Hello again. I'm working on a new project that will be the foundation for some transistor-based guitar effects. I have some old PNP germanium transistors that I've been playing with, but most of what I've read doesn't explain PNP operation very well. They typically go through great lengths to explain NPN transistor circuit analysis, then quickly summarize PNP circuits as simply "the opposite of everything". Well, it doesn't appear that simple to me.

Please reference the attached schematic. I've already built this on my breadboard and it's working fairly well with the following parameters:

Vs = +9 volts DC
Vin = 1 volt AC peak-to-peak referenced to ground

My question is since everything is the reverse of an NPN circuit, should Vin be placed across ground and Vin or should the ground of my input signal (guitar) be connected to Vin and the "hot" connected to Vs? Is the output correctly taken between ground and Vout or should the output be across Vout and Vs?

Please ask if I haven't been very clear on anything (very likely). I haven't played with transistors in decades (since college) and even then I mostly used them for switching.

 

The traditional method is to always inflict the signal from the base to the emitter because the emitter resistor is part of the input impedance. If you referenced the signal to the collector side, you would decrease the current that the signal gets through the base of the transistor.

 

Thanks #12. I see it more clearly now. If the input signal is placed across Vs and Vin, there will be a higher emitter-base current on account of the lower Re. So, to make sure I have this correct, the signal source should have its ground connected to Vin and its "hot" to Vs?

What about the output? Should that still be taken between ground and Vout? Thanks again.

 

No...Vs is AC coupled to ground with a capacitor. (A battery is a big capacitor.) Connect your signal to Vs and you just grounded it.

yes, your voltage gained signal will be at Vout compared to ground.

ps, you have way too much idle current going through the transistor.

 

So I should reverse the signal - place the "hot" at Vin and ground lead to Vs (seems counter-intuitive, but I can see that working)?

The Q-point is off? I believe I calculated a base current of 50 uA, which at a beta of around 80, gives me an Icq of 4 mA, roughly half my Ic max. Isn't that correct for a Class A amp?

See what I mean by these PNP's confusing the crap out of me? All the examples in the texts display an NPN circuit.

 

You can't just tell a transistor to have a DC gain of 80. They aren't that predictable.
You can set the voltage at the base with R2 and R3 then let R4 set the idle current,
or you can connect a base biasing resistor from the collector to the base,
or you can use a jfet or an op-amp if you need high impedance,
but your method doesn't work.

 

I didn't "tell" the transistor anything. I tested three of them and they all had a beta of around 80. I've read beta is a poor design parameter because it drifts with temperature and they vary from one device to another, but using a voltage divider for bias is supposed to eliminate that concern.

If I can't use Beta, then how do I determine what the base current should be? That's been my confusion from the beginning I believe, but as I said initially, this amp is working as pictured. Perhaps not optimumly, but it is amplifying a signal.

Thanks again for your help.

 

Hand sorting is the penalty for that kind of circuit and you did it, therefore it works. It's a miserable design for a production run because all the transistors would have to be sorted, but you can do it for a one-of-a-kind circuit.

The usual way is to use lower value resistors for R2 and R3. They set a voltage on the base and the emitter resistor sets the current.

 

My schematic has already appeared on Google images! That's some pretty quick web crawlers they have!

Anyways, I'll revisit this tomorrow when I'm not so tired and I can record all my measurements again.

I'm not too concerned about production runs or anything. I'm just a hobbyist trying to relearn electronics after a 25 year hiatus.

Have a good night.

 

Hand sorting is the penalty for that kind of circuit and you did it, therefore it works. It's a miserable design for a production run because all the transistors would have to be sorted, but you can do it for a one-of-a-kind circuit.

I'm not too concerned about production runs or anything. I'm just a hobbyist trying to relearn electronics after a 25 year hiatus.

Even a one-of-a-kind circuit should be made to work with off-the-shelf parts. Do you really want to hand test more transistors a year or two from now when the part you used fails?

 

Even a one-of-a-kind circuit should be made to work with off-the-shelf parts. Do you really want to hand test more transistors a year or two from now when the part you used fails?

Hardly something I'm concerned about at this point. I'm just trying to understand how to properly bias this thing. If you would like to explain the changes that need to be made so I can pop any ol' germanium PNP transistor in there, I'd love to hear it.

 

I fired up my toys again and took some readings. Fortunately, it works just as before when I put it away a month ago, so at least I have some element of stability here, but is it optimized? I'm going crazy trying to figure out how I can increase the current through the voltage divider (R2 & R3) without sending Ib through the ceiling! If I increase Re (R1) to several thousand ohms, I'm killing Ic. The only thing I can think of is raise the voltage across R3 to just a few tenths of a volt below Vs. Is that kind of precision truly necessary?

Anyways, here are the DC q-point values I measured with Vs at +9 volts:

Ic = 4.4 mA
Ib = 49 uA
Ir2 = 6.7 uA

And here is a scope shot: The lower trace is the input displaying 0.5 v/division and the upper is the output at 2 v/division. My signal generator is set to 1 volt p-p and inserted at Vin with the ground at ground. If I place the ground on Vs, then I'll short my supply through the scope's ground wire. How do I get around that?

Thanks again for all the help.

 

Ok, I bit the bullet and made some changes just to the voltage divider (R2 & R3) by lowering the values significantly. Fortunately, the circuit has remained stable and fairly close to before. Ic dropped a little, so I've lost a little headroom due to my mid-point bias point dropping, but I'll address that later.

Below is the new circuit with the following q-point values:

Vs = +9 volts
Ic = 3.9 mA
Ib = 27 uA
Ir2 = 0.8 mA

Does this look better?

 

Note that C2 is shorted.

Lowering the base bias impedance (R2 || R3) gives the circuit a more stable operating point (because the signal current is a lower percentage of the total current), which lowers harmonic distortion. But it also lowers the input impedance, possibly loading down the source and decreasing the effective overall circuit gain.

As to referencing either the input or output to Vs because this is a PNP rather than an NPN circuit - no. In classical circuit analysis, all equations for everything are based on the assumption that there is zero ohms impedance between all power sources and ground. This is impossible to achieve without infinitely large perfect components, but we try. In the real world, that translates into there being way fewer ohms than any other circuit impedance at all frequencies of interest, something that can be done with a few capacitors. Once that is achieved, all common emitter amplifiers operating within their linear active regions behave the same from an AC analysis point of view - BUT:

A common emitter amplifier has asymmetrical input and output impedances. That is, the current drawn from the source and the current available to the load are different depending on which way the voltage is changing. Using your circuit's output as an example, the PNP transistor can source current, but it can not sink any. So the output impedance when pulling the load up toward Vs could be quite low (maybe only 150 ohms if T1 saturates) depending on the signal frequency, value of C2, etc. Pulling the load down toward GND is done only by Rc, so the minimum output impedance is 1K if T1 turns off. So while it is true that you can adapt an NPN amp circuit to PNP by using "the opposite of everything", there is a difference between PNP and NPN common emitter amplifiers in how they behave in non-linear operation.

ak

 

Note that C2 is shorted.

Good eye! Yes, I inadvertently "wired" across C2 when I moved it, but I'm only using Qucs currently for schematic capture. I'll have to learn a lot more about it before using the software for simulation.

Lowering the base bias impedance (R2 || R3) gives the circuit a more stable operating point (because the signal current is a lower percentage of the total current), which lowers harmonic distortion. But it also lowers the input impedance, possibly loading down the source and decreasing the effective overall circuit gain.

That has been a concern of mine because I intend to use these simple circuits with an electric guitar, which have output impedances of several thousand ohms. Obviously, using an FET or op-amp would be a better way to create a high-impedance input, but I'm not looking for "better" at this point. I'm trying to learn how these things operate, warts and all, and figured they made these circuits work in the old days, so why not today? Am I basically looking at a trade-off between high-impedance input with low-stability and low-impendace with high stability? Is the input impedance largely determined by the Thevenin equivalent of R2 || R3? If so, I have it very low for a guitar signal.

As to referencing either the input or output to Vs because this is a PNP rather than an NPN circuit - no.

Ah ha! Thank you!

A common emitter amplifier has asymmetrical input and output impedances. That is, the current drawn from the source and the current available to the load are different depending on which way the voltage is changing...

Great point! I hadn't thought of that, but now I see it. Does that mean Class A amps make poor current sources, say a power amplifier? Fortunately, I'm mainly interested in small-signal voltage amplification with this circuit. The output of this will eventually go into a high-impedance guitar amp or another pedal.

Thanks for all the help!

 

Class A amps are poor efficiency and take more current, so are not used for power amps, class AB is better.

 

Is the input impedance largely determined by the Thevenin equivalent of R2 || R3?

R2||R3||βRe

Instead of calculating Ib and relying on a given value of β to predict the current (which isn't a good parameter to rely on in real transistors), try this:

1. Calculate Vb
2. Ve = Vb+.65 (approximately)
3. Ie = (Vs-Ve)/Re
4. Gain ~ Rc/Re

That's called "forced β", because the actual voltage gain is not dependent on the transistor current gain β (mostly).

 

Old stomp boxes were built with bipolar transistors because they were, "good enough" and a lot of designs simply discarded the high harmonics. 4 instance, the LBP-1 (or is that LPB?) used a 10 k input impedance and a 100k pot on the output, thus reinstating the impedance mis-match between the guitar and the guitar cable.

You want a simple amp that matches impedances and works across the audio frequency band? Try a j-fet.
Here's one I did, but it's way more sensitive and way more gain that most effects need. That means you can discard some of the parts.

Crutschow did one that resembles this, but I can't find it right now.

 

Always liked bootstrap circuits...

ak

 

You want a simple amp that matches impedances and works across the audio frequency band? Try a j-fet.

No, I want to learn about amplifier design. First using BJT's, then later FET's. Also, I want to learn the basics of Class A design before getting into Class B/AB design. After that, I'll probably just tinker around with op-amps, mostly in the audio frequency spectrum.

R2||R3||βRe

Instead of calculating Ib and relying on a given value of β to predict the current (which isn't a good parameter to rely on in real transistors), try this:

1. Calculate Vb
2. Ve = Vb+.65 (approximately)
3. Ie = (Vs-Ve)/Re
4. Gain ~ Rc/Re

That's called "forced β", because the actual voltage gain is not dependent on the transistor current gain β (mostly).

Thanks Veracohr. The dilemma I'm having is I don't have characteristic charts for these old transistors I'm playing with and I've read that using those charts and constructing a load line (the textbook method) isn't very practical in the real world, so where does that leave me?

Your 4 steps above make perfect sense except what value of Vb should I use to begin with? In a Class A amp, I'm trying to achieve a q-point where Vc is about half way between my rails, Vs and ground. Isn't that correct? To achieve that voltage across Rc, I need a certain Ic. Trying to figure out what value of Ib that gets me that Ic without using β or characteristic curves is where I'm lost.

Thanks again.