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| Hello, I was wondering how could a common collector bipolar transistor saturate in DC, if it uses the same power supply for colector and base. In saturation, Vbc = 0.5V, but if common collector, Vc = Vcc, and base cannont raise above that, so... what happends if so much current enters the base (enough to saturate the transistor)?? | |
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| Common collector can't saturate unless the collector-base junction is forward-biased. If collector=vcc, then the only way it can saturate is if the base is greater than vcc (by about 0.5V, as you said). | |
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| So, what happends if you raise base current so much?? Soon or later Ic = Hfe*Ib won't be able to be reach and that would mean saturation?? What would happend then?? | |
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| I know. But imagine you drive base with voltage source, and you make the base resistor very low. then, current trough base will be greater as you lower the resistor... | |
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Here is a quote from Wikipedia: Quote:
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| Here is a simple simulation, with the DC operating point calculated. I put GND on the collector for simplicity. Here is the operating point: Code: --- Operating Point --- V(b): -3.11236e-012 voltage V(e): -0.6535 voltage V(vee): -10 voltage Ic(Q1): 0.000931538 device_current Ib(Q1): 3.11236e-006 device_current Ie(Q1): -0.00093465 device_current I(R2): 3.11236e-006 device_current I(R1): 0.00093465 device_current I(V1): -0.00093465 device_current | |
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| Great! thank you very much | |
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| This is why you don't generally use a transistor in that way, for a saturated switch you would use common emitter, not common collector. It's also why push-pull audio amplifiers commonly use boot-strapping, in order to get sufficient base current for the top output transistor (usually the top one in most designs). | |
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