# Finding the component value of PNP transistor

This is my circuit

simulate this circuit – Schematic created using CircuitLab

This is the way I tackled this problem:

$$U_{e2} = 12 - I_e \cdot R_e = 12 - (2.02 \cdot 2) = 7.96 \text{ V}$$

$$I_e = \left( 1 + \beta \right) \left( \frac{ I_C }{ \beta } \right) = \left( 1 + 100 \right) \left( \frac{ 2 \text{ mA} }{ 100 } \right) = 2.02 \text{ mA}$$

This is where I'm struggling:
$$I_{be} = I_e - I_c = (2.02-2) \text{ mA} \rightarrow 0.02 \text{ mA}$$

I don't know how to find $$\U_{be}\$$ so I can find the resistance of $$\R_{be}\$$

I have been trying to find a tutorial and some help from several articles.

• What problem??? – Andy aka Feb 9 at 18:39
• @Andy aka, finding Ube so I have then find Rbe – be1995 Feb 9 at 18:44
• Why on earth did you write the question as an image? It's not searchable. – pipe Feb 9 at 18:45
• @pipe I have done it in word becuase I dont know how to paste equations here. sorry – be1995 Feb 9 at 18:48
• Ube=0.60V @Ic=1mA so use that – Tony Stewart Sunnyskyguy EE75 Feb 9 at 18:50

This configuration is very unstable with only 1 pull-up resistor for bias.

You can never predict Ve using assumptions of hFE due to the wide hFE tolerance and total dependence of hFE on bias voltage.

simulate this circuit – Schematic created using CircuitLab

Vbe will be around to 0.65V ± 0.05 @ 2mA, if you look up many datasheets shown as typical values in the curves section @ 25'C, so it can be neglected as the uncertainty is dominated by hFE tolerance error.

Checking from a simulation.
If hFE=100 Vbe=613mV Vcc=12V, Ie=2mA, Re=2k, Rb=370k
Using formula Schematic above: Vcc-Ve = 4V = 2k/(2k+Rb/100)*12V
For Rb=370k, Vcc-Ve= 2k/(2k+3.7k)*12V = 4.2V .. close enough
but if we apply tolerance error hFE=150 Ve= 2k/(2k+2.47k)*12V= 5.36V ... wide tolerance error.
For details on Vbe vs Ic Vbe vs Ic characteristics of NPN transistor at different Vce in active region

• The teacher usually buys thousands of transistors and tests each one hoping to find one with Vbe= 0.7V and a beta of 100. – Audioguru Feb 9 at 19:48

Vbe is a range of voltages depending on the transistor type, its current and on its part number. A transistor type is germanium (Vbe= 0.2V - 0.4V) or silicon (Vbe= 0.6V - 0.75V). Your circuit has a transistor with no part number so you cannot lookup its range of Vbe.

You assumed a beta of 100 which is wrong because it is also a range that could be from 20 to 600. Its datasheet would show its range of beta.

• in this case beta is 100 and lets say the transistor is 0.7V – be1995 Feb 9 at 18:57
• I like the superbeta transistors with hFE > 2000 – Tony Stewart Sunnyskyguy EE75 Feb 9 at 20:00