Output power VS Peak output power

Nordic · 23rd September 2005, 09:29 PM

When designing an amplifier
Say lets say something like the LM7835, wich is rated as 50W @8ohms, but guaranteed for 40W, and also has a Peak output figure of 100W.

In designing the circuit do you
Only supply enough voltage for 50W, but enough current handling capacity in your circuit to do 100W, or how would the amp hit those peaks?

I am trying to work out component capacities, things like how many watts each resistor would have to be rated for in a standerd inverted orientation.

Treefingers · 23rd September 2005, 10:13 PM

isn't the general rule when finding the watt rating of components to double what is required, thus taking into account the peaks?

wes-ninja250 · 24th September 2005, 12:19 AM

Personally, I take into account the amplifier output power, power dissipation (heat), and add in a fudge factor for the other components.

This is easy for chip amps, especially the Overture chips; NatSemi has done most of the work: http://www.national.com/appinfo/audi...gn_Guide13.xls

Let's say, 3875 with 35 volt rails... 1% THD Output is 57W, maximum Pd is 31W, total is 88W. 172W for stereo, drop in a ~50% fudge factor, call it VA, and go out and buy a 250VA transformer.

Of course, people have reported successfully using a 160VA transformer, so maybe I'm too conservative...

On the other hand, the spreadsheet says that Ipeak is 3.78A, you're running 70V total, 3.78 * 70 = 265W, so maybe I'm not totally out to lunch. A 250VA transformer should easily handle a 265W transient.

The other interesting thing to note with this sheet is that the TF package requires 33% more heat sink with this setup than the T.

Wes

Nordic · 24th September 2005, 08:26 AM

Hey ninja don't worry about being conservative, kinda what I'm after for this excercise...

I went through your xls sheet, it seems to be aimed at non inverted input , so I assume one will swap Ri and Rb's values ?

Anyway, the rateing I was after was not for those (still helpfull thanks)

I will attach image... If you or someone else could help me to calculate the blanks (a watt, b watt...) and maybe supply the formulas used for each, I would be ever so greatfull..

Nordic · 25th September 2005, 05:55 AM

Bump?

azira · 25th September 2005, 06:08 AM

Easy. None of those are high current areas so 1/4-watt all around.

Calculations:

rule #1 about opamps - Effectively, the inputs sink and source no current.

There for V(in+) = GND and c watt could be 1/8 or 1/16th watt resistor and wouldn't burn up.

rule #2 about opamps - The inputs of the opamps are always at the same potential. (It's like that's the job of the opamp).

Therefore V(in-) = GND. Power = Vo^2 / R. So 35^2 / 200k = 0.006. 1/4 watt will be fine.

Ok, since current in = current out and the opamp doesn't take any current at it's input, whatever flows through Rf must flow through Ri. So I = Vo/R = 35/200k = 0.175 mA. Power = i^2 * R so .175 mA^2 * 10k = still small so 1/4 watt is fine.

Lastly, d watt. Since your input signal is going to be in the 2-3 Vpp range, well you can do the math on that one. Yep.. 1/4watt is fine again.
--
Danny

24th September 2005, 12:19 AM	#3
wes-ninja250 diyAudio Member Join Date: Feb 2005 Location: Kingston, ON	Personally, I take into account the amplifier output power, power dissipation (heat), and add in a fudge factor for the other components. This is easy for chip amps, especially the Overture chips; NatSemi has done most of the work: http://www.national.com/appinfo/audi...gn_Guide13.xls Let's say, 3875 with 35 volt rails... 1% THD Output is 57W, maximum Pd is 31W, total is 88W. 172W for stereo, drop in a ~50% fudge factor, call it VA, and go out and buy a 250VA transformer. Of course, people have reported successfully using a 160VA transformer, so maybe I'm too conservative... On the other hand, the spreadsheet says that Ipeak is 3.78A, you're running 70V total, 3.78 * 70 = 265W, so maybe I'm not totally out to lunch. A 250VA transformer should easily handle a 265W transient. The other interesting thing to note with this sheet is that the TF package requires 33% more heat sink with this setup than the T. Wes __________________ Do daemons dream of electric sleep()?

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23rd September 2005, 09:29 PM	#1
Nordic diyAudio Member Join Date: Sep 2005	Output power VS Peak output power When designing an amplifier Say lets say something like the LM7835, wich is rated as 50W @8ohms, but guaranteed for 40W, and also has a Peak output figure of 100W. In designing the circuit do you Only supply enough voltage for 50W, but enough current handling capacity in your circuit to do 100W, or how would the amp hit those peaks? I am trying to work out component capacities, things like how many watts each resistor would have to be rated for in a standerd inverted orientation.

23rd September 2005, 10:13 PM	#2
Treefingers diyAudio Member Join Date: Jul 2005 Location: MK	isn't the general rule when finding the watt rating of components to double what is required, thus taking into account the peaks?

24th September 2005, 08:26 AM	#4
Nordic diyAudio Member Join Date: Sep 2005	Hey ninja don't worry about being conservative, kinda what I'm after for this excercise... I went through your xls sheet, it seems to be aimed at non inverted input , so I assume one will swap Ri and Rb's values ? Anyway, the rateing I was after was not for those (still helpfull thanks) I will attach image... If you or someone else could help me to calculate the blanks (a watt, b watt...) and maybe supply the formulas used for each, I would be ever so greatfull..

25th September 2005, 05:55 AM	#5
Nordic diyAudio Member Join Date: Sep 2005	Bump?

25th September 2005, 06:08 AM	#6
azira diyAudio Member Join Date: Dec 2003 Location: Near Seattle	Easy. None of those are high current areas so 1/4-watt all around. Calculations: rule #1 about opamps - Effectively, the inputs sink and source no current. There for V(in+) = GND and c watt could be 1/8 or 1/16th watt resistor and wouldn't burn up. rule #2 about opamps - The inputs of the opamps are always at the same potential. (It's like that's the job of the opamp). Therefore V(in-) = GND. Power = Vo^2 / R. So 35^2 / 200k = 0.006. 1/4 watt will be fine. Ok, since current in = current out and the opamp doesn't take any current at it's input, whatever flows through Rf must flow through Ri. So I = Vo/R = 35/200k = 0.175 mA. Power = i^2 * R so .175 mA^2 * 10k = still small so 1/4 watt is fine. Lastly, d watt. Since your input signal is going to be in the 2-3 Vpp range, well you can do the math on that one. Yep.. 1/4watt is fine again. -- Danny

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