BJT Class A/B Bias

[kurtlives]

So I have this power amplifier


It is being fed by a BJT buffer.
I am wondering how do you calculate the value of the two resistors (R10 & R12)?

Thanks!

[PRR]

http://www.diystompboxes.com/smfforum/index.php?topic=94494.msg815313#msg815313
"...anything that can get typed in here extemporaneously is a serious understatement"

Did you do the assigned reading?

Plagiarize/study the Lil Tiger.

Why do you need R10 R12?

How did you come up with "1 ohm"?

The two 1 ohm resistors is a 0.5 ohm load on your buffer; if it can drive that, it can surely drive the 4 ohms directly.

5V-0.6V across 1 ohm is 4.4 AMPERES; 1N4148 is rated 0.1 Amps.

Why do you have two 100uFd caps?

Why have caps at all?

These questions lead to more questions.

> how do you calculate the value of the two resistors (R10 & R12)?

Mindless cheat: if you don't need efficiency or high power, use a little less than Rl * hFE.

[kurtlives]

Yes I read most of that, thank you.

The 1R is default when I enter resistors into the simulation program.


This AB stage I am basing this design on had the caps. The input to the power amp is coming from a buffer so I figure I need to block the DC from the emitter.

Question 7: (similar but class B)
http://www.allaboutcircuits.com/worksheets/bjtamp_b.html

[PRR]

> 1R is default when I enter resistors into the simulation program.

The simulator is NOT a designer. You tell it what to do. The "default" values in the schematic editor are always not-what-you want. (And I have been burned a few times by forgetting to set values ASAP.)

> Question 7: http://www.allaboutcircuits.com/worksheets/bjtamp_b.html

Top of that page:

* choosing resistor values high enough to make damage to any active components unlikely.
* Mathematically analyze the circuit, solving for all voltage and current values.

Is 1r "high enough" to reduce damage? 5V/1r is 5 Amps, 5V at 5A is 25 Watts, these resistors will be large or burnt. If all that 5 Amps flows into a 2N3055 and is multiplied by hFE >20, the 3055 current seems to be 100 Amps, which at 5V is 500 Watts. Take a 500W work-light and jam it inside a TO-200 package. It will burst from the heat.

What do you "need"? 5V in a 4 ohm load needs about 1 Amp from the transistor. To make that happen you need 1A/hFE or 50mA into the base. Over-simplifying, take 5V at 50mA, try 100 ohm resistors.

"Problem" then is that as the output rises toward the power rail, the voltage across the resistor drops, current drops, until it is unable to supply the base current needed to flow load current. Assuming your '3055 actually has hFe as low as 20, it will pull up about 2V of the available 5V. Half of your precious power supply power goes un-used.

Question 9 hints a way to improve this.

It also shows the single coupling cap. If you assume the bias diodes are "constant voltage", then you can drive at either end or in the middle; driving both ends is redundant. (They are not truly constant-voltage to AC signals, and this opens some side-questions; however if you have a bi-polar supply you can design-out most coupling caps so I don't want to go down that path.)

[SISKO]

I think he just picked up the resistor from the library and connected them, then took a shoot of the screen and pasted it here to show us how they are connected. But in my opinion, the 1 ohm value should not be taken into account. In fact he is asking how do you calculate those values.
It may be better if he would had  rubbed off the 1ohm value, but its ok.

Plesas, if Im wrong, and Kurtlives, you DID wanted to put 1 ohm value, then dont listen to me.

Btw, I dont know how to calculate them, but you may need and medium current transistor between your low current transistor and the 3055s

[R.G.]

It is being fed by a BJT buffer.
I am wondering how do you calculate the value of the two resistors (R10 & R12)?

First, look at the transistor datasheet. Referring to the On Semi datasheet for the 2N3055/MJ2955:
- for a 4 ohm load, if the output transistors could saturate perfectly, that would put 5V across 4 ohms. The resulting current would be 5/4 = 1.2A.
- looking at page 3, figures 5 and 6, at 1A, the collector current only gets down to roughly 0.4V before base current starts skyrocketing, indicating entry into the beginnings of saturation. So you won't really get 5V out, but 4.6V. So the actual max output current would be 1.15A.
-  but looking at figures 7 and 8, the base-emitter voltages are about 0.8V at 1A, so you can't get to 4.6V out, but more like 4.2V. And that's only if your base drive can go all the way to the +(and -) power supply.
- From figures 4 an 5, gain is "typically" 75 at 1A. So you need 1A/75 = 13.3ma into the base to do this. Probably more, but we'll look. If the drive is from the + and - 5V supplies like you show through a resistor, then you have 13.3ma flowing through 0.8 ohms. That's 0.104 ohms.

So 1 ohm is optimistic. It will not drive the base to full output, because it can't supply the current at full output. You need something like 0.1 ohms. The problem is that the two 0.1ohm resistors are there all the time, so the current they let flow from +5 to -5 is 10V/0.2ohm = 50A. There's a problem here, letting 50A flow to get 13ma into a base.

But maybe the BJT buffer is really supplying all the base current into the two bases. In that case, the actual current from the resistors only needs to be enough to keep the diodes on. So maybe they can be (5V-0.6V ) divided by - uh, how much?

It's somewhat arbitrary. Call it the same as max base current, about 15ma, maybe. So that's 4.4V/0.015 = 293 ohms; go for 270. These resistors will dissipate about
4.4*4.4/270 = 72mW.

I'd start there.

But this moves the base current problem to the unshown BJT buffer. Can that buffer really put out 13ma to a base and 33ma into the unused bias resistor? That's a pretty stout buffer, especially if it is not itself a Class AB buffer.

The comments are largely correct - this approach has practical issues that mean it's going to be tough to get to work. Another stage of transistors between the signal and outputs is probably going to be needed as a practical matter.