Go Back   Home > Forums > >
Home Forums Rules Articles diyAudio Store Blogs Gallery Wiki Register Donations FAQ Calendar Search Today's Posts Mark Forums Read

Everything Else Anything related to audio / video / electronics etc) BUT remember- we have many new forums where your thread may now fit! .... Parts, Equipment & Tools, Construction Tips, Software Tools......

Need help in calculating power dissipation
Need help in calculating power dissipation
Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Thread Tools Search this Thread
Old 24th May 2004, 01:25 AM   #1
mclagett is offline mclagett  United States
diyAudio Member
Join Date: May 2004
Location: Yardley, PA
Default Need help in calculating power dissipation

Hi --

I'm trying to figure out how much power will likely be dissipated by some resistors that I have in the following circuit configuration: one 4-ohm resistor connected in series to two 8-ohm resistors connected in parallel. The device will allow the user either to tap into the two 8-ohm resistors connected in parallel for a total of 4 ohms of resistance or to tap in further upstream with the added 4-ohm resistor for a total of 8-ohms of resistance.

The signal I will be feeding this device will be from the speaker output of a guitar amp. I am a near total electronics newbie so I have no confidence that my understanding is actually correct, but I'm assuming that if the amp itself is rated at 50 watts, that will be the maximum power that could this circuit could need to dissipate (someone please correct me if this is not the case). If this is so, is it possible to know based on the configuration of the resistors that I've described above what the maximum power is that will be dissipated by any one of the three resistors?

I'm wondering if it might be that the power dissipation will be evenly divided between the first 4-ohm resistor and the second 4 ohms of resistance and that in turn the dissipation in this second part will be evenly divided across the two 8-ohm resistors in parallel. This would lead to something like 50% for the 4-ohm resistor and 25% for each of the 8-ohms. Based on a total of 50 watts of incoming power this would mean that the 4-ohm resistor would dissipate 25 watts and each of the 8-ohms would dissipate 12.5 watts. Does it work anything like this or is this guess totally off base? Can anyone help me come up with a more solid understanding of this.

The reason I wish to know this is that I am attempting to calculate the necessary heat sinks for each of the resistors, which I apparently can't really do with any sort of accuracy without knowing the power that will be dissipated by them.

Thanks very much for any help anyone can offer.


  Reply With Quote
Old 24th May 2004, 01:53 PM   #2
Jay is offline Jay  Indonesia
Join Date: Feb 2003
Location: Jakarta
Your guess is not off base. They fall in the scope of basic physics, which are P=VI, V=IR and Kirchoff law.

The non-basic things comprise of the understanding of what creates a voltage drop or an electron jump, so you know what comes first, I or V.

More advanced things fall in the scope of audio electronics, which are what, when and how the 50W number is derived.

50W amp (at 8 ohm load) could probably means that when the amp is connected to an 8 ohm load, the voltage across the load is such that V*V=P*R. That is when the amp is driven to certain level of power (near clipping? Certain THD?).

So it depends on your device and application (which I donít know what), the resistors donít have to be able to cope with the ďmaximumĒ load. 20W resistor is okay, and even 10W resistors will probably okay I guess (with some heat but it wonít be fried). So, why would you need a heatsink if you can get a 20W resistor easily? (Each of the 8 Ohm resistors wonít have to be rated as huge as that unless you will ever bypass the first 4 Ohm series resistor)

CMIIW. I really hate to apply theories and calculations in my audio hobby, but at least I had been always the best in electronic classes, if Iím not mistaken
  Reply With Quote
Old 24th May 2004, 02:08 PM   #3
jan.didden is online now jan.didden  Europe
diyAudio Member
jan.didden's Avatar
Join Date: May 2002
Location: The great city of Turnhout, BE
Your assumption is 100% correct, assuming that the 50W is a 'constant'. Assuming 50W in 8 ohms means, at that max level, 25W in the 4 ohms and 12.5w in each 8 ohms. BUT, if you connect that amp to the 4-ohms tap (the two 8 ohms parallel), it will probably deliver, at max power, more than 50W, like 70W. Now this 70W will evenly divide between the two 8 ohms, or for 35w each.
So, looking at it with these numbers, you would be needing a 25w 4 ohm and two 35w 8 omh resistors.
BUT (always those buts..) it is highly unlikely that your amp will be putting out max power for any appreciable time. Average power under music is probably not more than 10w or so, which means the resistors can be 5 times less in power handling capacity.
BUT if you want to do this for testing purposes, like running for half an hour at max power to see if it starts to smoke, then again you need the full power handling capacity of the resistors.
As noted above, there are many surplus resistors available that can handle that sort of power without a heat sink. Or do you already have these resistors? If so, what type are they?

Jan Didden
Music is dither to the brain; lets me think below the usual chaos - me
Get more Linear Audio for less! Check out my Autoranger and SilentSwitcher
  Reply With Quote
Old 24th May 2004, 11:52 PM   #4
mclagett is offline mclagett  United States
diyAudio Member
Join Date: May 2004
Location: Yardley, PA
Hi Jay and janneman --

Thank you so much for your kind responses. I am actually following a schematic I got off the web, which was for a dummy load for apmplifiers rated up to 100 watts. There is also a 16-ohm resistance option at another position of the switch that involves adding yet another 8-ohm resistor in series. But to simplify my first message, I didn't bother discussing that.

The person who posted this schematic also referenced some particular resistors, which I went out and bought (they weren't really that expensive). For the time being, I am planning to use exclusively the 8-ohm option with my guitar amp which is rated at a maximum of 50 watts. I bought all four resistors, however, (three 8-ohms and one 4-ohm) so that I can eventually build the device to accomodate other more powerful amps and ones which might have other values of resistance from their speakers.

So, to reply to your post, janneman, I will not really be using the the option that taps into the just 4-ohm part of the circuit (with the two 8-ohm resistors in parallel) -- at least not at this time. For the device I am currently building, I will exclusively use the 8-ohm resistance option with the configuration I described in my first post. So it sounds like I sort of had it right. I do plan to use this dummy load for numerous hours in a row, and it sounds like with the resistors I have in hand, that I am not likely to encounter a problem with their overheating.

Thinking I might need the full square footage of heat sink that Vishay recommends for their resistors when a full 50 watts of power is applied, I had resigned myself to cobbling together a makeshift device with two large aluminum cookie sheets connected together with some pieces of wood. But it now sounds like I may be able to get away with a more aesthetically pleasing aluminum enclosure and just use the enclosure itself as the heat sink. Because if I understand what you guys are saying, even in the worst case, my suspicion that the 4-ohm resistor will only have to dissipate at most 25 watts of power and the two 8-ohm resistors will each have to dissipate at most 12 watts of power is basically correct (assuming I always limit myself to the total 8-ohm resistance configuration I described earlier). It also sounds from what you are saying that it isn't likely they will even have to reach this level very often. The fact is, in my recording scenario (I basically want to be able to connect my speaker output via a direct-in box to my line mixer for recording the signal) I rarely turn the volume more than about an eigth of the way up.

Thanks so much for your help. You 've saved me from my cookie-sheet kludge. If you see anything wrong with my thinking, though, I would appreciate being corrected.


  Reply With Quote


Need help in calculating power dissipationHide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
What is power dissipation? geminni Everything Else 7 14th January 2007 08:39 AM
Q about Total power dissipation Bensen Everything Else 8 29th December 2005 11:02 AM
power dissipation darkfenriz Parts 2 14th December 2004 04:08 PM
Power dissipation control byteboy Pass Labs 14 16th December 2003 06:30 AM
power dissipation Opie Solid State 6 1st March 2002 08:54 AM

New To Site? Need Help?

All times are GMT. The time now is 02:34 PM.

Search Engine Optimisation provided by DragonByte SEO (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
Resources saved on this page: MySQL 16.67%
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.
Copyright ©1999-2017 diyAudio