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
Hi all,
Is anybody knows how to compute resistors RE (Q1 Q2) in this schematics
It's a 70W PowerAmp final stage. As anyone already noticed, RE are placed in Emitters nodes insuring thermal stability.
Many Thanks,
Habib
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
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.
Emitter swamping resistors are typically 0.1 or 0.2 ohms in that type of application.
What is the lowest load impedance that you want to drive 70 watts into? In a typical audio amplifier, that may be four ohms or even two ohms. Lets assume four ohms for the sake of discussion. The peak power is twice the average power = 2 X 70 = 140 watts. The peak current therfore is: Ip = sqr(P/R) = sqr(140/4) = 5.92 Amps. The reason the peak is considered is because that is what the amplifier has to deliver, anything less and the amp will clip if you want it to reach full power. The current in the emitter resistors will be even higher depending on the transistor beta. Lets assume 6 amps. Now if you let the voltage across those resistors be a couple of volts at max current, that voltage will totally dominate the little re in the transistor and the amp will be thermally stable. You can run the thermal calculations and prove that is true. Of course, the transistors have to be appropriatley heat sunk. Then, Re = 2/6 = 0.5 ohms. The power required for the resistors will be 2^2(.5)/2 = 1 watt. Use 2 watt or 5 watt resistors. Now, the peak voltage required on four ohms for 70 watts average is: Vp = sqr(Pp*R) = sqr(140*4) = 23.6 volts. For an 8 ohm load, the peak voltage is: Vp = 33.5 volts. These voltages imply that the 80 volt rails are excessive for a 70 watt amplifier, at least for an 8 ohm load. All of that extra voltage has to be dropped in the transistors and creates heat. You shouldn't need any more than +- 36 volts for a 70 watt amplifier. Bob
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*Icand 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.
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
placed
becomes:
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.
With a 50 ohm load- the peak Ic is only 1.7A for 70W load power.
In article , Fred Bloggs wrote: [...]
What link worked? Neither worked for me when I tried them.
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
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.
In article , Fred Bloggs wrote: [....]
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
In article , John Larkin wrote: [...]
You could thermally stablize the driver/output combination by using diodes to set the bias point of the driver and thermally coupling them to the output devices. Beware that a sloppy coupling can lead to either thermal run away or thermal bounce/oscillation.
Thermal bounce/oscillation is caused when the thermal lags get long enough that the stablization diodes are too late in correcting. If you suddenly load the output, both the drivers and output devices heat up. The diodes the diodes have to correct for both but the drivers being hot causes the output devices to continue to heat up a little bit longer. This means that the system will overshoot thermally. This can, in some cases, make the distortion numbers go bad and then recover. I've never actually seen thermal oscillation but in theory it could happen too.
-- -- kensmith@rahul.net forging knowledge
I don't believe the load is 50 ohms. That resistor is a minimum internal load when no external load is applied. The 80 volt rails are not high enough to deliver 70 watts into 50 ohms. It takes about 84 volts with no drops or droops. Furthermore, a 50 ohm load is not a typical audio power load so, I don't assume that is the load without more information. One can check the literature on power amplifiers and observe that these resistors are typically between about 0.2 ohms and one ohm at that power level. Above, I figured 0.5 ohm would do the job. Three point nine ohms would definitely be thermally stable but would dissipate way to much power for a four ohm load. Bob
Well you know it's way too much voltage for 70W into even 8 ohms which is 33V. But as you say, it's only 64W into 50 ohms. So who knows. Maybe he took that 0.1 ohm suggestion and ran with it- the transistors will run with that too-) The spread on quiescent collector current is
1+Rb/(Hfe,min*Re) so that will be something like 10:1 at Re=0.1 ohm, versus 1.25:1 at Re=3.9 ohms.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.
-- -- kensmith@rahul.net forging knowledge
If I have not made a mistake, I calculated your power in Zch to be 53 watts with 80V rails, 4 ohms emitter resistors, and allowing 1V drop across the transistor.
John
Le Sun, 23 Jan 2005 04:52:39 +0000, Bob Eldred a écrit :
Power ultrasonic transducer. That resistor is a minimum internal
mmhh 64W in theory ... I expect 60W in Zch. It takes about 84 volts with no drops or
This is not an audio PowerAmp
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