Voltage divider bias

Thread Starter


Joined Dec 6, 2010

I have a doubt in the section Volume III - Semiconductors » BIPOLAR JUNCTION TRANSISTORS » Biasing techniques.

The figure “Impractical base battery bias” where a dc battery is in series with the AC signal makes perfect sense to me. DC = 2.3V and AC = 1.5 pktopk. I understand how the combined base biasing is varying from 0.8 to 3.8.

But when the battery is replaced with a voltage divider in the figure “Voltage divider bias” DC voltage drop across the dividing resistor and the AC signal are in parallel separated by a capacitor. I could not understand how this arrangement can have the same effect as the earlier one.

Could any of you please help? I am learning electronics on my own and I don't know any electronics expert to clear this doubt. It would be great if any one here can help.



Joined Feb 17, 2009
Voltage divider provide the DC voltage. So without input AC signal the base voltage is equal to V1 *R3/(R3 + R2) = 2.3V.
Now the input AC voltage will change the base voltage from 3.8V to 0.8V in "rhythm" of input ac voltage.
It is possible thanks to input capacitor. The capacitor is charge to 2.3V also. And AC input signal is now in series with a capacitor but parallel to R3 resistor.



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panic mode

Joined Oct 10, 2011
the manual is correct. the whole idea is of replicating what we need using equivalent circuit.
for transistor to operate in first circuit it needed battery so that base is forward biased for any amplitude of AC signal. base current will vary in accordance with AC voltage.

note that voltage on the left side of the base resistor is sum of two components:
DC voltage from battery and
AC voltage from signal source.

result is DC voltage with varying amplitude (because battery is chosen so that magnitude of AC is always lower than battery voltage, which keeps base forward biased).

i guess it would be good to add one more picture (with voltage divider but no AC signal source) to explain how is this same as previous case (AC and battery) - when AC is zero.

then there could be one more picture with AC source between voltage divider and base resistor. this would work but it is not practical as input and output don't share common.

then the next picture (Voltage divider bias, no capacitor) should be used to mention that voltage divider simply has no effect because it is in parallel with AC and unlike AC it has impedance greater than zero so AC source would dominate (current from voltage divider would flow into AC source).

to fix this we insert capacitor. the thing about capacitors is that they isolate DC circuits but let AC signal through.

capacitor is any two conductors, separated by insulator. because they are separated, current cannot flow through it (there is an insulator in between). however, current CAN flow through capacitor leads. this is what happens when capacitor is charged or discharged (this takes time, and depends on capacitor size). current does NOT flow through insulator but it polarizes it. and those charges come and go through leads. if we connect capacitor into AC circuit, it will constantly charge/discharge and even though current does not go through insulator, there will be AC current through leads.
If you connect capacitor to DC source like battery, you will get initial current to charge capacitor, but after that there is no more current flowing through capacitor leads.

this is why we can say that to DC, capacitor is open circuit. to AC capacitor is like resistor (larger capacitor value means lower "resistance" - correct term is "impedance")

in circuit you have problem with, capacitor is sufficiently large value (for the frequency of AC signal) so that it does not have time to charge or discharge. that means that voltage across capacitor will be some average...

this is dynamic process so let's first consider "long term";

on the left side you have AC source. it is sometimes positive, sometimes negative but it is periodic and eventually it's average is zero (evaluated over long time, that is why we needed large capacitor).

on the right side you have voltage divider and we know that voltage (2.3V or whatever). so voltage across capacitor is the difference (2.3V).

now lets see "short term":

because voltage across capacitor is fixed (2.3V or whatever) because it does not have time to charge or discharge, we can consider that (at least short term) it acts as if it was a battery (whose voltage we fixed by choosing values of voltage divider). btw, compare this with original circuit, neglect voltage divider and you will see that this is EXACTLY like original circuit - except that battery and AC have swapped places, which is ok (series circuit).

what happens next should be easy to see:
when AC is zero, right side of capacitor is 2.3V.
when AC is 0.5V, right side of capacitor is 2.3V +0.5V (because capacitor voltage did't suddenly change, that takes time, long time in this case).

when AC is 1.0V, right side of capacitor is 2.3V +1.0V = 3.5V
etc. eventually AC changes polarity:

when AC is -0.5V, right side of capacitor is 2.3V + (-0.5V) = 1.7V

the voltage at voltage divider is not fixed, it is influenced by changes on other side of capacitor. if you compare step by step voltages at same point (left side of base resistor) with original circuit (one without voltage divider, just battery and AC signal source) you will see that they are the same. they may look different but they behave the same way - they are equivalent.
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Thread Starter


Joined Dec 6, 2010
Thanks a ton “Jony130” and “Panic mode”.

It is much clearer to me now.

Panic mode, you have explained the whole thing such that any one can understand the concept.

I would love to see those extra diagrams you mentioned added to the E-book.

Thanks a lot again, both of you.



Joined Jan 3, 2011
There's a simpler way of putting it in my opinion.

Assume you're already aware that a capacitor will pass AC and block DC. You should then agree that the voltage across the capacitor varies with the source AC voltage (ignore AC impedance for the moment). Since you know the capacitor can block DC you should be able to recognize that the two sides can be at different DC levels.

If your source is centered on 0V, and you assume it is an ideal source, then you know that midpoint won't change. Therefore the AC voltage across the capacitor must manifest as the voltage between the two resistors changing proportionally to the input voltage.