Resistor Computation

[habib bouaziz-viallet]

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

 

[Paul Burridge]

On Sat, 22 Jan 2005 13:11:59 +0100, habib bouaziz-viallet
<habib@mynewserverposfix.com> 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.

Can't get the page up for some reason, but if it's what I think you're
talking about, they're probably power resistors of between 0.25 - 1
Ohm in value.
--

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

 

[Charles Schuler]

Can't be. There's no way Re can be 12 ohms in a 70 watt amplifier or more
power would be dissipated there than in the load. The purpose of an
amplifier is to deliver power to a load, not heat rooms.
Bob

Emitter swamping resistors are typically 0.1 or 0.2 ohms in that type of
application.

 

[Fred Bloggs]

Bob Eldred wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41F2A187.3090100@nospam.com...


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

The main transistor parameters sensitive to junction temperature are
Hfe, Icbo, and Vbe- all of which contribute to thermal runaway of
collector current and transistor power dissipation. If you call that
base resistor Rb, the power transistor base current Ib and collector
current Ic, and the base drive current from the push-pull Id, then you
have the following mathematical relationships:

summing currents at base node:

Id + Icbo -(Vbe+ (Hfe +1)*Ib*Re)/Rb - Ib=0

and the collector current :

Ic= Icbo + Hfe*Ib

then substituting for Ib=(Id+Icbo-Vbe/Rb)/( (Hfe+1)*Re/Rb +1)

Ic= Icbo +Hfe*(Id+Icbo-Vbe/Rb)/( (Hfe+1)*Re/Rb +1)

or

Ic= Icbo + Hfe*( Rb*(Id+Icbo)-Vbe)/((Hfe+1)*Re+Rb)

This is differentiated wrt Icbo, Hfe, and Vbe to yield:


d(Ic)/d(Icbo)= 1 + Hfe*Rb/((Hfe+1)*Re+Rb)= S(Icbo)

d(Ic)/d(Vbe)= - Hfe/((Hfe+1)*Re+Rb)=S(Vbe)

d(Ic)/d(Hfe)=( Rb*(Id+Icbo)-Vbe)*(Re+Rb)/((Hfe+1)*Re+Rb)^2 =S(Hfe)

In most practical cases, the above sensitivities will simplify to:

S(Icbo)=Hfe

S(Vbe)=-1/(Re+Rb/Hfe)

S(Hfe)=Ic*(Re+Rb)/(Hfe*(Re*Hfe+Rb)) worst case.

If you now specify that a 100oC increase in junction temperature yields
no more than a fractional increase of Ic by F*Ic, then the equation

becomes:

1000*Icbo*S(Icbo)- 0.2*S(Vbe)+ Hfe*S(Hfe)< F*Ic

and in most practical cases ( for Si anyway), the 1000*Icbo*S(Icbo) is a
negligible fraction of the fractional Ic increase so that it can be
discarded. This allows the inequality to be simplified to :

Re> (0.2 +Ic*Rb*(1-F)/Hfe)/(Ic*(F-1/Hfe))

Then for example if Ic=1 Amp, Hfe=10, and Rb=100 ohms with F specified
as 0.5, you have Re must be greater than Re> ~12 ohms. Double check the
equations, and consult a discrete transistor amplifier design text to
see where all this is coming from.



Can't be. There's no way Re can be 12 ohms in a 70 watt amplifier or more
power would be dissipated there than in the load. The purpose of an
amplifier is to deliver power to a load, not heat rooms.
Bob

I pulled those example numbers out of the air and did not mean they
should apply to the actual circuit. I did some calcs on his circuit
which looks like 70W peak into 50 ohms. Then assuming an Hfe of 50,
Icbo(25oC)=100nA, Ft=150Mhz,and absolutely no more than Ta=70oC, the
circuit is absolutely thermally stable at Re=3.9 ohm, and maximum per
transistor power dissipation of 15W, allowing for IcQ=10% xIpk at
Vout=0VDC. The 3.9 ohms should be 10W rating. Rb should be 47 ohms or so
for these Ft and 100KHz bandwidth.

 

[Fred Bloggs]

Charles Schuler wrote:

Can't be. There's no way Re can be 12 ohms in a 70 watt amplifier or more
power would be dissipated there than in the load. The purpose of an
amplifier is to deliver power to a load, not heat rooms.
Bob


Emitter swamping resistors are typically 0.1 or 0.2 ohms in that type of
application.

With a 50 ohm load- the peak Ic is only 1.7A for 70W load power.

 

[John Larkin]

On Sat, 22 Jan 2005 18:55:11 GMT, Fred Bloggs <nospam@nospam.com>
wrote:

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

The main transistor parameters sensitive to junction temperature are
Hfe, Icbo, and Vbe- all of which contribute to thermal runaway of
collector current and transistor power dissipation. If you call that
base resistor Rb, the power transistor base current Ib and collector
current Ic, and the base drive current from the push-pull Id, then you
have the following mathematical relationships:

summing currents at base node:

Id + Icbo -(Vbe+ (Hfe +1)*Ib*Re)/Rb - Ib=0

and the collector current :

Ic= Icbo + Hfe*Ib

then substituting for Ib=(Id+Icbo-Vbe/Rb)/( (Hfe+1)*Re/Rb +1)

Ic= Icbo +Hfe*(Id+Icbo-Vbe/Rb)/( (Hfe+1)*Re/Rb +1)

or

Ic= Icbo + Hfe*( Rb*(Id+Icbo)-Vbe)/((Hfe+1)*Re+Rb)

This is differentiated wrt Icbo, Hfe, and Vbe to yield:


d(Ic)/d(Icbo)= 1 + Hfe*Rb/((Hfe+1)*Re+Rb)= S(Icbo)

d(Ic)/d(Vbe)= - Hfe/((Hfe+1)*Re+Rb)=S(Vbe)

d(Ic)/d(Hfe)=( Rb*(Id+Icbo)-Vbe)*(Re+Rb)/((Hfe+1)*Re+Rb)^2 =S(Hfe)

In most practical cases, the above sensitivities will simplify to:

S(Icbo)=Hfe

S(Vbe)=-1/(Re+Rb/Hfe)

S(Hfe)=Ic*(Re+Rb)/(Hfe*(Re*Hfe+Rb)) worst case.

If you now specify that a 100oC increase in junction temperature yields
no more than a fractional increase of Ic by F*Ic, then the equation becomes:

1000*Icbo*S(Icbo)- 0.2*S(Vbe)+ Hfe*S(Hfe)< F*Ic

and in most practical cases ( for Si anyway), the 1000*Icbo*S(Icbo) is a
negligible fraction of the fractional Ic increase so that it can be
discarded. This allows the inequality to be simplified to :

Re> (0.2 +Ic*Rb*(1-F)/Hfe)/(Ic*(F-1/Hfe))

Then for example if Ic=1 Amp, Hfe=10, and Rb=100 ohms with F specified
as 0.5, you have Re must be greater than Re> ~12 ohms. Double check the
equations, and consult a discrete transistor amplifier design text to
see where all this is coming from.

Thermal runaway is usually driven by the Vbe tempco, not by beta and
especially not by leakage current. Beta doesn't change that much with
temperature. Ken's math is right.

John

 

[Fred Bloggs]

Ken Smith wrote:

In article <41F2AE5B.1040600@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[...]

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

Brilliant Ken ! Thank you very much.

Habib

Yeah- I get Re>0.3*80*0.0022=53 milliohms - hmmmm, seems kinda low.


0.3 degrees per watt sounds a bit optimistic. It takes quite a lot of
surface area to get the thermal resistance that low for just a heat sink.
The junction to case of a TO-220 typically runs something like 0.7 degrees
per Watt. What sort of package are you assuming?

Most TO-3 are 0.3 Rj,c- okay so add 0.2 for the other interfaces and a
big heat sink- that makes 0.5*80*0.0022=0.09 ohm Re. Ic,Q will be nearly
indeterminate - I am assuming this is a linear amp.

 

[Ken Smith]

In article <41F2AE5B.1040600@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[...]

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

Brilliant Ken ! Thank you very much.

Habib

Yeah- I get Re>0.3*80*0.0022=53 milliohms - hmmmm, seems kinda low.

0.3 degrees per watt sounds a bit optimistic. It takes quite a lot of
surface area to get the thermal resistance that low for just a heat sink.
The junction to case of a TO-220 typically runs something like 0.7 degrees
per Watt. What sort of package are you assuming?



--
--
kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <41F2DDF0.6090300@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[...]

If you took the time to look at that circuit, which is a

What link worked? Neither worked for me when I tried them.

complementary current drive with voltage feedback- you would see there
is no way in hell the collector current can be determined.

My posting made no reference to the actual circuit because I couldn't see
it. I assumed, perhaps incorrectly that it was a complimentry emitter
follower boosted with a pair of current mirrors. This was based on my
reading between the lines of the OP's text. If I assumed the wrong stuff,
my calculations may not be worth the electrons they were painted on.



--
--
kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <csuv5a$he7$2@blue.rahul.net>,
Ken Smith <kensmith@green.rahul.net> wrote:
[...]

0.3 degrees per watt sounds a bit optimistic. It takes quite a lot of
surface area to get the thermal resistance that low for just a heat sink.
The junction to case of a TO-220 typically runs something like 0.7 degrees
..................^^^^ Sink ooops
per Watt. What sort of package are you assuming?
--

--
kensmith@rahul.net forging knowledge

 

[John Larkin]

On Sun, 23 Jan 2005 01:44:20 +0000 (UTC), kensmith@green.rahul.net
(Ken Smith) wrote:

In article <41F2DDF0.6090300@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[...]
If you took the time to look at that circuit, which is a

What link worked? Neither worked for me when I tried them.


complementary current drive with voltage feedback- you would see there
is no way in hell the collector current can be determined.

My posting made no reference to the actual circuit because I couldn't see
it. I assumed, perhaps incorrectly that it was a complimentry emitter
follower boosted with a pair of current mirrors. This was based on my
reading between the lines of the OP's text. If I assumed the wrong stuff,
my calculations may not be worth the electrons they were painted on.

No, you got it right. But it's a bit unusual in that the output q's
don't feed back into the drivers emitters, so it has the potential for
thermal runaway in both the driver and the output stages.

It's embarassing to design a nice stable output stage and then have a
driver stage blow up on you.

John

 

[Ken Smith]

In article <41F303C8.8030001@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[....]

His output is complementary CE with emitter degeneration- collectors
tied to GND'ed load. The CE are driven by complementary CE with emitter
degeneration also- voltage feedback through divider from output to
junction of input emitter resistor junction. It was the output CE Re's
the OP wanted. The original link worked fine in my browser.

If I'm imagining it right, yuck. Without the diodes to make the CE
outputs into something like current mirrors, the circuit will have bad
distortion and stability troubles. Without good stablization of the
driver stage, there can be thermal run away because it gets heated.

--
--
kensmith@rahul.net forging knowledge

 

[habib bouaziz-viallet]

Le Sat, 22 Jan 2005 19:49:51 +0000, Fred Bloggs a écrit :

habib bouaziz-viallet wrote:
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

Yeah- I get Re>0.3*80*0.0022=53 milliohms - hmmmm, seems kinda low.

Hi Fred,

I'm quite disagreed with that calculation.

Assuming 70W dissipated in Zch (load) and 20W in Transistors Q1, Q2 (10W
each) then

Rthja(Q1) = (100-30)/10 ; 100 jonction temp, 30 amb temp, 10W dissipated
in Q1 ==>
Rthja(Q1) = 7oC/W (Rthja = Rthjonction-case-dissipator-air)

By Ken (formula verified by me after a couple of hours)
RE > 2.2mV x V x Rthja ; V=80V , Rthja=7
RE > 1.2 Ohm

Actually RE = 2 Ohm/3W is the choice.

Thank you all good guys !

Habib

 

[Ken Smith]

In article <41F3E188.8030305@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
[...]

But he also said that the rise will be greater by a factor of
1/(1-K*V(Q1)*2.2mV) due to Vbe variation, where I presume K=Rtheta/Re,
making your assumed 70oC rise now 70* 1/(1-(7/2)*80*0.022)=180oC?

Yes, you have to watch the peak temperature not just that in theory it
will stop siring somewhere.

--
--
kensmith@rahul.net forging knowledge

 

[John Larkin]

On Sun, 23 Jan 2005 13:49:22 +0100, habib bouaziz-viallet
<habib@mynewserverposfix.com> wrote:

Actually RE = 2 Ohm/3W is the choice.

Thank you all good guys !

Habib

Since you're driving a 50 ohm load, there's not a big efficiency
penalty for using a relatively large emitter resistor. 2 ohms sounds
safe.

But one caution: if you want to run class AB, with some modest
quiescent current, changes in the output transistor Vbe will change
the quiescent current. Your circuit has no Iq feedback or temperature
compensation. So if you set Iq to some nice value, and there's a small
corresponding quiescent voltage drop in the last emitter resistors,
then you run this thing at high power for a minute or whatever and
heat up the transistors, Iq may increase. That's worth a quick
numerical analysis.

John

 

[betula electronique]

Le Sun, 23 Jan 2005 11:57:30 -0800, John Larkin a écrit :

On Sun, 23 Jan 2005 13:49:22 +0100, habib bouaziz-viallet
habib@mynewserverposfix.com> wrote:

Actually RE = 2 Ohm/3W is the choice.

Thank you all good guys !

Habib

Since you're driving a 50 ohm load, there's not a big efficiency
penalty for using a relatively large emitter resistor. 2 ohms sounds
safe.

But one caution: if you want to run class AB, with some modest
quiescent current, changes in the output transistor Vbe will change
the quiescent current. Your circuit has no Iq feedback or temperature
compensation. So if you set Iq to some nice value, and there's a small
corresponding quiescent voltage drop in the last emitter resistors,
then you run this thing at high power for a minute or whatever and
heat up the transistors, Iq may increase. That's worth a quick
numerical analysis.

John

Hi John,

You are in right John. Temperature compensation is not represented in this
schematic but it exist. AA' potentiel decrease with rise of T (oC) (and
increase with fall of T (oC) ).

Habib

 

[Rich Grise]

On Sun, 23 Jan 2005 04:52:39 +0000, Bob Eldred wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:41F2C9A5.6020109@nospam.com...



I pulled those example numbers out of the air and did not mean they
should apply to the actual circuit. I did some calcs on his circuit
which looks like 70W peak into 50 ohms. Then assuming an Hfe of 50,
Icbo(25oC)=100nA, Ft=150Mhz,and absolutely no more than Ta=70oC, the
circuit is absolutely thermally stable at Re=3.9 ohm, and maximum per
transistor power dissipation of 15W, allowing for IcQ=10% xIpk at
Vout=0VDC. The 3.9 ohms should be 10W rating. Rb should be 47 ohms or so
for these Ft and 100KHz bandwidth.

I don't believe the load is 50 ohms.

Why not? That's what it says on his schematic:
http://cjoint.com/data/bwjsabJB2F.htm

Cheers!
Rich

 

[Ken Smith]

In article <41F43597.90700@nospam.com>, Fred Bloggs <nospam@nospam.com> wrote:
[...]

I simply mentioned that Rthaj in consideration in the formula
RE > 2.2mV x V x Rthja
come from the power dissipation in Q1 and not in Load Zch, that's all.

That formula is a very rough inequality. At the lower limit of equality,
Tj is run right up to the maximum. Normally you would have
Tj-Ta=Rtheta*VCE*(IQ +2.2mV*(Tj-Ta)/Re) which solves for
Tj-Ta=Rtheta*VCE*IQ/(1-2.2mV*Rtheta/Re) and that is where that formula
comes from.

I guess I should have been more clear about my explaination. Fred has the
right take on it. My RE > 2.2mV... thing is the boundary between where
the system will go into thermal run away without being provoked and where
it will stabilize to some temperature. It tells you the minimum size of
RE that is allowed not the one you must use.



--
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kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <pan.2005.01.23.22.47.03.926435@example.net>,
Rich Grise <richgrise@example.net> wrote:

On Sun, 23 Jan 2005 01:54:18 +0000, Fred Bloggs wrote:
[...]
The original is a .png - I've saved it as a .gif, and uploaded it:
http://www.neodruid.com/images/bwjsabJB2F_ClasseAB.gif

Thanks.


If distortion isn't an issue, this circuit could run the output stage in
class-C. If distortion is a worry, making Q1's and Q2's circuit more like
a current mirror by adding a couple of diodes looks worth it.

Do we know what the driver chip does to the A and A' lines?

If you place a series RC in parallel with B.R, the circuit can be more
well behaved. It works to lower the gain at high frequencies and then
provide some feedforward from the B point to the load at really high
frequencies when Q1, and Q2 are not making useful gain. This lets you
have a high bandwidth in the enclosing loop.

--
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kensmith@rahul.net forging knowledge

 

[habib bouaziz-viallet]

Le Mon, 24 Jan 2005 01:10:03 +0000, Bob Eldred a écrit :

"Rich Grise" <richgrise@example.net> wrote in message
newsan.2005.01.23.22.22.01.578566@example.net...
On Sun, 23 Jan 2005 04:52:39 +0000, Bob Eldred wrote:
I don't believe the load is 50 ohms.

Why not? That's what it says on his schematic:
http://cjoint.com/data/bwjsabJB2F.htm

Cheers!
Rich

The original poster somewhere above claims that the load is 50 ohms, though
that's not clear on the schematic. Absent other information, my reasoning
was like this: There is not sufficient voltage to develop 70 watts into 50
ohms. In all of my experience I have never heard of driving 50 ohms at
power. Signal level yes, but not 70 Watts. RF amplifiers also yes but not
audio which this clearly is. Audio power amplifiers are for speaker loads,
typically 4 or 8 ohms or for 70.7 volt constant voltage lines, not 50 ohms.
More importantly, this kind of an amplifier has an open loop gain that is a
strong function of load. Such a design is a disaster waiting to happen.
Everybody was concerned about thermal stablitity but feedback stability,
gain-phase margin is just as important and this particular circuit is a
troublemaker. It adds couple of poles to the transfer function that is a
function of load and may cause oscillation if not properly handled depending
on what's driving it. I figured the 50ohm load on the schematic was a
pre-load to reduce the variation normal loading might produce. But I give
up. It's a crappy amplifier an that's it.
Bob

Ooooh no Bob don't give up ! Carry on designing perfect Audio PowerAmp's
which is not at all my actual problem.

Habib