when writing comments. \$\endgroup\$ – pipe May 8 '16 at 14:59
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As Olin mentions, the only thing completing the circuit is capacitive coupling, which is a very high impedance ,specially at low frequencies.

However there's one case where it might matter, that is well worth mentioning. And that is where a huge DC voltage between the two independent circuits. Carry V1 across a nylon carpet (or rub it on a cat) and it may be at a potential 50,000 volts away from Q1's base.

Now the actual voltage of V1 itself doesn't matter, but the 50,000V difference connected to Q1 base, with the circuit completed by a few picofarads of capacitance, can be enough to ...

no, not light D1 ...

but destroy Q1. (Especially -50,000V which could reverse bias the base and break the base-emitter junction).

A bipolar transistor like a 2N3904 is fairly tough and may survive this treatment : a MOSFET or an expensive microcontroller may not.

So use electrostatic safe working techniques when assembling and testing circuits!

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