Transistor Biasing BJT

[John Lyons]

Does anyone know of any reading material for transistor biasing with pull up/down resistors or base/collector feedback resistors as in the big muff make up (last) gain stage. I under stand it a bit and can make it work but I'd like to know theory more.
Layman's terms would be nice as well

thanks

john

[km-r]

http://web.telia.com/~u85920178/begin/bias-00.htm
http://web.telia.com/~u85920178/begin/bias-10.htm

[R.G.]

Which one did you want to start with? The final stage with a voltage divider bias network is easy. The voltage network sets the base voltage, the base voltage sets the emitter voltage, and the emitter voltage sets the emitter current.

[John Lyons]

Thanks for the links KMR, I'll look at those...

RG, ok, I should explain.
The last stage is my main curiosity.
I initially was wondering how to set the operating point and gain of the last stage of the big muff.
So the (typically 470k/100K) resistors off the transistors base are a voltage divider affecting the base voltage.
But this affects the collector voltage as well ultimately. That's why i'm a bit confused.
That voltage divider sets and keeps the base voltage steady correct?
+9v is AC ground correct? So the 470K is a pull up and the 100K is pulling the base down...holding the base at a
steady voltage. Now, how does the emitter to ground and collector come into play with this voltage divider?

I'm interested also to know how to set the output gain with in this configuration.

Thanks in advance. Showing the holes in my theory here 

john

[Cliff Schecht]

The gain of a common emitter amplifier like the one in a BMP can be calculated simply by the resistor ratio of Rcollector/Remitter. Since this configuration is inverting, the actual gain is -(Rc/Re).

[davidallancole]

If you use the formula for a simple voltage divider, which the 470k/100k forms, you will find the voltage at the base is 1.58V.  The formula using the values in the circuit would be (9V * 100k)/(470k+100K).  Next if we subtract the voltage drop of a forward biased diode (0.3 for Ge and 0.7 for Si), which the base to emitter acts like in this circuit, you end up with 0.88V.  Next we can divide the 0.88V by the 2k resistor to get the emitter current, which is 440uA.  Assuming emitter current = collector current since the base current will be so small anyways, the voltage drop on the 10k resistor is 4.4V.  So the voltage at the collector/resistor point is 9V-4.4V=4.6V which is the mid point voltage of the power supply.

This output stage is a textbook voltage amplification stage.  Using what I have just written down should help you in designing stages similar to this with different transistors and resistors.  Hopefully this sheds a little light on why the designer choose the values of resistors that they did.

[mac]

John,

Go to
http://www.diystompboxes.com/biascalc

or download a little app I wrote for winboxes and macs
http://geocities.com/guitarfxs/exes/biascalculator.zip

Hope you find it helpful.

mac

[R.G.]

I initially was wondering how to set the operating point and gain of the last stage of the big muff.
So the (typically 470k/100K) resistors off the transistors base are a voltage divider affecting the base voltage.
But this affects the collector voltage as well ultimately. That's why i'm a bit confused.

I've typed this into this forum several times.

The standard bipolar transistor is most easily biased by noting that the base-emitter junction is always a forward-biased semiconductor junction when the transistor is active (amplifying). If it's amplifying, the base MUST be one diode drop away from the emitter. It can't be anything else.

So you put the base one diode drop higher (for NPN) than you want the emitter.  The emitter, um, follows.   

This happens every time as long as the currents involved are such that the transistor can supply/conduct them, and the current gain can supply enough gain above the available base current to bring the emitter up. This is why high-gain bipolars are good. You can use high-resistance (and impedance!) biasing resistors and still get whatever emitter voltage you want.

Why is emitter voltage important? Because the emitter voltage across the emitter resistor sets the current flowing through the transistor. The same collector current (that is, emitter current minus base current) flows no matter what resistor is between the collector and power supply [caveat: as long as the power supply is big enough to supply enough voltage to make that current flow with the collector resistor you choose; ohm's law still applies] .

If you choose a base voltage and emitter resistor to make 1ma flow in the emitter and the transistor gain is 100, then the collector current is 990uA for all collector resistors from zero up to a megohm if you have the power supply to support it. The collector voltage is set by choosing the collector resistor to put it where you want it. And yes, Mother Nature is watching. You cannot "put it" where things like unsupportable power supply voltages and currents are needed, or where the transistor can't support the collector current, or power dissipation, etc.

In effect, the base voltage has nothing to do with collector voltage; it's the emitter current and collector resistor that do it. Once you choose the resistors and start things running, then yes, the base voltage changes the base current, which changes the collector voltage through the change in emitter current.

That voltage divider sets and keeps the base voltage steady correct?

Yes.

I'm interested also to know how to set the output gain with in this configuration.

A rule of thumb is that the gain of a bipolar common emitter stage is Zc/Ze, where Zc is the impedance the collector sees and Ze is the impedance the emitter sees. Like all things Mother sets up for us, there are some subtleties. Notice I said Zc and Ze, not Rc and Re. For AC gain, not DC biasing, the load impedance on the collector gets into the gain calculation. If you have an NPN with a 100K load resistor driving a 10K load resistance through a 10uF cap, the 100K sets the DC position, the parallel combination of the 10K and 100K sets the AC gain once you get over the capacitor being significant.

If you have a 10K collector resistor and a 1K emitter resistor, then the voltage gain is 10, as long as you don't load the collector. Or as long as you don't bypass the emitter with a capacitor. If we bypass the emitter with a capacitor big enough to act like a short at the frequencies of interest, then the emitter 1K sets the DC bias, but the capacitor runs the gain up when we look at AC gain.

What is the AC gain if the cap acts like a short? Rc/Re = infinity? No; remember Mother is watching. There are very few infinite anythings, and we usually don't get to play with them. There is an internal emitter resistor inside the transistor where you can't get at it. This is the Shockley resistance, named for the guy who helped invent the bipolar transistor. It acts like a small *current dependent* resistor of about 26mV/Ie is in the emitter circuit, where it can't be bypassed. This is a small resistor, smaller at low currents, and it is always there. You can of course use two emitter resistors and bypass only one of them to get intermediate AC gains without changing the DC biasing.

This stuff, complete with diagrams and equations, takes about three or four junior-level college lectures with attached homework; there are some fine points in there that you won't hit unless you work some examples.

[John Lyons]

Thanks for the replies folks.
I need to digest this and play with those calculators Mac...thanks

john

[R.G.]

Remember that using a calculator before you understand what underlies the calculations will help prevent you from gaining the understanding.

[mac]

True.

Just a hint. If you open the html version of my calculator in your browser, view the source page. IIRC the javascript code is at the bottom. By looking at the graphics you can follow the code step by step.

mac