I don't think so. When the load is small or zero, the current is at its highest, and all the supply voltage is across the transistor. Ouch! It gets hot. I am thinking of the pass transistor of a linear power supply....Generally speaking, a BJT's dissipation is greatest when its resistance is equal to that of the load.
I don't think so. When the load is very small, the current will be at the highest for a given voltage, and the voltage across the transistor will be at the highest. That is because the transistor has to drop the voltage that the load does not, and the small resistance value of the load cannot drop a lot of voltage So high current and small load voltage in relation to Vcc stresses the transistor the most. The ultimate stress to a transistor is a zero load or short circuit.But in "normal" condition (not short-circuit) BJT dissipation is greatest when Vce=0.5Vcc
Ptot=(Vcc^2)/(4*RL)
If you say so.I don't think so.
I think I do. After all, the same current is going through both the transistor and the load. It is just a matter of voltage distribution between the two....Since you don't understand those basic principles.
You are assuming a constant load amplifier. I made it clear in post #3 that I was referring to the pass transistor in a linear power supply. The load can change in that situation.1. Emitter follower with Re=8Ω; Vcc=12V and inputs signal change from 0 to 12V
2. CE amplifier with Rc=8Ω; Vcc=12V and input signal from 0V to voltage that saturate the BJT.
When you finish you calculation, then you will know that
Ptot_max=Vcc^2/(4*RL) is correct answer.
So was the OP's question: 4Ω or 8Ω.You are assuming a constant load amplifier.
A 100% increase in load resistance tells me that the load is variable.So was the OP's question: 4Ω or 8Ω.
Do you do this often?kdillinger,
A 100% increase in load resistance tells me that the load is variable.
Ratch
yes, it is all about output transistor power dissipation...Are you talking about the output transistors in an audio power amplifier?
If the amplifier is class-AB then they heat with the amount of output power. A 4 ohm speaker will try to draw up to double the power than an 8 ohm speaker so the transistors will heat with up to double the dissipation. But if you turn down the volume control so that the output power into 4 ohms is the same as the output power into 8 ohms then the transistors' heat is the same.
by Jerry Twomey
by Aaron Carman
by Jake Hertz
by Jake Hertz