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Old 8th April 2013, 12:52 AM   #1
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Default power handling of output stage emittrer resistors

How do I calculate the power that this resister will need to handle? Do I just figure the full rail voltage across the voltage divider formed by this resistor and the load?

Such that:

rails are 20 volts
load is 8 ohms
emitter resistor is .33ohms

8 / (8 + 8.33) = .96

.96 * 20 = 19.2v across load

leaving .8v across the emitter resistor

(.8 * .8) / .33 = ~ 2 watts

since this is peak watts, I imagine this is a pretty conservative value in practice, right? Of course 8 ohms is a light load compared to what may actually be encountered too.

What is the correct way to compute this, and what is a safe margin to use?
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Old 8th April 2013, 06:09 AM   #2
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Quote:
Originally Posted by Dan Moos View Post
How do I calculate the power that this resister will need to handle? Do I just figure the full rail voltage across the voltage divider formed by this resistor and the load?

Such that:

rails are 20 volts
load is 8 ohms
emitter resistor is .33ohms

8 / (8 + 8.33) = .96

.96 * 20 = 19.2v across load

leaving .8v across the emitter resistor

(.8 * .8) / .33 = ~ 2 watts

since this is peak watts, I imagine this is a pretty conservative value in practice, right? Of course 8 ohms is a light load compared to what may actually be encountered too.

What is the correct way to compute this, and what is a safe margin to use?
Yes that's pretty conservative as it assumes continuous full power, which of course is not realistic.

I would do it like this, assume continuous power of say 5W (which is also already conservative), and from that calculate the load current assuming lowest load impedance like 4 ohms.
This current also goes through the Re's so that gives you the dissipation.
Since there are two Re's each takes half of the calculated power.
Then add a safety factor of your liking.

But aside from the power spec there is also the max current spec and that's really the one you need to select those resistors. Even with low continuous power, they should be able to withstand at least a current of Vsupply/Vload, plus a safety margin.

So it will be the current requirement that is more critical than the power dissipation requirement.

jan
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Old 8th April 2013, 12:58 PM   #3
AndrewT is offline AndrewT  Scotland
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Dan,
the emitter resistors in a psuh-pull amplifier pass current in alternate halves of the AC cycle that passes to the load. That is the same as saying the Re operate on 50% duty cycle.
Next the output is AC waveform and you rightly point out that the average power dissipated in each Re is half the Peak Instantaneous Power.
Apply both factors to your Re and you find that the 0r33 resistor will dissipate a maximum of 500mW when feeding max power to an 8r0 test load from +-20Vdc supply rails.

But if your average output power is -10dB relative to the maximum capability of the amplifier you will find that even a 500mW resistor does not get hot.

Peak current capability is what matters, as Jan mentioned.

If the transient output to an 8ohms speaker is 6Apk, then that 6Apk has to pass through the Re.
Peak power dissipation is 6^2*0r33 ~ 12W
But that is a transient.
You must look up the resistor datasheet to find the overload capability for a single short term event.
Many resistors can pass double the maximum current for a single short term event (that is a peak dissipation of 4times the rated maximum for the device).
Most can dissipate double the maximum power for a single medium term event.
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Last edited by AndrewT; 8th April 2013 at 01:15 PM.
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