Off Topic: NPN Transistor Amplifier Configuration Help

[leeeeee]

Let's see if I can remember this stuff...

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                    Vcc o
                        |
                        <
                        > Rc
                        <
                 Rf     |
              +--\/\/\--+----> Vout
              |         |
              |       | /
      Ri      |       |/
 o---\/\/\----+-------|
                Vb    |\
 Vin                  | v
                        |
 o----------------------+---->

For a 1/2 Vcc settling point

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Rf = Rc * Hfe

E.g. If Rc = 1K and Hfe = 100 then Rf = 1K * 100 = 100K

Ri is usually the impedance of the driving circuit and Vb is assumed to be constant.

So for any change in Vin, IRi changes by dVin / Ri

As Vb and hence Ib are considered fixed so this change in IRi must be matched by an opposite change in IRf so Vout must change by ..

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dVin / Ri = -dVout / Rf

From this you can get the gain which is approximately ..

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-dVout / dVin = Rf / Ri

.. as long as Rf / Ri is less than or equal to Hfe.

Advantages, needs fewer components than auto biased stages and is more stable than simple biased stages.

Disadvantages, effected more by component variations than auto biased stage.

That looks about right.

Lee.

[kc5tja]

THANK YOU!

This information ROCKS, and allowed me to implement an amplifier on the breadboard with predictable results.

Now, my next step is to figure out why it works (e.g., why it self-biases the way it does), but that can wait for another day.

[leeeeee]

It has to settle where it does, it's Ohms law. If you draw it out like a voltage divider and put in the currents it looks like this.

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 Vcc o-+
       |
       v IR1
       |
       <
       > R1
       <
       |
       |  Ic
       +--->---> Vc
       |
       v Ib
       |
       <
       > R2
       <
       |
       |
  Vb o-+

  0V o---------> 0V

We know that Ic / Hfe = Ib and we can see that IR1 = Ic + Ib so from this we can get IR1 / (Hfe + 1) = Ib.

Putting this value into the standard voltage divider formula of ..

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Vc = Vcc * R2 / (R1 + R2)

.. gives ..

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Vc = Vcc * R2 / ((R1 * (Hfe + 1)) + R2)

This is all well and good for germanium transistor circuits with a 6V or more supply and a relatively small Vb of ~0.3V but for silicon taking into account Vb gives..

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Vc = ((Vcc - Vb) * R2 / ((R1 * (Hfe + 1)) + R2)) + Vb

But usually for an Hfe of more than 10 and ignoring Vb you'd use ..

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Vc = Vcc * R2 / ((R1 * Hfe) + R2)

And, just to tie it in to my previous post of finding Rf values for Vc = Vcc / 2, rearranging you get ..

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Vc / Vcc = R2 / ((R1 * Hfe) + R2)

Put in the value for Vc / Vcc

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0.5 = R2 / ((R1 * Hfe) + R2)

Multiply both sides by 2

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1 = 2 * R2 / ((R1 * Hfe) + R2)

Multiply both sides by (R1 * Hfe) + R2

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(R1 * Hfe) + R2 = 2 * R2

Subtract R2

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(R1 * Hfe) = R2

Which is the formula I started out with yesterday.

I think that's right. Ok, that's enough disturbing the dusty corners of my brain for now.

Lee.

[GARTHWILSON]

The biggest problem with the circuit in the last few posts is that it does not allow for variations in β. You can sort of get away with that if you're just making one of something and you can take the time to measure exactly what it will be under the conditions of interest; but in production, the β for transistors of a given type can vary wildly.

[kc5tja]

Well, if push came to shove, a 500K trimmer pot would work to compensate for that.

For example, the 2N3904 has a range of nominal hFEs from 100 to 300, with most sitting at 200. Trimming the feedback resistor would work to compensate for this range, assuming your amplifier gain is 100 or less.

While not the cheapest solution, it would work fine for moderate production runs. Most analog electronics need some form of calibration work anyway. I remember taking apart old (and some not-so-old) TV sets and the guts were just packed with trimmer pots, coils, and capacitors. It's amazing anything worked at all when things got out of alignment.

The biggest concern with this kind of circuit, frankly, is thermal run-away. But since these circuits seem to find their greatest use as pre-amplifiers, I doubt signal strengths are strong enough for this to be of much concern.

[leeeeee]

While it won't be exactly as designed this type of circuit will still work adequately with a +/- 30% variation in transistor gain. This is evidenced by the milllions of transistor radios that used just this form of biasing for a lot of stages. It also has the advantages of requiring only half the passives of an auto biased stage and not losing the emitter feedback voltage from the available output swing, something that may be important in a low cost, low voltage design.

Lee.