Author Topic: Transistor Biasing Question  (Read 215 times)

iamthesoyman

Transistor Biasing Question
« on: January 30, 2021, 08:50:24 AM »
If I was looking to adjust the bias of the transistors in the schematic here:



Which resistors should I look at adjusting ?


iainpunk

Re: Transistor Biasing Question
« Reply #1 on: January 30, 2021, 09:50:27 AM »
mainly R1 and R2, since they determine Q1 bias, which has a large influence on the Q2 bias as well.
so changing them to bias Q1 right, and if that's correct, Q2 should be correct, if its not correct, try changing R10.

cheers, Iain
half man - half snail - 6 feet to scale - Snail man's - not frail - He's been - to jail - snail man is fuccing real
『snailpilled』

R.G.

Re: Transistor Biasing Question
« Reply #2 on: January 30, 2021, 12:04:46 PM »
Iain is right - mostly the Q1 biasing string.

Here's a bit more on the topic of what to look at when biasing.

Biasing is a DC affair - only DC-connected parts matter. So to analyze biasing, you can (conceptually at least) take the schematic and erase every capacitor. Caps are DC blocks and do not matter at all to biasing. Once that is done, look at the remaining resistors. Some of them went only to caps, so one end is not connected now. Erase any one-ended resistors. Now you have just the parts that could affect biasing.

Biasing bipolars when the circuit has an emitter resistor involved is easy. You decide what voltage you want across the collector load resistor; Ohm's law tells you the collector current that must happen. That current must flow in both the collector and emitter resistors. (With modern bipolars, the gain is so high and the leakage so low that you can neglect the base current.)

Ohm's law then tells you the voltages across both collector resistor and emitter resistor, and what is left from the total power supply voltage is what is across the transistor from collector to emitter.

So far, this is simply stating what you want to come true. Next, you have to choose the base conditions to make that happen. Bipolar transistors require that the base be one forward-diode-drop away from the emitter. If it's not, there is no base current and nothing happens. So you arrange all the resistors and whatever else on the base to make the base voltage be correct for the combination of voltages on the emitter and collector.

It is nearly always better to start at the output side of things and determine what conditions need to be met as you work your way back through the circuit. In this case, the OC71 is the output. The emitter and collector resistor are close - 4.7k and 5.6k. For maximum linear swing on the transistor, you would probably choose the voltage from collector to emitter to be about half the available power supply, so for a 9V supply, half on the transistor, there's 4.5V across the transistor, and the remaining 4.5 across 4.7k and 5.6k. A quick calculator excursion says the current in the OC71 must be 4.5V/(4.7k+5.6k) or 437uA. The original designer got this by deciding what voltages and currents he wanted and choosing the resistors to make it true.

The emitter is sitting 437uA*4700= 2.05V lower than the 9V power supply, at 6.95V. To make this come true, the base must be one germanium diode drop lower. Germanium leaks, and has a very low turn on voltage at low currents, so it's a bit harder to just guess at what it will be, but it definitely will be down in the 0.2V or lower range. It's OK to just guess as reasonable a value as you can, then run back through the calculations to correct for any minor mis-guesses. In this case, I'd call it 0.2V. So the base of the OC71 could reasonably be at 6.75V.

This voltage must be made by the collector resistors of the BC549, so we already know that the collector of the BC549 has to be at 6.75V. IF the bias string isn't there, then there must be a current of (9V - 6.75V) / (1K +5.6K) = 340uA in the 1K and 5.6k collector resistor. But the bias string is there, and it pulls ... hmmm, several equations in several unknowns. AAACKKK!!!
Math!!!

Actually, it's not that bad. We can get close enough with "numerical methods", which is what the computation guys call guessing an answer and then seeing how bad they did, then guessing again. The bias string is a total of 89k. If you ignore the 1K for a moment, then the bias string eats 101uA. Just hold that for a moment. The collector of the BC is at 6.75V, and that runs through the 1k and the 5.6k collector resistor. 2.25V of drop and 6.6k = 341uA. Hmmm. That's close. The 1K is actually carrying upwards of 440uA.

First guess wasn't too good. If we guess again, and assume that 100uA goes to the bias string and 300 goes to the transistor,  then the 1k drops 400uA * 1000 = 0.4V. Grinding back through the calcs, the bias string current is (9V-0.4V)/89k = 97uA. the 100uA guess was only 3% high! On the collector resistor, the guess would result in (9V - 0.4V - 6.75V)/5.6k = 330uA. The guess of 340uA was only 3% high too! If you want to continue to home in on just the right values, you can, but with the resistors being +/-5% tolerance, that probably isn't a productive pastime. I would stop with 97uA and 330uA as a good estimate.

We're getting close. With 330uA through the collector of the BC, that same current (plus a gnat's eyelash of base current) flows in the emitter resistor. That resistance is effectively 5.6k + 1K + 68 = 6668 ohms. The emitter voltage is then 330uA times 6668 ohms, or 2.2V. THAT is, finally, what the base of the BC has to sit at to make our earlier calculations come true.

There are a couple of ways to make this necessity come true. The bias string resistors, 33k and 56k, can be adjusted, or the series base resistor can be adjusted. A BC549 has a typical gain of 200 (from the ON Semi datasheet typical hfe curves) but it may be as low as 100 or as high as 800. For 330uA of collector current, the base current might be as high as 3.3uA or as little as 0.4uA. The drop across that 220k resistor could be from 0.7V to 0.088V.

What to do???


I suggest: ignore the variation. The relatively large emitter resistors will make the circuit self adjust quite a bit. Build it or simulate it, and then tinker the bias string a bit. You're almost certain to be slightly off in the real circuit, but only slightly.

Now that you know how to work it, you can tinker. Changing the collector and emitter resistors change things a bit, but it's the emitter and base voltages that really matter. Changing the bias point of, say, the OC71 without changing the bias of the BC requires just changing the OC71's emitter and resistor and accounting for the change in the collector resistor value. Changing the bias point of the BC requires changing the BC's emitter resistor, and then changing the emitter and collector resistors of the OC71.

I really wish I'd had a spreadsheet program (or a personal computer of any kind) available to me when I was tinkering transistor biases back in my engineerling days. I did all this over and over and over on paper with a calculator. This stuff can be set up in a string of calculations on a spreadsheet and almost automated.
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

Rob Strand

Re: Transistor Biasing Question
« Reply #3 on: January 30, 2021, 06:10:21 PM »
Adjusting R4 is a good way to set the bias of the second transistor without messing up gains.   For a wide range of adjustment It pretty much *only* changes the biasing of Q2.  However, if you end-up with a very low resistor, say less than 470 ohm to 1k, the gains will be affected and the you might consider other options. 

When you set Q2's bias Q1's bias should be reasonable by nature because Q1's collector is pinned to Vcc-Vbe2-V_drop_on_4k7.  Vdrop on 4k7 is related to the Q2's bias if you fix Q2's collector voltage you fix the drop across the 4k7 resistor.
« Last Edit: January 30, 2021, 07:08:10 PM by Rob Strand »
The internet:  answers without the need for understanding.