# analysis of BJT PNP type - why they can use voltage divider?

Hi everyone, I try to understand why some books used in a voltage divider to find the Base Voltage? or maybe the right question is HOW they can use voltage divider if there is base current? what I miss here? what is the best way if I want to find what is the Base Voltage? this is some example : https://www.eeweb.com/quizzes/voltage-divider-biased-pnp-transistor Thanks ahead.

You are correct that base current will result in deviation from the voltage divider equation. However, if the following condition is satisfied the deviation will be small.

$$\ R_1 || R_2 << (\beta + 1)R_E \$$

This can be derived by considering the Thevenin equivalent of the one-port network of the base bias resistors. The terminals of this one-port network are the base of the transistor and ground.

The Thevenin voltage is given by a voltage divider, as this is the voltage with the terminals open-circuited i.e. no base current.

$$\ V_{th} = V_{cc} * \frac{R_1}{R_1 + R_2}\$$

And the Thevenin resistance is the two resistors in parallel (they are in parallel if you zero the DC source). So the equivalent circuit looks like this

simulate this circuit – Schematic created using CircuitLab

Assuming the transistor is biased in its linear region we can write a KVL equation that gives (assuming I got my directions right)

$$\ V_{th} + V_{be} + I_b(R_1 || R_2 + R_E(\beta + 1)) = V_{cc} \$$

Rearranging for the base current gives

$$\ I_b = \frac{V_{cc} - V_{th} - V_{be}}{R_1 || R_2 + (\beta + 1)R_E}\$$

And the base voltage equals the emitter voltage minus a diode drop. The emitter voltage is

$$\ V_E = V_{cc} - (\beta + 1) I_b R_E\$$

So $$\ V_B = V_{cc} - (\beta + 1) I_b R_E - V_{be}\$$

Substituting

$$\ V_B = V_{cc} - (\beta + 1) R_E * \frac{V_{cc} - V_{th} - V_{be}}{R_1 || R_2 + (\beta + 1)R_E} - V_{be}\$$

Now if $$\ R_1 || R_2 << (\beta + 1)R_E\$$ then,

$$\ V_B = V_{cc} - (V_{cc} - V_{th} - V_{be}) - V_{be}\$$

Which simplifies to

$$\ V_B = V_{th}\$$!

So provided that condition is satisfied the base voltage is the same as if it was a voltage divider. It is almost always a good idea to satisfy this condition as it makes design easier and better as you will have trouble making a real circuit have the bias point you want without it.

By looking at the circuit you can see that the base divider is reasonably well-designed— by that I mean that the voltage across the emitter resistor is not too small in relation to a typical Vbe and the base current is a small fraction (< 10%) of the divider current (given $$\\beta =150\$$ and assuming the transistor is not saturated, we'll check that later).

So you can either write an equation and solve it or use an iterative approach.

The base is being supplied by a Thevenin-equivalent 6.875V source with 6.875K in series. Base current flows out of the base so the actual voltage at the base will be a bit higher than 6.875V.

Say you ignore the base current and find the resulting emitter current. Then you know the approximate base current (divide by 151). Recalculate with the slight change and you’ll be within 1% of the correct answer. Even without the correction you are close enough for some purposes.

More satisfyingly, the emitter current is just (10-Vb-0.7V)/1K, so the base current is (10-Vb-0.7V)/151K, and Vb is 6.875V + 6.875K*Ib so we can solve for Ib and thus Ie. Ic is (150/151)*Ie so that gives you the collector voltage (about 5V) and you can confirm that the transistor is not saturated.

Since $$\\beta\$$ is specified in this case, using the second approach in a homework or quiz situation would be more likely to get full marks.

Hi everyone, I try to understand why some books used in a voltage divider to find the Base Voltage? or maybe the right question is HOW they can use voltage divider if there is base current? what I miss here? what is the best way if I want to find what is the Base Voltage?

The circuit biasing technique is called voltage divider bias. This biasing is the most widely used because it offers base voltage stability that is independent to the changes in the transistor's beta. This also sets the operating (Q) point to achieve the desired amplification class.

Some educators will use this circuit type first to bridge the ohm's law knowledge they previously taught with introduction to transistor biasing.

The ohm's law calculations could be easily applied to the voltage divider to find the Q bias (base) voltage.

If you want to use a BJT (pnp type) as a signal amplifier it is necessary to provide an external DC voltage between the emitter and the base of app. Veb=0.7 volts. Because you need a larger voltage between the collector and the emitter (supply voltage Vcc=10v) it is best to use a resistive voltage divider for biasing the emitter-base path with the required voltage Veb.

For some good reasons we have an emitter resistance Re which lowers the emitter voltage Ve < Vcc. In this case, of course, the required base voltage Vb (against ground) must be also lowered by the same amount (in order to keep Veb=0.7 volts).

For calculating the corresponding resistor values R1 and R2 it is of course necessary to consider the base current. Hence, the current through R1 is somewhat larger than the current through R2. That`s all.