Transistor base voltage calculation

[marz]

TL;DR Summary

I am using voltage divider to find base voltage of transistor. I am getting two different results in two different circuits.

Hello,
I am trying to find base voltage. In attached picture (using Multisim) there are two separate circuits. Here is how I am doing the voltage divider:
Circuit on right:
Vb = (Vcc - .7)*R6/(R6 + R5) = 3.2V which matches Multisim result. No problem here.
Circuit on left:
Vb = (Vcc - .7)*R1/(R1 + R2) = 1.99V. Multisim giving result = 1.5V. My question is why ?
I checked with another simulator, and got same results as Multisim.
Am I doing something wrong ?
Thanks for any help.

 

[berkeman]

What are the emitter voltages in the two circuits?

 

[marz]

What are the emitter voltages in the two circuits?

Emitter voltgaes:
The circuit on Right: Ve = 2.45V
The circuit on Left: Ve = .7V

 

[LvW]

In both cases, your calculation is not correct.
Why do you tink that the calculation for the circuit on the right would be in accordance with multisim?
Emitter voltage: Ve=2.45 V (according to your information)
Base voltage: Vb=Ve+0.7 (your value); Vb=3.15V.
However, your calculation gives 3.2V.
Something wrong? YES !
It is wrong to start the calculation with (Vcc-0.7)V.
I cannot see where this difference is coming from.
The only thing you know is that Ve=(Vb-0.7).

For a calculation as exact as possible you need the base current (resp. the factor B=Ic/Ib).
In case you have no further information, you must use a "good guess": B=100...200.

 

[Baluncore]

Solve the circuits by successive approximation. Each trial will bring it closer to the correct values.

You must allow for reduction in base voltage due to the base current flowing through the base bias circuit. The thevenin resistance of the base bias circuit is Rt. The simple divider voltage is Vt.

Beta is specified as 200A/A.
The base voltage of 0.7 volts is approximate. It could be computed.

What follows is a quick way of approximating the bias conditions.

For the first circuit it gives Vb = 1.451 volt, compare with model 1.5015 volt.
For the second circuit it gives Vb = 3.228 volt, compare with model 3.2324 volt.

' FreeBASIC
' first circuit
Dim As Double R1 = 220e3, R2 = 47e3, R3 = 510, R4 = 220, Rt
Dim As Double Vs = 12, Vt, beta = 200
Dim As Double Vb, Ve, Vc
Dim As Double Ie, Ib, Ic

Rt = 1 / ( 1 / R1 + 1 / R2 )    ' thevenin resistance of base bias circuit
Vt = Vs * R2 / ( R1 + R2 )      ' voltage on base due to divider
Vb = Vt
For trial As Integer = 1 To 100
    Ve = Vb - 0.7
    Ie = Ve / R4
    Ib = Ie / 200
    Vb = Vt - ( Ib * Rt )
Next trial
Vc = Vs - ( Ie * R3 )

Print Vb    '  1.451 volt
Print Vc    ' 10.259 volt, so not saturated

' second circuit

R1 = 50e3   ' new values, reuse the old names
R2 = 10e3
R3 = 3e3
R4 = 1e3

Vs = 20
beta = 200

Vt = Vs * R2 / ( R1 + R2 )      ' voltage on base due to divider
Rt = 1 / ( 1 / R1 + 1 / R2 )    ' thevenin resistance of base bias circuit
Vb = Vt
For trial As Integer = 1 To 20
    Ve = Vb - 0.7
    Ie = Ve / R4
    Ib = Ie / 200
    Vb = Vt - ( Ib * Rt )
Next trial
Vc = Vs - ( Ie * R3 )

Print Vb    '  3.228 volt
Print Vc    ' 12.416 volt, so not saturated

 

[LvW]

Solve the circuits by successive approximation. Each trial will bring it closer to the correct values.

OK, that´s possible - nevertheless, such a stepwise procedure is not necessary. Based on certain assumptions for B (e.g. 100 or 200) and Vbe (e.g. 0.7V) it is, of course, possible to analyze the circuit within one calculation step.

 

[Baluncore]

... it is, of course, possible to analyze the circuit within one calculation step.

I can see how a couple of simultaneous equations could solve it.
Can you write out a one step process ?

 

[LvW]

We can reduce the task simply to a system of two equations for two unknown quantities:
* Base voltage Vb=Vcc-I1R1=Vcc-R1(I2+Ib)=Vcc-[(Vb/R2)+Ib]R1
* Emitter voltage Ve=IeR4=Ib(B+1)R4=(Vb-0.7V).

Now we have two equations for the unknown values Vb and Ib (with estimated values for B and Vbe)

Comment: As an alternative we could apply the superposition theorem. At first, we find the voltage at the base Vb1 caused by the supply voltage Vcc only (set Vbe=0) and - as a second step - we calculate the portion Vb2 caused by a voltage source of Vbe=0.7 volts (set Vcc=0).
Then we have Vb=Vb1+Vb2.

 

[alan123hk]

I hope to provide a more intuitive, concise and easy-to-understand derivation method for your reference.

 

[DaveE]

This would be a great case for substituting the "T" model for BJTs (the simplest DC model), as shown below. Then you follow what @alan123hk did above. It's also a good place to use superposition. The base voltage will be the sum of the 2 cases: Vb with Vbe=0, and Vb with Vcc=0, which allows you to use the simple voltage divider rule (twice).