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In a common emitter of silicon BJT, .find the emitter current (IE) 1k! 5ko Ax= 100 E10 v 10 v 200 mA O 170 mA O zero 187.86 mA O

In a common emitter of silicon BJT, .find the emitter current (IE) 1k! 5ko Ax= 100 E10 v 10 v 200 mA O 170 mA O zero 187.86 mA O

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o)t AT gl S @ i B © U ®© ¢ H Asiacell infinity () In a common emitter of silicon BJT, * find the emitter current (IE) 1k0 fe=100 =10V 10v = 200mA O 170mA O zero O 187.86mA O A common emitter silicon BJT has V_CB=2.5V, thenV_CE is equal to
O + A Asiacell infinity O In a common emitter of silicon BJT, .find the emitter current (IE) 1k! 5kO { Pa= 100 10 v 200 mA 170 mA zero 187.86 mA A common emitter silicon BJT has V_CB=2.5 V, then V_CE is equal to 2.8 V O
Transcribed Image Text:O + A Asiacell infinity O In a common emitter of silicon BJT, .find the emitter current (IE) 1k! 5kO { Pa= 100 10 v 200 mA 170 mA zero 187.86 mA A common emitter silicon BJT has V_CB=2.5 V, then V_CE is equal to 2.8 V O
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