
Home  Forums  Rules  Articles  The diyAudio Store  Gallery  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Solid State Talk all about solid state amplification. 

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 
8th April 2013, 12:52 AM  #1 
diyAudio Member
Join Date: Mar 2013
Location: Lynden, WA USA

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? 
8th April 2013, 06:09 AM  #2  
diyAudio Member

Quote:
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
__________________
If you don't change your beliefs, your life will be like this forever. Is that good news?  W S Maugham Check out Linear Audio! 

8th April 2013, 12:58 PM  #3 
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Dan,
the emitter resistors in a psuhpull 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.
__________________
regards Andrew T. Last edited by AndrewT; 8th April 2013 at 01:15 PM. 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Class (A)B output stage: why 2 emitter resistors, why not 1?  deleveld  Solid State  23  13th December 2006 06:10 PM 
Power output vs speaker power handling  Bob0513  Chip Amps  4  5th November 2006 10:41 AM 
power handling for resistors?  tdemont  MultiWay  3  14th January 2005 01:20 PM 
Resistors in the output stage  Solid Snake  Solid State  42  18th January 2004 08:56 PM 
New To Site?  Need Help? 