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Old 7th July 2012, 05:17 PM   #11
tsiros is offline tsiros  Greece
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well, the current flowing into the p doped semi between the n doped semi, controls the flow of current, i'd rather not get into the QM behind it, but i never thought it can work in reverse?

how can i impose a current?

this can't work with only one transistor, right? not the current mirror, the appearance of a voltage at the base when forcing a current through C/E
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Old 7th July 2012, 05:51 PM   #12
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The voltage appears at the base because it is connected to the collector. A BJT is not a superconductor, so if a current flows it will cause a voltage drop. The voltage drop is across both base-emitter and collector-emitter as they are connected together.

You appear to be thinking about a BJT in current terms. As I said, you will never understand current mirrors (or many other BJT circuits) if you think in this way. Think in terms of applying a voltage to base-emitter.

Imposing a current is easy (approximately). You just need a battery and a resistor. If the resistor is the dominant resistance in the circuit then you have (approximate) control of the current. Remember, Ohm's Law says nothing about causality: does voltage cause current or current cause voltage?
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Old 7th July 2012, 06:09 PM   #13
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Quote:
Originally Posted by tsiros View Post
well, the current flowing into the p doped semi between the n doped semi, controls the flow of current, i'd rather not get into the QM behind it, but i never thought it can work in reverse?[snip]
If you have a transistor, B connected to C, the E grounded and the C/B through 1k to +15V, what do you think will happen?

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Old 7th July 2012, 06:18 PM   #14
tsiros is offline tsiros  Greece
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i will experiment and see how it goes

thank you all for your patience, effort and time!
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Old 7th July 2012, 06:20 PM   #15
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What will be the voltage across the 1k you think?

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Old 7th July 2012, 07:05 PM   #16
tsiros is offline tsiros  Greece
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15 V minus the Vbc drop?
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Old 7th July 2012, 07:24 PM   #17
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No, B is connected to C, remember? So Vbc = 0...
But you're close. You really must draw it, then it's clear. Remember, pencil and paper ;-)
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Last edited by jan.didden; 7th July 2012 at 07:26 PM.
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Old 7th July 2012, 07:26 PM   #18
tsiros is offline tsiros  Greece
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npn, right?

15 V minus the Vbe drop, usually 0.6 V

edit: i really should stop relying on mental images for things with which i haven't got much experience :/

Last edited by tsiros; 7th July 2012 at 07:30 PM.
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Old 7th July 2012, 07:32 PM   #19
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You're doing fine. So we have +15V on one side and Vbe (0.6V) on the other side. Ohms law says there will be 14.4mA through that 1k.
Where does that go?
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Old 7th July 2012, 07:36 PM   #20
tsiros is offline tsiros  Greece
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doesn't have anywhere else to go, to ground.

is the transistor operating in saturation right now?
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