What's the point of biasing any transistor?

mt1200

Hi guys.

I learned that you need to apply DC biasing to establish your transistor's amplification zone.

Some weeks ago I was struggling (as almost we all do with anything related to E.E.) with transistor amplifiers.

I had to make 3 amplifiers with 3 different biasing configurations. I read a lot using Malvino's book, and I think I learned how to make the DC model.

But when It comes to putting the AC input signal, it becomes almost useless, I had to modify almost all the resistor values, and I still could never get the gain that I had calculated ( I calculated 150X, all I got was 126).

I started to play a bit with the resistors until I could get my 150X gain. But my only conclussion was that the biasing is pointless because you'll end up replacing playing with the resistor values to fit your desired output.

So, what's the secret?, why would you calculate your DC biasing model if you'll end up with different resistor values to fit your design requirements. I wasted nights asleep but I just couldn't figure out how to get the exact values to fit your desired output.

yungman

For one, you try to get too too too too much gain out of one stage. Your Re must be very small, your r'e is getting in the way. I won't even try to get more than 30 or 40 with one transistor without really careful about it.

Malvino book works, I did years of transistor design using that book.

vk6kro

Transistors vary in current gain by an incredible amount, so simple circuits will always give variable gain.

It seems like it would be possible to produce transistors that have exact current gain even if the device is actually an integrated circuit with several components in it.

However they can cost just a few cents each and often an exact gain is not important as long as it is high.

There is a trick where you put an unbypassed emitter resistor in the circuit and then the gain becomes close to the ratio of collector resistance to emitter resistance.

So, in this case, you could have two stages with a gain of 12.25 and together they could give a gain of 150.

yungman

Yes, if you split to two with gain of about 12, you can get pretty accurate result.

This is more the matter of practical implementation. you need current to lower the r'e. But increase current will have more voltage drop on the collector resistor. Say you use 1mA to get r'e to 25Ω. Say you use a 10V supply. you want gain of 150, you need Rc=3750Ω, that drop 3.75V on the resistor. This is close to the mid point and you cannot use any external Re to degenerate the variation of the r'e. Yes you can add a little bit of Re, but the r'e is going to dominate and all the variables come in. You run into practical limit in this case.

Two stage, drop gain on each stage and you'll have no problem and Malvino works.

davenn

Another form of biasing is for RF transistors in linear amplification class AB mode (for SSB modulation use) as different from class C mode ( for FM and other contineous carrier modulation modes)

for Class AB we need to supply a small biasing voltage to make sure the transistor is already turned on before appling any drive.
have a look here for a longer explanation.

Dave

skeptic2

Is this homework?

If the output impedance can be higher than the input impedance, then your voltage gain can be somewhat higher than the transistor beta. Are you measuring gain with or without a load attached?

It is possible that you have a low gain transistor but I think it more likely that your amplifier configuration is not optimized. Perhaps if you could post your circuit we might be able to find reasons why you aren't getting more gain.

carlgrace

Your gain is greatly affect by Beta (hfe) which varies a lot even in well-made transistors.

If you need a specific gain you have two choices. Assuming a common-emitter amplifier you can put a resistor in the emitter loop and one in the collector loop, and your gain (assuming large but imprecise Beta) is Re/Rc. Don't expect to get a lot of gain in one stage like this. Something on the order of 10 is reasonable.

In practice, people would usually use an op amp in negative feedback for this type of thing. It's gain can be incredibly accurate (to the accuracy of the resistors) because the open-loop gain of the op amp can be so high.

mt1200

Sorry guys I don't have my circuit's snapshot here, I'll upload it as soon as I can.

Well, yes, I remember that first I used the common-base biasing, and got frustrated due to the 130X gain even if my calculations pointed to 150.

Then I moved to the common-emitter amplifier and the same thing happened, I got abour 125X only.

I used a 2N2222.

Op amps?, that would be fine but I still haven't used op amps in college. Probably next semester.

I mean come on, how can designers create THESE circuits?.

three W's point electronica-pt.com/circuitos/images/100W-amplifier.gif

I can't even think how that works, how did they biased every transistor without screwing up the input impedances?, or without relying on the unstable Beta?.

skeptic2

Is this homework or a lab exercise? I'd like to help you but the rules of the forum dictate how I can help you based your answer.

mt1200

It was a lab excercise, I don't know if that counts as homework.

Anyways, that was some weeks ago, I only have some questions in my head because I could never get the output gain I wanted.

carlgrace

The circuit you uploaded uses negative feedback. In fact, its design is quite similar to an op-amp's internal design. Three key things are going on here with respect to biasing:

first, R2 is biasing the base of T1. The current through the differential pair of T1 and T2 is set by D1 (D1 is a reverse-biased zener diode so its voltage is well controlled).

second, and this is key, R16 is feedback back some of the output to the input. If you follow the signal inversions in the circuit you will see this is negative feedback. In essence, the circuit is trading on a large, but uncontrolled gain for a smaller, but better controlled gain.

third, the base of T3 (and therefore its biasing condition) is trimmed using a variable resistor (a pot).

T5-T8 aren't providing voltage gain, they are configured as a push-pull output stage... they enable the amplifier to drive a heavy load without losing gain.

A common-emitter amplifier alone (like you did in your lab) will never give accurate gain without trimming. It just can't.

skeptic2

mt1200,

After I saw your post I put together a circuit in LTSpice using a 2N2222 and got a voltage gain of about 330. Furthermore it would be easy to modify the circuit to get higher gains than that. I'm not claiming it is a good circuit, only that it illustrates a way to bias a transistor for high gain. Don't forget that the higher the gain, the more easily the circuit will oscillate. The advice to design a 2 stage circuit with each stage having a gain of 12.25 is good advice. My example is for demonstrating a biasing for high gain.

So even though such a high gain stage is risky because it will oscillate if you're not careful, it does have some advantages over the configuration most frequently taught in the schools. First it is biased from the collector voltage. This means the collector voltage will stay the same regardless of your supply voltage. So if your supply has ripple or noise on it, the ripple or noise at the output of your amp will be much better than with the other configuration. Also for RF applications this is the preferred configuration as using an emitter resistor will often cause an RF amp to oscillate.

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carlgrace

What you got there is the standard textbook biasing scheme for a common-emitter amplifier. However, it doesn't solve the situation the OP had in that the measured gain was different from the calculated gain. The gain in your circuit is proportional to dc current gain beta (or hfe) and according to the BJT datasheet:

http://www.fairchildsemi.com/ds/PN/PN2222A.pdf

hfe can vary from 100 to 300! No wonder this circuit's gain is all over the place! Use negative feedback and repent!

Kurinn

Did you have a load at the output? The calculations for gain with a load involved are different.

Aside from that, even if you're careful with your calculations, you won't necessarily get what you expect. I just finished a term project consisting of a two stage amplifier. A self-bias JFET followed by a voltage divider biased BJT. Both of these in common-emitter configurations. My numbers looked solid, and indicated that I'd be looking at a combined, loaded mid-band voltage gain of about 625, and I ended up with 500 on the nose. 125% of the requirements for the project, with minimal distortion at the specified input signal voltage.

I could have accidentally missed some variable, I mapped out the DC and AC functions to a spreadsheet, and I skipped a good night's sleep playing with that before I built it, so who knows. In general, I suspect that no matter what, your calculations won't account for all of the transistor's behavior all of the time.

Then again, my JFET was heavily used. The specs for the MPF102 say it can have a drain-source sat current of as much as 20 mA and a pinch-off voltage of -8 V. My JFET, the one with the leads that look like curly fries? 1.6 mA, with a pinch-off where V_GS is somewhere around -0.8 volts... But that didn't matter too much. The JFET was to get high input impedance, and I didn't need a gain higher than 2 for that side. Anyway, it worked, and I met the requirements for the project with minimal distortion. It didn't hurt that I designed for as much gain as I could get, and worked back from there.

vk6kro

The output impedance of the transistor appears in parallel with the load, too, so if you don't include it, the real load will be a lower impedance than your calculated one, so you will get lower gain.

Another other source of error is if you use the DC current gain of a transistor (Hfe, as measured in a transistor tester) for your calculations. This is almost unrelated to hfe which is the real current gain for small signals.

"Biasing" really only refers to the process where you adjust the base current of the transistor so that it is operated in linear mode. Voltage gain is mostly concerned with the collector and emitter resistors and the current gain of the transistor.

Kurinn

Ah, that's probably what got me. I realize now that I had used the figure I got off of the DMM's tester to make my calculations. Thank you.