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i was solving questions on bjt, and i came across a question where i had to find voltage across emitter and collector and the answer was 2V, but i have read it's approx. 0.6v across both emitter base and collector base, then how does it amounts to 2v ?

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  • \$\begingroup\$ Are you sure that it's 0.6V across collector-base? \$\endgroup\$ – Chupacabras Aug 15 '16 at 16:33
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    \$\begingroup\$ Please use proper grammar and punctuation, it makes your question readable and it makes you look smart. Thanks \$\endgroup\$ – Voltage Spike Aug 15 '16 at 20:01
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Your answer was for a specific load (voltage and resistance) and drive current. Depending on the circuit, the collector-base voltage can go from the collector breakdown voltage to about -0.6 volts or less. The first level occurs when the input base current is zero and the collector is at breakdown, and the second occurs when the transistor is in hard saturation, with a Vce on the order of 0.1 volts. Since in some high base-drive situations the base-emitter voltage can get larger than the nominal 0.7 volts, the voltage difference can be greater than 0.6 volts.

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When a transistor is operating in its linear region, the base-emitter junction is biased on so it will develop a voltage of about 0.6-0.7 volts. However, the base-collector junction is normally biased off. Thus the collector-emitter voltage can be considerably higher than 2 junction drops and depends on the external circuitry.

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The emitter-base junction is forward-biased (about 0.6V) but the collector-base junction is reverse-biased in normal linear operation.

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