
Home  Forums  Rules  Articles  diyAudio Store  Blogs  Gallery  Wiki  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Instruments and Amps Everything that makes music, Especially including instrument amps. 

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 
25th June 2013, 08:53 PM  #11  
diyAudio Member
Join Date: Apr 2010
Location: Limeira, São Paulo, Brazil

Quote:


25th June 2013, 09:34 PM  #12 
diyAudio Member
Join Date: Apr 2008
Location: Carlisle, England

100watt resistors usually need a heatsink.
__________________
PCBCAD51 pcb design software. 2018 version out now with lower prices >> http://www.murtonpikesystems.co.uk 
25th June 2013, 09:39 PM  #13 
diyAudio Member
Join Date: Aug 2012

Could it be that you like the effect of the HT sagging when playing loud? Just like using valve rectifiers instead of silicon? The current goes from maybe 120mA idle to 500mA at full power.

25th June 2013, 09:58 PM  #14 
diyAudio Member
Join Date: Jun 2008
Location: SoCal

The '67 Plexis had a B+ of 460VDC, those are the magical ones. EVH comes to mind if that's your thing.
Craig 
26th June 2013, 12:53 AM  #15 
diyAudio Member
Join Date: Apr 2010
Location: Limeira, São Paulo, Brazil


26th June 2013, 08:46 AM  #16 
diyAudio Member
Join Date: Sep 2011
Location: Groningen

Just use the 100W resistor, mounted on a suitable heatsink to get the ful 100W capability. You should be able to even drop 134V across it!
Using the resistor is this way will cause sag, as many have stated above already. Sag depends of the amount of current drawn from the PSU, so B+ will vary accordingly and the amp sounds different depending on how hard it is driven. 
26th June 2013, 02:37 PM  #17 
diyAudio Moderator

All musical instrument amplifier related threads belong in Instruments & Amplifiers. Please read the forum subheadings for guidance on where to post your thread. Thank You.
__________________
"To argue with a person who has renounced the use of reason is like administering medicine to the dead."  Thomas Paine 
26th June 2013, 04:53 PM  #18 
diyAudio Member
Join Date: Nov 2007
Location: Near Dallas Texas USA

A sag resistor will not accomplish what you want. If the B+ drops 20V at idle, it will drop way too much at full power. Power Scaling or Zener diodes are the way you need to go.
Because transformer current flows in pulses where the RMS value is much higher than average value measured by your meter, the power dissipated by the sag resisitor can not be calculated by simple ohms law. Consider how this works. If 1 Amp flows for 1 second in a 1 ohm resistor, 1 Watt is dissipated. If 2 Amps flows for 1/2 a second and falls to zero for 1/2 a second, is that the same? NO!!! 2 Amps causes the resistor to dissipate 4 Watts for 1/2 a second and zero for half a second so averaged over the whole second, the dissipation is 2 Watts. If 4 Amps flows for 1/4 second and falls to zero for 3/4 second, the dissipation is 16W for 1/4 the time and averages out to 4 Watts. In this case the RMS current is 2 Amps. The RMS current squared times the resistance equals the power dissipation. Pop quiz: What is the RMS current when 2 Amps flows for half the time and falls to zero for half the time? Off the top of my head 180 ohms is way too high for a sag resistor. The best thing to do is to model the circuit with a SPICE like progran. Duncan's PSU Designer II works well and is simple to use. You will need to know: 1) What is the rectifier configuration? Full Wave (using a center tap) or the Full Wave Bridge 2) What is the open circuit (no load) voltage of the power transformer secondary 3) What is the resistance of the power transformer secondary winding(s)? 4) What is the power line frequency (50Hz?) 5) What is the current draw at idle? 6) What is the current draw at full power (rail to rail clipping) 
27th June 2013, 11:17 AM  #19  
diyAudio Member
Join Date: Apr 2010
Location: Limeira, São Paulo, Brazil

Quote:
I_rms=SQRT{ 1/T [ (2^2)*0.5T + [0^2]*0.5T ] } = 1.41 amps This is the value that the meter measure Because of the peak dissipation I thank 100W could support the effect of peak dissipation. I'm using 118_Oms 80W (4x20W ceramic resistors 470_Ohms in paralell), and It goes well. Quote:
Quote:
Quote:
Quote:
A: 60Hz A: 100mA Quote:
Thanks the hints Loudthud FAB 

27th June 2013, 01:56 PM  #20 
diyAudio Member
Join Date: Dec 2012
Location: SF Bay Area

Hey... There is a MUCH simpler way to do this than resistors!
1N5339  a 5 watt, 5.6volt zener diode. (industry uses 5 watt zeners a lot in motor and panel work). 850 ma working current. Can take quite a bit more than that 'pulsed.' So you want 20 volts of voltage drop? 4 of them in series. 50 volts? 9 of them in series. And ... they can "handle" more power collectively 'cuz of the series connection. And here's the best part: $0.36 each from Mouser, "quantity 1 to 25". Buy 20 of 'em and save on shipping! The cool thing about zener voltage is that it is almost constant over a HUGE range of current flow. This is exactly what the "resistor idea" was supposed to accomplish, but has not. ALL analysis above that points to the resistor getting hot 'cuz of the very large difference between idle current, and highoutput driven current ... is correct. Go with the zeners, and get a triplegood solution. Constant voltage drop. Current independence. More than adequate power dissipation over full operating range. Cheap... oops, a 4th factor. GoatGuy 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Driver sag  Studley  Subwoofers  33  14th March 2011 10:13 AM 
Driver sag Worries?  Toast_Master  Subwoofers  3  29th August 2009 11:30 PM 
Distortion from heater sag?  hihopes  Tubes / Valves  33  7th October 2007 05:33 PM 
Voltage sag  JonnySwitchblade  Solid State  6  8th July 2007 10:20 AM 
driver sag  xstephanx  Subwoofers  23  30th November 2004 07:32 AM 
New To Site?  Need Help? 