commom emitter amplifier

[shaw]

8) Is Av [ voltage gain ] depend on beta [current gain]?

if yes then how?

how can we relate them?

 

[Russlk]

Iin = ein/hie
Io = B*Iin
eo = Io*Rl
eo/ein (gain) = B*Rl/hie

 

[audioguru]

You need to bias the transistor so that the circuit works with any beta. A transistor without negative feedback is so distorted at high levels anyway that beta doesn't matter.
You can get high gain and pretty good linearity from a transistor that has a high impedance current source for a collector resistor. Bootstrapping the collector resistor is nearly the same. Then the output impedance is high possibly requiring an emitter-follower as a buffer.

The sim on the left is a transistor without negative feedback. Its 40% distortion is so high that its high gain can't be measured, unless its input was reduced to its noise level.
The sim on the right has negative feedback due to its unbypassed emitter resistor and has a gain of slightly less than 10 and fairly low distortion.

 

[ljcox]

8) Is Av [ voltage gain ] depend on beta [current gain]?

if yes then how?

how can we relate them?

You are asking a question that is difficult to answer in general. It depends upon the configuration, eg. common emitter, common base, etc. and whether there is voltage feedback, current feedback (or no feedback).

The aim of transistor circuit design is to minimise the effect of beta on the gain and DC operating point. This is bacause beta varies widely between transistors of the same type (buy 10 and measure their betas) and because Beta varies with temperature.

 

[DigiTan]

I think he's referring to the type of amp audioguru posted where you have the base biased and resistors on the collector and emitter. Russlk's equations look dead-on, just don't forget to also factor in Rc after you apply superposition.

 

[shaw]

thanks

but i'm still confused
can any of you tell in easy words

 

[Russlk]

Output current multiplied by load resistance = output voltage
Input current multiplied by beta = output current
input voltage divided by input impedance = input current
gain = output voltage/ input voltage

 

[audioguru]

Output current multiplied by load resistance = output voltage
Input current multiplied by beta = output current
input voltage divided by input impedance = input current
gain = output voltage/ input voltage

Yes, but higher beta also causes a higher input impedance, resulting in lower input current which cancels any increase of gain.

Ages ago, John Lindsey Hood (a Brit dude) invented an extremely high gain (20,000?) simple transistor circuit called "The Liniac". It is very linear. It has a transistor with its collector resistor bootstrapped, and a darlington emitter-follower transistor at the output. I can't remember if John or someone else later replaced the bootstrapped collector resistor with a high impedance current source transistor. It was in Wireless World magazine in about 1974 and is copywritten so I can't copy it and post it here.

 

[Nigel Goodwin]

Ages ago, John Lindsey Hood (a Brit dude) invented an extremely high gain (20,000?) simple transistor circuit called "The Liniac".

He's invented LOADS of stuff, he's one of the most influential designers on the audio scene.

 

[audioguru]

Ages ago, John Lindsey Hood (a Brit dude) invented .....

He's invented LOADS of stuff, he's one of the most influential designers on the audio scene.

He even heated his home in winter with a nice class-A valve amp. :lol:

 

[Russlk]

Audioguru wrote:
Yes, but higher beta also causes a higher input impedance, resulting in lower input current which cancels any increase of gain.

I don't think that is so, because hie = rb + re/(1-a) Where a=B/(1+B) ~ 1 therefore hie = rb, not significantly a function of B.

 

[Roff]

Audioguru wrote:
Yes, but higher beta also causes a higher input impedance, resulting in lower input current which cancels any increase of gain.

I don't think that is so, because hie = rb + re/(1-a) Where a=B/(1+B) ~ 1 therefore hie = rb, not significantly a function of B.

B=a/(1-a)
a/(1-a) ~ 1/(1-a) ~ B when a ~ 1 (B>>1)
hie = rb + re/(1-a)
therefore hie ~ rb+B*re
rb is usually << B*re
therefore hie~B*re

 

[Russlk]

Well, I guess that disproves the old adage "figures don't lie"!

 

[Roff]

Well, I guess that disproves the old adage "figures don't lie"!

I guess. I think you had a brain fart going from a ~ 1 to concluding that re/(1-a) ~ 0 (or re?), when if fact re/(1-a) ~ re*B.
But I still don't know what this has to do with voltage gain. With a zero impedance source (implied when measuring voltage gain), input resistance doesn't matter. The dependence of voltage gain on B is basically only due to the fact that Ic = a*Ie = Ie*B/(B+1).