|
Audio Asylum Thread Printer Get a view of an entire thread on one page |
For Sale Ads |
72.73.124.68
I just finished putting the RH Designs SEP EL84 amp. It works and sounds very good, except for some 120 Hz hum which I think is due to a loose power supply (Eli, etc. you were right - if I regulated my screens this wouldn't be a problem). I had to make a few compromises given the parts at hand - I'm going to rebuild the power supply and make it a little stiffer sometime this week. My power supply is especially weak, as I had to use a CRC filter instead of a CLC, and my B+ is around 270VDC instead of the designed 300VDC. I've thought about replacing the 6CA4/EZ81 I'm currently using with diodes (should gain another 30V), but haven't decided if I want to go that route yet.In any event, the heater voltage that I am measuring is ~7VAC which is too high. I'd like to step this down to the rating of 6.3VAC and am interested to know if I should use two equal value dropping resistors in either leg of the heater winding, or if a single resistor in one leg will work. The reason I ask is because I already have the parts necessary to put a single resistor in one leg (I need ~.24ohms, and I have two .47ohm resistors in my junk box). I'd rather not have to place an order for one resistor.
Follow Ups:
Hi.For AC heater decimal voltage trimming, sometimes it is hard to get resistors of the right resistance (plus tolerance %) to do the job precisely.
That's what I did & it works fine ever since. I would DIY wind 2 small inductors, with a few feet of magnet wires of SAME length & SAME size, say #18, BIfilar run, on a small non-ferrous core (say wood or plastic).
The leads of one inductor coil hooked up to in series of one leg of the heater PS circuit. Try to install this small inductor former closest to the tube heater socket lugs. Make sure to trim the lengths of the magnet wire of each coil to obtain the exact voltage drop you need.
The way both coils are hooked up is like a common mode noise suppressor. With a 0.1uF polyester AC bleeder cap across the O/P legs of the coils into the heater circuit, it forms a common mode suppressor while serving as a micro voltage dropper.
Here you will kill two birds with one stone.
Copper has a temperature coefficient of resistance that is much larger than the nichrome wire use to make commercial wirewound resistors. What this means is that the resistance will increase as the unit heats up. This is not a deal breaker, it's just that you need to design for a somewhat lower cold resistance and then check the voltages after everything has been running for a while.Of course, this also ruins the whole precision angle, but again, it's not necessarily a deal breaker as line voltages usually vary by several percent anyway.
Hi.Like everything else, overheating is no good which includes the glass body of a power tube - something we concern the least.
Yes, the temp coeff of NiCr wire (say common 80/20) is 0.00011R/R/C. (20C ref). Copper is 0.00393R/R/C which is 35.7 times of 80/20 NiCr resistance wire.
But for a couple or more temp C increase across a few feet of magnic wires (min 200C rated), as I found out from my installation (finger feeling only), & with voltage drop monitoring for a few hours on a dummy resistive load (IDH tube heaters I measured pretty linear & behave pretty pretty close on AC power to resistance load), there would not be an issue.
Of course, proper ventilation always help. That why I installed the inductor former close to the tube sockets & tightly against the chassis bottom surface with small ventilation pores drilled ontop.
I'm using this two circuits:Pic. 1
Case 1: If the heater windings have a middle tap.
Case 2: I force symetry.
In both cases I use 2 resistors.
In your Case 2, what wattage are the 100 ohm resistors? 1/2 watt?
The formula is N = U * U / R. in the case of 6.3V filament voltage this will give 6.3 * 6.3 / 100 = 0.3969. I always take 1 Watt. In case of 12.6 V it's twice the value, about 0.8W.The reduction of voltage in his example 0.1R and an estimated current of 2.5A will be 0.25 V. So you must align this value for your specific situation. The load then will be 0.25 * 0.25 / 0.1 = 0.625 W. I take some more, because the nominal load of a resistor is defined for a certain temperature (e.g. 50 - 60 °C).
_______________________________
Long Live Dr.Gizmo
_______________________________
Long Live Dr.Gizmo
This post is made possible by the generous support of people like you and our sponsors: