Transistor biasing help

You can really get tangled up in your analysis. How about thinking about it this way.

You know if it's a silicon transistor it's going to drop about 700 mV base to emitter. Any reasonable beta will mean emitter current will be much greater than base current. You probably know the dynamic resistance base to emitter is about 30 divided by the emitter current in mA.

SO, choose an emitter current. Say 1 mA for our purposes.

Choose a resistor, emitter to ground. Suppose you want the emitter at say 5 volts, so you'll put 5 K there. Now the base has to be 5.7 volts, just do a divider to generate that, with resistors say no more than 5 times the emitter current so you need not worry about base current.

Now, let's just say we want the collector at 7.5 volts or so. If you've a 10 volt supply the collector is taking about a mA, you'll need 2.5 K.

DC gain is going to be pretty much the ratio of collector to emitter resistance. think about it this way. With a small signal going in, the emitter voltage will pretty much follow the base voltage, it looks like an emitter follower.

So 1 mV swing at the base gives about 1 mV swing at the emitter, 1 mV across 5 K is .2 microamps. 0.2 microamps across the 2.5 K in the collector is 0.5 mV.

Bypass the emitter resistor with a cap, you have a dynamic emitter resistance of about 30 ohms, and the AC gain is 2.5K/30.

Start there, what I've offered is pretty much independent of the transistor parameters for reasonable transistors. Beta and other factors come into play when you're worried about power consumption or other factors. The back of the envelope stuff here will give reasonably accurate results, and the good news is, you'll be able to look at a circuit and estimate what should be happening before you get lost in analysis.

Hope this helped.

Thanks to all - this whole discussion helped me. Playing with the ideas on a breadboard today!

Perion

Well I think this helped. I guess you just have to do this all the time in order to get ideas.

I'm still not sure how to judge what the resistor from Vcc to the base and the resistor from base to ground should be. I thought I knew to make the resistor from Vcc to the base 20 times higher than the Ib, but it doesn't seem to be the case from what the teacher is saying. But he just plugs numbers in and assumes and then finished off the problem that way.

There is no hard fast rule that says the resistor should be 20 times the other. What is common is that the divider will pass about 10 times more current than the base is actually using, and the voltage at the base or junction of these two resistors needs to be about .6 volts higher than the emitter voltage. Some other pointers ---- A silicon transistor begins to conduct collector current when the base to emitter forward bias is about .5 volts. At .7 volts the transistor should be conducting heavily. Maximum beta occurs in the area of .6 volts. The relationship between Vbe and IC is logarithmic. A tenfold increase in base current produces only a 60 mv change in base to emitter voltage. So by knowing what Vbe is, does not tell you where Ic is, it's just a figure that gets you in the ballpark. After your circuit is breadboarded, then you can tweak the ratio of those two resistors to get the collector voltage exactly where you want it. There is nothing wrong with this kind of approach because it greatly simplifies the calculations needed to make a working circuit.

Here is an example for you--- Let's assume that we need an extra 20db of gain for a microphone preamp. The input z needs to be at least 2.5k and the output Z needs to be about 10k. We have a wharehouse full of 2n3904's and a supply voltage of +18vdc. We are not too concerned about distortion, noise or power consumption and any frequency response covering 200hz to 10khz will suffice. It will be hard not to succeed with these constraints. If the microphone outputs 10 mv peak, then our output will have a 100mv peak signal. Compare

Ra could be made a 27k which is a standard value, without any adverse effect on the Q point. If you fire this up with your favorite software you will see that it meets our requirements. The ratio of Rb/Re, our circuits current gain, is only 2.5. This means that any transistor with a beta of 10 or better capable of operatiing with a collector current of 1ma should bias up correctly.

The voltage gain is basically 40 times the supply voltage if there is no external re./ This occurs near saturation of the transistor.

Nothing else matters.

I forgot to put my sig in the subject.

I once had to do a similer problem in class.

the gain of a transistor is simply 40 times Vcc and it occurs close to saturation.

Nothing else really matters.

It there is large external resistors, the gain is Rc over Re.

I seem to remember this to be the the small signal voltage gain: 40 Ic Rc, where R is the Ic is the collector current and Rc is the collector load resistance. Ic Rc is voltage across the load resistor, which might be said to approximate to Vcc when the emitter resistor is small and the transistor near saturation. The value 40 is temperature dependent, and is the approximate value for room temperature. The calculation only makes sense if the emitter resistance is decoupled, since otherwise most of the input signal voltage variation appears across the emitter resistor, and is not amplified.

Sylvia.

SO WHAT, now design it so it works over temperature -40C to +85C

(see how your transistor characteristics very over temp. Ain't 0.7 anymore Alice.)

The derating figures tell me that at 85 degrees, the power dissapation is derated to about 100mw. We are at 7+ mw. So Pd is not an issue. Using circuitmaker, the collector Q point drifts from less than 11 vdc to a bit more than 8 vdc. So Q point is not an issue. The voltage gain still sits at 20db. At those temps, I'd be more worried about the person speaking into the mic than the circuit failing. Just don't expect DC coupling.

Riight on sylvia.

If you write the equations for the gain of a transistor and take the internal Re of the transistor as being 25ohms/Ire(in miliamps) then Re and Rc drop out or the equation completely and the gain becomes 40 X Vcc,

Ass Re at a given current can vary with temperature if temperature is a consideration then as you say the gain can also vary.

But Re does not really vary much with temp unless you go to extreems. Of course "very much " is open to individual thoughts on the matter.

At close to saturation IcRc is equal to Vcc. So you can drop them from the equation.

The example was given for a circuit with out an external emmitter resistor.

I got an A in the course for that one.

The Instructor just wanted to know the trick.

Thanks for everyone's help. It is difficult to try and match all these parameters. Right now I'm trying to work on a project for class.

Common emitter circuit, gain of -10, input impedance minimum 10k, output impedance of 5k and the output has to be 5v rms.

I can get just about everything except matching the input impedance. If I change a resistor to get that, then I change the gain, the I need to change something else then I ruin the output impedance.

Thanks again

Yeah, life sux sometimes, and sometimes it sux all the time...