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Transistor calculations help ?

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curry87

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Hi i have some questions about the two different configurations of transistor in the pic.

I have the following figures for a 200 hfe transistor which was run in a simulator.

[A]
ve = 7.3v
vc = 8v
vb = 8v
Ib = .06ma
Ic = 13.2ma
Ie = 13.2ma


ve = 0.3v
vc = 0.4v
vb = 0.8v
Ib = 7.32ma
Ic = 14ma
Ie = 27ma


How does the sim calculate the following:
How is the vc calculated in both examples ?
How is the vb calculated in both examples
How is the ve calculated in both examples ?
How do i calculate the vcc voltage with these figures ?
How do i calculate the emitter or collector resistor load ?


Thanks
 

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In case A we have emitter follower configuration.
So from KVL
Vcc = Ib*RB + Vbe + Ie*Re

Ie = Ib + Ic = Ib + hfe*Ib = (hfe+1)*Ib

Vcc = Ib*Rb + Vbe + (hfe+1)*Ib*Re
then
Ib = (Vcc - Vbe ) / (Rb + (hfe+1) ) = (8.1V - 0.7V) / (1KΩ + 201*560Ω) = 7.4V/113.56KΩ = 65.16uA

Ie = (hfe+1)*Ib = 13.09mA
Ic = hfe * Ib = 13mA


For circuit B we can write

Vcc = Ib*Rb + Vbe

Ib = ( Vcc - Vbe) / Rb = (8.1V - 0.7V) /1K = 7.4mA
Ic = hfe * Ib = 1.48A and Vce = Vcc - Ic*Rc = -820V (off course this is impassible)
This means that BJT in case B is in saturation and Ic = hfe*Ib don't hold anymore.
Now we can only apply KVL and Ie = Ib+Ic
So Ib = 7.4mA
but for Ic
Vcc = Ic*Rc + Vce
Ic_max = (Vcc - Vce_sat)/Rc = (8.1V - 0.2V)/560Ω= 14.1mA
and Ie = 7.4mA + 14.1mA = 21.5mA
 

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Thanks for the clear answer i still have some questions to ask.

How do i calculate the RB, RC, RE ?


Is the max safe BJT base current 1/10 of the max safe collector current usually?


Does the transistor cause a 0.7 voltage drop between c and e or is it just the base emitter junction ?


For a transistor with 100hfe to switch a load of 50ma @9v on its collector on what should the base current and voltage be to do the following:


To fully saturate the base emitter junction ?
Put the transistor just enough in the active region to prevent overheating ?
 
Last edited:
When a transistor is turned on hard it is said to saturate. The maximum saturation voltage loss between its collector and emitter is listed on the transistor's datasheet which might be 0.3V and the typical saturation voltage loss is shown on a graph and might be 0.08V. The numbers are on the datasheet for a 2N3904 transistor at 50mA.

The hFE of a transistor is used only when the transistor is a linear amplifier with plenty of collector to emitter voltage.
When a transistor is used as a switch then hFE is not used, the datasheet shows the maximum saturation voltage loss when the base current is 1/10th the collector current.
 
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