Current Mirror Explanation Needed.
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 7th July 2012, 05:17 PM #11 tsiros   diyAudio Member     Join Date: Jun 2012 Location: Patra, Greece 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
 7th July 2012, 05:51 PM #12 DF96   diyAudio Member   Join Date: May 2007 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?
jan.didden
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Join Date: May 2002
Location: The great city of Turnhout, BE
Quote:
 Originally Posted by tsiros 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?

jan
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 7th July 2012, 06:18 PM #14 tsiros   diyAudio Member     Join Date: Jun 2012 Location: Patra, Greece i will experiment and see how it goes thank you all for your patience, effort and time!
 7th July 2012, 06:20 PM #15 jan.didden   diyAudio Member     Join Date: May 2002 Location: The great city of Turnhout, BE What will be the voltage across the 1k you think? jan __________________ Music is dither to the brain; lets me think below the usual chaos - me Linear Audio Vol 13 is out! Check out my Autoranger and SilentSwitcher
 7th July 2012, 07:05 PM #16 tsiros   diyAudio Member     Join Date: Jun 2012 Location: Patra, Greece 15 V minus the Vbc drop?
 7th July 2012, 07:24 PM #17 jan.didden   diyAudio Member     Join Date: May 2002 Location: The great city of Turnhout, BE 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 ;-) __________________ Music is dither to the brain; lets me think below the usual chaos - me Linear Audio Vol 13 is out! Check out my Autoranger and SilentSwitcher Last edited by jan.didden; 7th July 2012 at 07:26 PM.
 7th July 2012, 07:26 PM #18 tsiros   diyAudio Member     Join Date: Jun 2012 Location: Patra, Greece 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.
 7th July 2012, 07:32 PM #19 jan.didden   diyAudio Member     Join Date: May 2002 Location: The great city of Turnhout, BE 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? __________________ Music is dither to the brain; lets me think below the usual chaos - me Linear Audio Vol 13 is out! Check out my Autoranger and SilentSwitcher
 7th July 2012, 07:36 PM #20 tsiros   diyAudio Member     Join Date: Jun 2012 Location: Patra, Greece doesn't have anywhere else to go, to ground. is the transistor operating in saturation right now?

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