OK, with Mooly's welcome help, I have simulated the Quad 303 circuit in Spice. Using this solves the question discussed earlier of the discrepancy between the Vrms^2 / R calculation (70 Watts) and Quad's specification of 45 Watts. Here goes -
Total dissipation in output transistors and emitter resistors,goes like this: Q1=11.928W, Re1=0.8785W, Q2=11.822W, Re2=0.81318W. This lot totals 25.44168W. Power in the 8 ohm load is as Quad states - 44.627W. Adding the non-output dissipation to the load dissipation gives the formula's predicted total of 70.07W (formula gave 70.15W).
I now understand what's going on. The experts were right (as they were bound to be!). A lesson for me about the caution to be used when predicting with that formula. Thanks to all of them for their patience - and especially to Mooly for his help.
Total dissipation in output transistors and emitter resistors,goes like this: Q1=11.928W, Re1=0.8785W, Q2=11.822W, Re2=0.81318W. This lot totals 25.44168W. Power in the 8 ohm load is as Quad states - 44.627W. Adding the non-output dissipation to the load dissipation gives the formula's predicted total of 70.07W (formula gave 70.15W).
I now understand what's going on. The experts were right (as they were bound to be!). A lesson for me about the caution to be used when predicting with that formula. Thanks to all of them for their patience - and especially to Mooly for his help.
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Hello
greetings i have a chinese amp using 2sc5200 2sa1943 has 9 pairs i pair as drivers
problem is it heats a lot as dc supply is very high 120 volts +/- can mjl 21193 94 be used
to make the amp more reliable base resistors are 33 ohms emitter resistors .33 ohms can anyone please give me more advice
warm regards
andrew lebon
greetings i have a chinese amp using 2sc5200 2sa1943 has 9 pairs i pair as drivers
problem is it heats a lot as dc supply is very high 120 volts +/- can mjl 21193 94 be used
to make the amp more reliable base resistors are 33 ohms emitter resistors .33 ohms can anyone please give me more advice
warm regards
andrew lebon
The amount of heat will be exactly the same regardless of the transistors you use as this is purely a function of voltage across transistors x current through transistors, and this will hardly change by output transistor substitution. The real question is, can your transistors take the amount of heat they make. The answer is only partially in the transistor choice, and much more in heatsink choice, airflow etc. Amount of heat transforms into temperature dependent on hw good the heatsink is at transferring the heat to the surrounding air. The better, the lower the temperature of the heatsink - and it's the temperature that actually degrades and destroys components. That being said, the components themselves have their own capability of heat transfer. If you make more heat inside a component than it can transfer to the heatsink or surroundings, the temperature will rise (quickly) to dangerous levels, and the component will be destroyed.
The short version - it's not the transistors fault, the heatsink is too small or the airflow too low (use bigger heatsink and/or a fan)
Chimed in to answer the doubt I saw in my mail, but ilimzn was first with the accurate answer: power dissipated/lost will be the same, actual temperature depends both on amount of calories generated and how you dispose of them.
What I wanted to add is that at this power level it starts being justified going to a more efficient "Class", be it F/G/H (all still analog).
Splitting rail voltages in two will roughly halve dissipation.
A 1200W or larger Class AB2 amp *can* be built, it's physically possible, simply it ceases being practical.
What I wanted to add is that at this power level it starts being justified going to a more efficient "Class", be it F/G/H (all still analog).
Splitting rail voltages in two will roughly halve dissipation.
A 1200W or larger Class AB2 amp *can* be built, it's physically possible, simply it ceases being practical.
Looks like the "Chinese maker" grabbed the "inspiration" around here, or at "Red Circuits".
He was lucky
, he might have built:An externally hosted image should be here but it was not working when we last tested it.
bias that amp to classB and it will run cool at standby....why insist on running the 2sc5200 2sa1943 at those rails? n+/-55v dc is what i will use them at...
if SS then classAB, classAB2 is for tube amps with g1 drive voltage turning positive...
That's the point : all bipolar power amps are Class AB2.
By definition.
That it isn't often written with all the letters simply comes out of the fact that it's implicit, but that doesn't make it less true.
That's the point : all bipolar power amps are Class AB2.
By definition.
That it isn't often written with all the letters simply comes out of the fact that it's implicit, but that doesn't make it less true.
same as classB? it should be class B2? well, whatever, okey with me....
Now I know why AB and AB2 exists.
The "2" tells us that the output sinks current from the driver stage.
What happens when the complementary SS devices are used and there is no equivalent to AB2 in the valve/tube scenario? i.e. the output device sources current into the driver stage.
How does the rule get massaged so that it can be applied to common emitter or common source output devices?
Ah, mosFETs have an effective zero gate current, it flows in and out again to/from the gate capacitance !
Does that mean that BJT are AB2 and FET are AB?
The "2" tells us that the output sinks current from the driver stage.
What happens when the complementary SS devices are used and there is no equivalent to AB2 in the valve/tube scenario? i.e. the output device sources current into the driver stage.
How does the rule get massaged so that it can be applied to common emitter or common source output devices?
Ah, mosFETs have an effective zero gate current, it flows in and out again to/from the gate capacitance !
Does that mean that BJT are AB2 and FET are AB?
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Yes, of course.
It's just not explicitly mentioned because it's presumed known.
PD: and tubes also have some capacitive derived current flow , you must charge and discharge that Miller capacitance.
Doesn't look like much, but 100pF at high voltage and impedance may be as tough to drive as 1000/2000 pF typical of MosFets at lower ones.
It's just not explicitly mentioned because it's presumed known.
PD: and tubes also have some capacitive derived current flow , you must charge and discharge that Miller capacitance.
Doesn't look like much, but 100pF at high voltage and impedance may be as tough to drive as 1000/2000 pF typical of MosFets at lower ones.
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