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Question, transistor

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ericb

New Member
Have a question

Heres the situation

Its an NPN bipolar transistor



I want to calculate I_BQ, I_CQ and U_CEQ

I know R_1, R_E, R_C, U_BEQ,E and B=100

My first question is, how do I incorporate U_BEQ in my calculations?

Because my strategy is to calculate I_EQ first, then I can use the two equations in the picture to calculate I_B and I_C and then U_CEQ follows by using KVL.

Thanks for the help!
 
Last edited:

audioguru

Well-Known Member
Most Helpful Member
The spec's on the datasheet for the transistor and the currents determine VBE.
It is about 0.5V at low currents for a little transistor and is 1.5V at high currents for a power transistor.
 

MrAl

Well-Known Member
Most Helpful Member
Hi there,


One thing to keep in mind when dealing with real world transistor
circuits like this is that it is not always possible to calculate
exact voltages like V_CE because that will depend on the beta, and
the beta is often a spread like 50 to 150 or something like that.
This means when you do the analysis you first use maybe 100 to get
the circuit working the way you want it too, and then check to
make sure the other values like 50 and 150 dont cause the circuit
steady state to vary too much away from what you started with,
or at least that the circuit will still function for the given
application that way.

For the V_BE, often a guess is made of 0.6 to 0.7 volts and from
there on in V_BE is taken to be a voltage source (yes like a battery)
of maybe 0.65 volts or so. This simplifies the circuit right away.
Also, the terminals C and E are taken to be a current generator
who's current is determined by the base current I_B. Taking all
this together you end up with equations that can be solved using
the usual methods normally used in algebra.

For example, knowing (or assuming) a base emitter voltage of 0.6v
means the base current is:
I_B=(Vcc-V_BE-I_E*R_E)/R1
where Vcc is 6v in your circuit.

You can then write the equation for I_C, then perhaps use substitution
or whatever means you want to solve the two equations.

Finally, you should check the circuit using various values for beta
and for V_BE to make sure it will work even if there is some temperature
variation.
 

barf

New Member
If you know V_BEQ then I_BQ = (E-V_BEQ)/R1

then I_CQ = B*I_BQ

then the voltage drop across RC would be I_CQ*RC

then since I_CQ is in series with I_EQ, the current is exactly the same so I_EQ = I_CQ

then the voltage drop across RE would be I_EQ*RE

and finally since you now know the two resistor (RC & RE) voltage drops, V_CEQ = E - VRC - VRE

Barf
 

Roff

Well-Known Member
If you know V_BEQ then I_BQ = (E-V_BEQ)/R1

then I_CQ = B*I_BQ
You have ignored the voltage on the emitter.

then the voltage drop across RC would be I_CQ*RC

then since I_CQ is in series with I_EQ, the current is exactly the same so I_EQ = I_CQ
I_EQ I_CQ. You have not accounted for base current.

then the voltage drop across RE would be I_EQ*RE

and finally since you now know the two resistor (RC & RE) voltage drops, V_CEQ = E - VRC - VRE

Barf
You need to rethink this.
 

barf

New Member
You have ignored the voltage on the emitter.

I_EQ I_CQ. You have not accounted for base current.

You need to rethink this.


He states he knows the base voltage (V_BEQ), so he can easily calculate base current. He also states he knows the transistor gain (B), so he can easily calculate collector current, and collector voltage.

Oops, you are right - Emitter current is the sum of base and emitter current. He can now CORRECTLY calculate voltage at emitter.

Barf
 

Roff

Well-Known Member
He states he knows the base voltage (V_BEQ), so he can easily calculate base current. He also states he knows the transistor gain (B), so he can easily calculate collector current, and collector voltage.

Oops, you are right - Emitter current is the sum of base and emitter current. He can now CORRECTLY calculate voltage at emitter.

Barf
V_BEQ is not base voltage - it is base-emitter voltage.
 

barf

New Member
V_BEQ is not base voltage - it is base-emitter voltage.
Yikes - my bad.

Okay so we have to assume upon startup that transistor is off, meaning that emitter voltage is initially 0v. Now since we know the base-emitter voltage, we can calculate base current and collector current. Emitter current is the sum of base and collector currents, so now we can calculate emitter voltage.

Base voltage now changes to emitter voltage plus base-emitter voltage, so base current also changes along with collector current.... and so on.

Barf
 

Roff

Well-Known Member
You need to find Vb. You can solve for Vb with these 4 equations:

Vb=Ve+Vbe
Ve=IeRe+Vbe
Ie=(β+1)Ib
Ib=(Vcc-Vb)/Rb (Our OP called it R1).

Everything else is a piece of cake.
 
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