ElectronicsI_L23 - Lecture 23 Example 4 Now we will look at...

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Lecture 23
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Example 4 • Now we will look at the same circuit, but with a higher base voltage of 6 volts. • Also, we will assume the beta to be 50 or larger. • Once again we can determine that the emitter junction is forward biased. • This means the emitter voltage is 6- 0.7=5.3V • Thus the current through the emitter is 5.3V/3.3k =1.6mA
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Example 4 Now we want to calculate the collector current by once again utilizing alpha: The collector voltage from this is: Now we have a problem, this collector voltage will clearly put the CBJ into forward bias, so our assumption that the BJT was in active mode is incorrect We could also have done a quick and dirty approximation by taking the collector current to be equal to the emitter current. – We can see that alpha is very close to 1 anyway. – The difference between the emitter and collector current is very small, so either way we would have determined the BJT was in saturation. 98 . 0 1 min min min = + = β α mA mA I I E C 57 . 1 6 . 1 98 . 0 = × = = α V R I V C C C 63 . 2 10 = = V V V V B C CB 37 . 3 = =
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Example 4 • Armed with this knowledge, we now must solve the circuit assuming it is in saturation – This is the only other state it could be in because the EBJ is forward biased. • The emitter voltage is still correct
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This note was uploaded on 02/13/2012 for the course PHYSICS 16.365 taught by Professor Staff during the Summer '10 term at UMass Lowell.

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ElectronicsI_L23 - Lecture 23 Example 4 Now we will look at...

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