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how to hookup DC heaters?


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8/9/1999 8:29 PM
Will
how to hookup DC heaters?
I was looking at the Matchless Hotbox/Vibrobox schems, which have fullwave rectified heaters, but how are these hooked up exactly? Say you're using 12.6VDC for the heaters (series arrangement), then pin 9 is unused, so pin 4 gets -12.6VDC, pin 5 gets 12.6VDC, which would make the Matchless schems wrong (single +ve DC only). Anyone?
 
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8/10/1999 2:28 AM
Steve A.

Will:  
 
 
 
    There are instructions on how to wire up heaters to 6.3vdc in the Pignose G40V article on my site. As for a 12.6vdc filament supply, one lead would be the 12.6vdc and the other lead would be "almost" ground ("almost" because it is isolated from the chassis ground by a 0.1uF cap- this value is from the Torres book).  
 
 
 
    For a 6.3vdc heater circuit, Torres say that you should connect the plus (+) side to pins 4&5 and the minus (-) side to pin #9. (FWIW, Peavey uses the opposite polarity in their Classic 20 amp.)  
 
 
 
    Peavey wires up their Classic 30 and 50 amp preamp heaters in series to the minus supply that also feeds the bias circuit... They connect the minus voltage to pin #5 and it is pin #4 that eventually connects to ground (nothing is connected to the #9 pins- the internal connection between the two heaters).  
 
 
 
Steve Ahola  
 
 
 
http://www.techaccessinc.com/blueguitar/
 
8/10/1999 4:50 PM
ken gilbert
quote:
"Say you're using 12.6VDC for the heaters (series arrangement), then pin 9 is unused, so pin 4 gets -12.6VDC, pin 5 gets 12.6VDC, which would make the Matchless schems wrong (single +ve DC only). Anyone?"
 
 
 
 
Umm, if you hook up your socket like that, then the tube will get 12.6+12.6=25.2VDC across its heater. It won't like that.  
 
 
 
12.6V means either:  
 
 
 
a) from +12.6VDC to 0VDC (ground)  
 
b) from +6.3VDC to -6.3VDC  
 
c) from 0VDC (ground) to +12.6VDC  
 
d) from 0VDC (ground) to -12.6VDC  
 
e) from +8.6VDC to -4.0VDC  
 
 
 
...OR somewhere in between. Do you see what I am getting at here?  
 
 
 
It's the same way with using the 6.3VDC arrangment--you wouldn't want to use +6.3VDC and -6.3VDC, because that would net you 12.6VDC total.  
 
 
 
I hope this is more clear.  
 
 
 
~KG~
 
8/10/1999 7:45 PM
Brian

Bill:  
 
 
 
You could use a 9v supply transformer. Full wave recifying it would give 9*1.4 = 12.6v, which is exactly what you need to a series connection in the 12ax7 heaters. (power to pin 4&5, pin 9 unconnected)  
 
 
 
Brian
 
8/10/1999 8:31 PM
ken gilbert Don't forget Vf
An oft overlooked point with rectifying AC heaters is the diode's forward voltage drop. When dealing in the hundreds of volts necessary for plate supplies, they can be forgotten, but with the low voltages we're talking about here, they're a bit more of an issue.  
 
 
 
IF you had diodes with no forward voltage drop, then indeed 9VAC would net 9 * 1.41 = 12.6VDC. The next question is, though, how many diodes does the current have to pass through, and what is their Vf?  
 
 
 
Most likely you will use a non-center tapped winding, and a full wave bridge diode circuit. That means that the current must pass through two diodes in order to charge the smoothing caps. If you were to use a center tapped tranny, then you'd only pass through one diode, not two.  
 
 
 
Most diodes have a Vf somewhere between 0.5 and 1V. The ubiquitous 1N400X diodes have a rated Vf of 0.9V @ 1A continuous. To limit this loss of voltage in low voltage rectification systems, you can use Schottky, or "hot carrier" diodes. These have Vf's about half of the standard silicon junction, which gets you a Vf of approx. 0.4V. They will probably be more expensive, but sometimes you need them.  
 
 
 
Note that as the junction temperature increases, the Vf will drop somewhat--but as current through the junction increase, the Vf will increase (the diode does have a finite conductance). These characteristics tend to offset one another, since the temperature of the junction is related to the current which passes through it.  
 
 
 
So, in our original example of the 9VAC winding and a full wave diode bridge, you'd have to subtract an average of 1.8V for the plain-jane diodes, and 0.8 V for the Schottky versions, netting you:  
 
 
 
plain jane:  
 
(9VAC * 1.41) - 1.8V = 10.89VDC  
 
 
 
Schottky:  
 
(9VAC * 1.41) - 0.8V = 11.89VDC  
 
 
 
Which are low (but just about useable) for 12V6 heater operations (+/- 10% nominally).  
 
 
 
Interestingly enough, the gain of a tube usually rises with lower heater power. Don't understand that myself, yet.  
 
 
 
~KG~
 
8/10/1999 11:24 PM
Steve A.

ken:  
 
 
 
    That's a good point you bring up, but to dispense with the math, the dc filament supply mod in the Torres book puts out a DC voltage slightly under the 6.3VAC going in. (You can check out the same design in my Pignose G40V mods article.) However you may get different results from different amps.  
 
 
 
    According to Dan you can safely run the filaments at 6.0 or 6.1vdc, and that it is better to have the heater voltage too low than too high. On my Pignose the heater supply was like 7.1vac and dc rectified it was maybe 6.8vdc. I had picked up some small resistors (0.18R 0.22R 0.33R 0.47R) and experimented with them in series with the dc voltage to bring it down to to ~6.3vdc.  
 
 
 
    So where does the missing voltage go? In a full wave bridge rectifier there are 4 diodes but I think you'd only count the Vf for 2 of them. So 6.3vac * 1.4 minus 0.9 * 2 comes out to 7.1vdc. So where did the other volt go? (Maybe something to do with the filter cap??? Just guessing!)  
 
 
 
Thanks!  
 
 
 
Steve Ahola  
 
 
 
http://www.techaccessinc.com/blueguitar/g40v_mod.pdf
 
8/11/1999 11:06 AM
Dave H.

quote:
"So where does the missing voltage go?"
 
 
 
 
Itís the filter cap. 6.3V is the RMS voltage of the sine wave. 6.3 * 1.4 gives the peak voltage. The cap will charge up to almost the peak voltage - 2 * Vf at the peak of the waveform but between the peaks the cap has to supply the load current so it loses charge and the voltage drops.  
 
 
 
Dave  
 
 
 
 

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