Transistors biasing qestion

[soggybag]

Is it possible to bias two transistors from the collector output of a preceding transistor stage? Seems like it should work. In the schematic below omit Q3 and you have a common arrangement where Q2 is biased by Q1c. If I understand BJTs correctly very little current flows into the base. I would guess that Q1 should be able to supply enough current to bias both.

[R.G.]

Is it possible to bias two transistors from the collector output of a preceding transistor stage? Seems like it should work. In the schematic below omit Q3 and you have a common arrangement where Q2 is biased by Q1c. If I understand BJTs correctly very little current flows into the base. I would guess that Q1 should be able to supply enough current to bias both.

Good question.

First, Q1 supplies *zero* current. The current all comes from R3 to the power supply. Q1 can steal all that away, if turned on hard. But first things first.

Let's look at what voltages and currents flow in this circuit.

Q1's base is held at 9V *(10K/110K) = 0.818V. The base-emitter voltage of Q1 is between 0.5 and 0.7V. The 2N3904 data sheet says Vbe is 0.6V at 100uA of collector current.

So there is 0.818-0.6 = 0.218V across R4. The current in R4 then must be 0.218/1800 = 121uA. Notice that the difference between the bias voltage and the base-emitter voltage is small; this leads to higher variations in the finished circuit with Vbe variations.

If the transistor gain is at least 100, then the base current is 1.2uA, and the base looks like a resistor of 0.818V/1.21uA = 682k in parallel with the 10K. Yep, it was probably OK for us to ignore the base current.

The current through the collector is then 121uA minus the base current of 1.2uA; let's ignore the base current and call the collector current 121uA. The voltage dropped across R3 is then 121uA times 47k or 5.687V.  The collector voltage on Q1 is then 9.0V - 5.687 =3.313V. And that's the voltage on the following bases of Q2 and Q3.

Assuming the same 0.6V in Vbe on the next two transistors, you get emitter voltages of 3.313V minus 0.6, or 2.713. Happens on both of them, because the transistor current gain pulls up the emitter to make that come true.

Q2 has an emitter (and approximately collector) current of 2.713/10K, or 271uA, and a collector voltage of 9.0-(10k*271uA) = 6.287V. Q3 has an emitter current of 2.713V/1800 = 1.51ma. It's collector voltage is 9 - (47k*1.51ma) = -61.8V.

Ooops. That can't happen. The most Q3 can do is make its collector-emitter voltage go to (nearly) zero if it saturates. What really happens is that the transistor acts like a switch, pulling R7 and R8 together. If Q3 was shorted, the current in R7 and R8 would go to 9v/(47K/1.8K) = 184uA and the voltage at R8 would be 184uA *1800 = 0.331V. The base voltage would be 0.931V if this could happen. But R3 is trying to hold it up at 3.313V. It gets no help from Q3, which is saturated, so Q3 base pulls R3 down, and the base of Q2 at the same time.

So - it won't work as shown. Not because Q1 can't drive Q2 and Q3, but because R7 lets Q3 saturate and foul up the situation on R3 because a transistor's gain is what makes the base voltage approximations work, and the gain goes to nearly zero in saturation.

[soggybag]

Thanks R.G. I was really hoping you would reply.

Sounds like I need to adjust R7 and R8.

I want to use Q2 as a phase splitter by taking the output here from both the collector and emitter. I'm guessing I need to keep R5 and R6 the same value for this to work.

If worse comes to worse I could always a caps before Q2 and Q3 and then bias with a voltage divider like Q1?

[R.G.]

I want to use Q2 as a phase splitter by taking the output here from both the collector and emitter. I'm guessing I need to keep R5 and R6 the same value for this to work.

That part would work fine. You'll need to pay some detail to the collector and emitter voltages on Q2, because for a phase splitter, the transistor can't be more off than off, so the maximum voltage across the two outputs is the power supply voltage, and the transistor can only saturate, which means the two outputs meet in the middle, so for maximum undistorted signal, you must bias the emitter at 1/4 of the power supply and the collector at 3/4. (There are slight refinements, but this is the way to think about it first.)

To make that happen with direct coupling, the collector of Q1 now *must* be at 1/4 of the power supply plus one Vbe, or about 2.25V + 0.6V = 2.85V. Otherwise, you get less signal at the phase splitter outputs. And notice that if the collector of Q1 is fixed at 2.25V, it can only go down to about zero, but up to 9V if it cuts off. So it's giving you gain, but it's voltage swing is limited by where it has to bias because it provides the bias for Q2.

Even if that worked, you're stuck with Q3, because it wants still a different bias point on its base. This is the bane of all direct biasing - the first time you try to do more than one thing at once, your DC voltage requirements start conflicting. You always wind up either trying to bias for one thing (if you can!) and tossing multiple requirements overboard, or having to replicate things or give up direct connect biasing.

If worse comes to worse I could always a caps before Q2 and Q3 and then bias with a voltage divider like Q1?

Yes. That solves the bias voltage problems.

However, you still run into the signal level and output voltages requirements. In a circuit with Q2 being a phase splitter, it's just not possible to get output signals any bigger than 2.25V peak, and usually significantly less, before they start distorting. A split load phase inverter like that always has a gain of 1 (to a first approximation) to both outputs, so with a gain of 47k/1.8K ~ 26 on Q1, the input to Q1 can't be any bigger than 86mV and not have distortion on the output of Q2 as a phase inverter. That may be OK if distortion is what you're after, but you ought to be aware of it. That's down in the single-coil region, and some of those will cause distortion.

Q3, if biased independently, will give another 20+ in gain, and will be solidly distorting with a single coil at the input of Q1 - if that's even how you use it.

Most of the first order circuits stuff is not magic, and not writing huge equations. It's all ohm's law, estimating gains and currents, and figuring out whether you even have a chance with the available power supply and devices.