Calculating base voltage of an npn transistor

  • Thread starter jearls74
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  • #1
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i have been trying to solve this problem for about 14 hours and have gotten nowhere, im trying to figure out the voltage going into the base of an npn transistor, im trying to build a circuit where i can take the current from the battery and build a circiut from 7npn transistors in a common emitter mode and a voltage regulator, up to 20 amps of steady current at output. i cant seem to figure out the the voltage going into the base of the first transistor. i know it needs to be atleast .7 volts to turn on the transistor. ive used ohms law several times and only end up with what i started with, how do you figure the voltage drop through the resistor? how many volts are dropped when 12 volts are passed through a 470 ohm resistor? the 470 ohm resistor is connected to the base. any explanations would be appreciated. thanks
 

Answers and Replies

  • #2
Averagesupernova
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You need to post a schematic.
 
  • #3
NoTime
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how many volts are dropped when 12 volts are passed through a 470 ohm resistor?
Since you already gave the answer, it's 11.3v to good approximation, assuming grounded emitter.

A bipolar transistor is a current device not a voltage device.
Ic = Ib Hfe over the linear range.

Hope that helps some.
 

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