power calculation...correct?

[lt cdr data]

Could someone please tell me if I have this right?

I have a schematic of a rather well know and respected japanese power amplifier

with

70 volt rails, dc.

equals 49.49 volts available for ac swing, rms

assuming 5 volts dropped across transistor, std emitter follower bipolar output stage with

2sc3858 transistors

that makes 44 volts say.

which is 44v squared /8 ohms

gives 242 watts into 8 ohms?

now the feedback resistors are 28,890 ohms series and 560 ohms shunt

gives a gain of 52.5 which makes 34.4 dbs?

the sensitivity is .775 volts rms to full power, from the specifications listed, so....

a signal of .775 volts times 52.5 will give the full voltage ac out which is...

40.68 volts...

so based on that, we get a power output of 206 watts into 8 ohms?

so it will clip at around 242 watts? ie never?

its biased at 10 mv across a 0.22 ohm emitter resistor, which makes 45.45 milliamps,

and there are 4 pairs of devices, so class A output will be 45.45milliamps squared, times 4 devices, times 2

times 8 ohms?

if its biased hotter, will that reduce the total power out?

any way to easily calculate the db's of feedback without scoping the thing?

finally, will reducing the feedback by 3-6 dbs increase the sensitivity and the power out?

thanks in advance

[janneman]

Originally posted by lt cdr data
Could someone please tell me if I have this right?

power amplifier, 70 volt rails, dc.

equals 49.49 volts available for ac swing, rms

assuming 5 volts dropped across transistor, std emitter follower output stage

that makes 44 volts say.

which is 44v squared /8 ohms

gives 242 watts into 8 ohms?

now the feedback resistors are 28,890 ohms series and 560 ohms shunt

gives a gain of 52.5 which makes 34.4 dbs?

the sensitivity is .775 volts rms to full power, from the specifications listed, so....

a signal of .775 volts times 52.5 will give the full voltage ac out which is...

40.68 volts...

so based on that, we get a power output of 206 watts into 8 ohms?

so it will clip at around 242 watts? ie never?

[snip]

I'm with you so far. It will not clip provided you limit Vin to 0.775V as stated.
One minor point: I think you should take the 5V loss from the peak voltage, not from the RMS voltage. That will give you a bit more max output power.

Originally posted by lt cdr data
[snip]its biased at 10 mv across a 0.22 ohm emitter resistor, which makes 45.45 milliamps,

and there are 4 pairs of devices, so class A output will be 45.45milliamps squared, times 4 devices, times 2

times 8 ohms?

if its biased hotter, will that reduce the total power out?

any way to easily calculate the db's of feedback without scoping the thing?

thanks in advance

In class A, the Ie can vary from zero to 90mA per transistor or 90mA peak, or 360mA peak per side in 4 transistors. Analogous for the Vout, the peak Iout (in class A) is 360mA, so the RMS Iout is about 250mA RMS in 8 ohms gives some 500mW from the top of my head (please check this last number).

If you bias it higher, nothing changes for the max Vout and thus the max Pout, although the part of the output in class A will of course increase.

What exactly do you mean by the feedback dB's? The feedback is the difference between the open loop gain and the closed loop gain. The latter is (approximately) what you calculated above 34.4 dB (I say approx because it assumes infinite open loop gain which is not what you have).

If, say, your open loop gain (without feedback) is 80dB, your closed loop gain 34.4, then your feedback is 45.6dB. But that ol gain will vary with frequency, so your feedback varies with frequency, and so will your closed loop gain (although much less so, if the ol gain is high enough wrt the closed loop gain).

Edit: this may also help:
http://www-k.ext.ti.com/srvs/cgi-bin...i.exe?Company={5761bcd8-11f5-4e08-84e0-8167176a4ed9},kb=analog,case=obj(32624),new

Jan Didden

[sreten]

Originally posted by lt cdr data

finally, will reducing the feedback by 3-6 dbs increase the sensitivity and the power out?

thanks in advance

Hi,

It will increase sensitivity and stability margins but not the power output.

/sreten.

[janneman]

Originally posted by sreten


Hi,

It will increase sensitivity and stability margins but not the power output.

/sreten.

Thanks, I missed that

Jan Didden