# transistor amplifier questions

#### disantlor

Joined Jun 21, 2006
20
So I've read through the allaboutcircuits.com, art of electronics, and a few other treatments on transistor amplifiers, and while I generally understand them, a few questions are sticking out in my mind. I've tried my best to come up with my own solutions before I turn to you guys/girls, but still a few remain...

One thing that is confusing me is the base input current. The way I understand it is that in a common-emitter amplifier configuration, like the one a quarter down http://www.allaboutcircuits.com/vol_3/chpt_4/5.html, the resistor is used becuase the mic input might go above the forward voltage drop of the diode. If the resistor was not there, then the excess voltage would have to drop over a short circuit (since the emitter is connected to ground). In the previous example with just a solar cell, the resistor is omitted, however if the solar cell can not put out a voltage greater than the diode drop, wouldnt the transistor never even turn on?

Lastly, when biasing, two resistors are used in a divider network. However in calculations these resistors are considered to be in parallel eventhough both their ends do not share common points. One side does, but the other is connected across a battery. I must be misunderstanding something about the definition of a parallel circuit.

As always, many many thanks for any insight that you can provide. Like I said, I really do try to answer my own questions, but sometimes (ok, often) the answers elude me.

#### mrmeval

Joined Jun 30, 2006
833
• mahmoodsani

#### disantlor

Joined Jun 21, 2006
20
mrmeval said:
Funny and informative, shows what not having the resistor does.
Quite a number of animations.

http://williamson-labs.com/480_xtor.htm
wow this site is so great! it's startling how much that one animation with various VU meters explains what dozens of pages couldn't (to me at least).

#### Dcrunkilton

Joined Jul 31, 2004
422
Voltage divider bias of a BJT (bias junction transistor) involves a simplification using Thevenin's theorem. The divider is replaced by a single R(th) Thevenin equivalent resistance which replaces the two resistors in the divider. This R(th) is arrived at by replacing all voltage sources in by a short circuit, part of Thevenin's theorem. Thus, the upper resistor top terminal in the divider is shorted to ground, which puts it in parallel with the lower resistor for the R(th) calculation.

Also the original voltage, at the top of the divider is replaced by a V(th) Thevenin equivalent voltage in series with the R(th) which connects to the base of the BJT. This is Just the center point voltage of the two resistor voltage divider based on the original voltage using the voltage divider ration method--- or any other valid method. Vth = V[R(b) / (R(a) + R(b))], where R(b) is the bottom resistor, R(a) the top resistor, V the voltage at the top of the divider.

Unless you are an EE or ET student having been exposed to Thevenin's theorem, this may not be very clear. Furthermore, just reading about Thevenins therorem in most any text will only do a little bit of good, as they do not emphasize practical applications. But it will get you an official definition of Thevenin's theorem.

The bottom line is that we replace the two resistors by a simpler circuit with one resistor which looks like base bias. That is, a single resistor and voltage source in series with the base of the BJT. The "base bias" is usually described in texts a page before "voltage divider bias".

#### mahmoodsani

Joined Jun 3, 2010
43