Resistor Computation

[Paul Burridge]

On Sat, 22 Jan 2005 12:41:45 -0500, John Popelish <jpopelish@rica.net>
wrote:

There is no single, best answer. Higher resistance improves
stability, but uses up supply voltage (limits output swing) and
consumes power, alters the frequency response, raises the output
impedance and lowers the loop gain inside the final voltage feedback
loop. The best resistance depends on how you value several other
factors. I would guess that you would want to consider values well
below 50 ohms.

It's been a long while since I studied this type of amp, but isn't it
quite common to overcome these problems by incorporating current
sources at these points in the circuit? :-|

--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Paul Burridge]

On Sun, 23 Jan 2005 03:47:11 +0000 (UTC), kensmith@green.rahul.net
(Ken Smith) wrote:

[snip]

It's an unpredictable world in audio power amp design! Where's Phil
Allison when you need him? ;->
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[John Larkin]

On Tue, 25 Jan 2005 17:44:04 +0000, Paul Burridge
<pb@notthisbit.osiris1.co.uk> wrote:

On Sun, 23 Jan 2005 03:47:11 +0000 (UTC), kensmith@green.rahul.net
(Ken Smith) wrote:

[snip]

It's an unpredictable world in audio power amp design!

The amps are relatively simple. It's the audio designers that are
unpredictable.

John

 

[John Popelish]

habib bouaziz-viallet wrote:

Hi all,

Is anybody knows how to compute resistors RE (Q1 Q2) in this schematics
http://cjoint.com/?bwjsabJB2F

It's a 70W PowerAmp final stage. As anyone already noticed, RE are placed
in Emitters nodes insuring thermal stability.

Many Thanks,

Habib

There is no single, best answer. Higher resistance improves
stability, but uses up supply voltage (limits output swing) and
consumes power, alters the frequency response, raises the output
impedance and lowers the loop gain inside the final voltage feedback
loop. The best resistance depends on how you value several other
factors. I would guess that you would want to consider values well
below 50 ohms.

--
John Popelish

 

[habib bouaziz-viallet]

Le Sat, 22 Jan 2005 17:36:16 +0000, Ken Smith a écrit :

In article <pan.2005.01.22.17.12.45.733277@aldebaran.systems>,
betula electronique <betula@aldebaran.systems> wrote:
[...]
Is anybody knows how to compute resistors RE (Q1 Q2) in this schematics
http://cjoint.com/?bwjsabJB2F
[...]
try this pointer instead : http://cjoint.com/data/bwjsabJB2F.htm

Neither link worked, but here's what I think you want to know.

Vbe of a transistor changes 2.2mV per degree C.

Assume that the Vbe of Q1 has just changed by 2.2mV.
The current in Q1 will increase by 2.2mV/Re.
i.e dIc/dtheta = 2.2mV/Re OK for that.
The power in Q1 will increase by V(Q1)*(2.2mV/Re)

You can work out the thermal resistance from Q1's junction to the air by
adding up all the thermal resistances along the way. Lets call this total
"K" so we can say:


rise = K * V(Q1)*(2.2mV/Re)

For things to be stable, a decrease in Vbe of 2.2mV must not cause an
increase in temperature of another degree thereby causing another 2.2mV
and so on.


1 > K * V(Q1)*(2.2mV/Re)


Re > K * V(Q1)*(2.2mV/1)


The thermally caused reduction in Vbe still will cause the rise due to
temperature to be larger than it would otherwise be. It increases by a
factor of:


1/(1 - K * V(Q1)*2.2mV)



I hope this helps.

--

RE > Rtherm(jonction-case-air) * Vce(Q1) * 2.2mV

Brilliant Ken ! Thank you very much.

Habib

 

[Ken Smith]

In article <pan.2005.01.22.17.12.45.733277@aldebaran.systems>,
betula electronique <betula@aldebaran.systems> wrote:
[...]

Is anybody knows how to compute resistors RE (Q1 Q2) in this schematics
http://cjoint.com/?bwjsabJB2F
[...]
try this pointer instead : http://cjoint.com/data/bwjsabJB2F.htm

Neither link worked, but here's what I think you want to know.

Vbe of a transistor changes 2.2mV per degree C.

Assume that the Vbe of Q1 has just changed by 2.2mV.
The current in Q1 will increase by 2.2mV/Re.
The power in Q1 will increase by V(Q1)*(2.2mV/Re)

You can work out the thermal resistance from Q1's junction to the air by
adding up all the thermal resistances along the way. Lets call this total
"K" so we can say:


rise = K * V(Q1)*(2.2mV/Re)

For things to be stable, a decrease in Vbe of 2.2mV must not cause an
increase in temperature of another degree thereby causing another 2.2mV
and so on.


1 > K * V(Q1)*(2.2mV/Re)


Re > K * V(Q1)*(2.2mV/1)


The thermally caused reduction in Vbe still will cause the rise due to
temperature to be larger than it would otherwise be. It increases by a
factor of:


1/(1 - K * V(Q1)*2.2mV)



I hope this helps.

--
--
kensmith@rahul.net forging knowledge