Transistor Amplifier Design Methods

[Dylfish]

Hey Guys,

When designing single stage amplifiers (lets say class A with a NPN) are you able to start off by using a load line / operating point chart from the datasheet and then derive the rest of the details and values from that? (obviously if you know how you want it biased and such)

I'm not 100% on all the maths yet, just a question that popped to mind since you can basically derive Ic,Ib and use Vcc as a starting point? Im still brushing up on all my Ohms law and KVL (been a while)

Cheers

Sorry if i'm vague.

[Resynthesis]

Quick overview of one approach

[url]http://www.electronics-tutorials.ws/amplifier/amp_2.html[url]

[Joe]

Scratched this out at work today:

r1 = lower bias resistor
r2 = upper bias resistor
r3 = emitter resistor
r4 = collector resistor

How to get the results of a given resistor arrangement:
Alpha = Beta / (Beta + 1)

Vbias = (r1 / (r1 + r2)) * Vcc
Rbias = (r1 * r2) / (r1 + r2)

Ie = (Vbias - 0.7) / (r3 + Rbias / (Beta + 1))
Ib = Ie / (Beta + 1)
Ic = Alpha * Ie

Vc = Vcc - (Ic * r4)
Vb = 0.7 + (Ie * r3)
Ve = Vb - 0.7


Example going backwards, choosing a 10k collector resistor and 2.2k emitter resistor:
Vcc = 9
Beta = 500
Alpha = .998

Vc = 4.5

Ic = (Vcc - Vc) / 10k = .45mA
Ie = Ic / Alpha = .4509mA
Ib = Ie / (Beta + 1) = 0.0009mA

Ve = (Ie * r3) = .9919V
Vb = 0.7 + Ve = 1.6919V

Now the biasing details can be estimated, there is no single solution however:

Vbias = Vb + 1.0V = 2.6919V (Can be anything above Vb, just adding a volt for the example.)
Rbias = ((Beta + 1) * ((Vbias - 0.7) - (Ie * r3))) / Ie = 1111K

Then extrapolate to 2 resistor values that when in parallel would equal 1111K and also provide the correct bias point:
r2 = 1111K / (Vc / Vcc) = 3714K
r1 = 1111K / ((Vcc - Vc) / Vcc) = 1585K

The problem is that datasheets will give some wide beta range, however some resistor arrangements are more tolerant to these differences than others.

[PRR]

Plagiarize.

Then "design" is in _understanding_ what each plan does, and stealingselecting the best for your use.

Design starts from REQUIREMENTS.

What output do you need? Load impedance, voltage, power?

What input do you have? Impedance, voltage, power?

What power supply is available? How much does it cost?

Will it "work" with any part you pluck from the bin? Or is your labor so cheap you can afford to trim every build?

Space and flavor may matter.

> single stage amplifiers

The requirements may be easier to meet with a 2-stage design. This is especially true at guitar-cord impedances.

Load-lines are for learning. Once you have the concept you don't draw lines.

[Dylfish]

Thanks Guys,

This might be a silly question but how did you come up with the resistor values for the Collector resistor and the Emmiter Resistor before anything else? Vc would be about 1.6v, but wouldn't we want it to sit around 4.5v so the AC can swing + & - without clipping?

To be honest im at work at the moment so i need to sit down after and go through this closer.

p.s am i getting vc and vce mixed up?

Cheers

[Joe]

If you run the resistor values from the example back into the first set of equations, you get 4.5V at the collector where it needs to be. The choice of Rc/Re is pretty close to the voltage gain (i think...) so 10K/2.2K = ~4.5x gain (no bypass cap).

[Dylfish]

ahh ok sweet cheers,

so you can roughly gauge your gain by chosing the ratio between Rc & Re and then work other values around there.

There seems so many ways of being able to work it out that i'm seeing differnt methods and it's been confusing me somewhat. 

Thanks for all your help so far guys. any other info that might help is very much appriciated

Cheers!

[Dylfish]

Can anybody explain why most designs i've seen for a NPN single stage usually start with 10k resistor on Rc?

"Most" designs are using 9v leaving Ic at 0.9ma. From what I've read Ic should be chosen depending on Noise, Heat and Beta being dependent on Ic. Doesn't choosing 10k as a (from what I can I can assume) a default/starting value limit options and the biasing of the output operating point?

Thanks 

[Joe]

Correction:

Then extrapolate to 2 resistor values that when in parallel would equal 1111K and also provide the correct bias point:
r2 = 1111K / (Vbias / Vcc) = 3714K
r1 = 1111K / ((Vcc - Vbias) / Vcc) = 1585K

Sorry!

About you're last question I'm not sure, old habits I guess. Maybe 10K is a decent balance between power draw, gain, and output impedance? I think a lot of the temperature effects are minimized when there is a fairly large emitter resistor, which is why you often see at least 1K in that position and a bypass cap for more gain rather than grounding the emitter.

[Dylfish]

yeah I wasn't sure if it was a habit thing everyone seems to do or if theres a reason behind it. If anyone else has a theory i'm more than open to it

[Keppy]

I think a lot of the temperature effects are minimized when there is a fairly large emitter resistor, which is why you often see at least 1K in that position and a bypass cap for more gain rather than grounding the emitter.

Unlike tubes, transistors that are amplifying conduct through the base-emitter junction. Without a larg-ish emitter resistor, the input impedance drops dramatically. In fact, the input impedance of the transistor (neglecting the biasing) is equal to the hfe of the transistor times the emitter resistance. A grounded emitter still has some emitter resistance within the transistor, but the impedance still gets low enough to load down many signals.

A 10k resistor x 300hfe = 3M input impedance. This is high enough to ensure that the input impedance becomes more a function of the bias resistors. In other words, the input impedance of the transistor itself has been made irrelevant.

Someone please make sure I got all that right.

[dwmorrin]

Can anybody explain why most designs i've seen for a NPN single stage usually start with 10k resistor on Rc?

Depends on what circuit were talking about, I'd venture to say 10k is on the high side and it's for stages with high gain.  And it's not necessarily the "start" of the design.

"Most" designs are using 9v leaving Ic at 0.9ma. From what I've read Ic should be chosen depending on Noise, Heat and Beta being dependent on Ic.

If the transistor is "active" (not cutoff or saturated) then Ic should be set by the emitter current.  That is, it's somewhat backwards to consider the Ic=9V/Rc equation... the Ic=(Vb-0.6V)/Re equation comes "first" in my mind.  The design limitations for Rc are that it is large enough to develop the right amount of voltage gain, but small enough to keep the transistor from saturating (at least at idle conditions).

Noise and beta become less of an issue if you use more current, with lower value resistors and higher voltage.  That's how you'd design a premium, discrete pro audio/hi-fi circuit, but not necessarily issues that are so important for stompbox purposes.  Maximum gain seems to be the primary goal of many stomp circuits.

Doesn't choosing 10k as a (from what I can I can assume) a default/starting value limit options and the biasing of the output operating point?

You're limiting your options primarily by choosing a single transistor stage.  There's only so much poking and prodding of those 4 transistors (assuming the circuit in the 2nd post) you can do.  It's inherently restricted to a narrow range of values that "work."

[PRR]

Here is an extract from an early (1951) paper on junction transistor circuit properties.



http://www3.alcatel-lucent.com/bstj/vol30-1951/bstj-vol30-issue03.html

Even at the beginning, 10K is a happy load for first-scratch design with junction transistors. In other work I've used from 15 ohms to 2Meg, "dictated... by power output required". 

[Dylfish]

Wow, Great post! thanks