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24th May 2004, 12:25 AM  #1 
diyAudio Member
Join Date: May 2004
Location: Yardley, PA

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 4ohm resistor connected in series to two 8ohm resistors connected in parallel. The device will allow the user either to tap into the two 8ohm resistors connected in parallel for a total of 4 ohms of resistance or to tap in further upstream with the added 4ohm resistor for a total of 8ohms 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 4ohm 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 8ohm resistors in parallel. This would lead to something like 50% for the 4ohm resistor and 25% for each of the 8ohms. Based on a total of 50 watts of incoming power this would mean that the 4ohm resistor would dissipate 25 watts and each of the 8ohms 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. Regards, Mike 
24th May 2004, 12:53 PM  #2 
Banned
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 nonbasic 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 don’t know what), the resistors don’t have to be able to cope with the “maximum” load. 20W resistor is okay, and even 10W resistors will probably okay I guess (with some heat but it won’t be fried). So, why would you need a heatsink if you can get a 20W resistor easily? (Each of the 8 Ohm resistors won’t 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 I’m not mistaken 
24th May 2004, 01:08 PM  #3 
diyAudio Member

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 4ohms 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
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24th May 2004, 10:52 PM  #4 
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 16ohm resistance option at another position of the switch that involves adding yet another 8ohm 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 8ohm option with my guitar amp which is rated at a maximum of 50 watts. I bought all four resistors, however, (three 8ohms and one 4ohm) 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 4ohm part of the circuit (with the two 8ohm resistors in parallel)  at least not at this time. For the device I am currently building, I will exclusively use the 8ohm 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 4ohm resistor will only have to dissipate at most 25 watts of power and the two 8ohm resistors will each have to dissipate at most 12 watts of power is basically correct (assuming I always limit myself to the total 8ohm 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 directin 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 cookiesheet kludge. If you see anything wrong with my thinking, though, I would appreciate being corrected. Regards, Mike 
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