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Simple Questions: Diodes and Heaters

luvdunhill

Member
2006-07-09 6:59 pm
hey, I have a few questions that seem so basic, I just wanted to be sure of myself. First, when calculating the power dissipated by a rectifier diode (Schottky), for full-wave rectification, is it simply Vf * current consumed by the load for each of the four diodes? Here's an example:

500px-Gratz.rectifier.en.png


Is this the same, or different power dissipated for two diodes in FWCT, i.e:

500px-Fullwave.rectifier.en.png


Also, what is the purpose of a heater center-tap on a tube? Does this come into play for tubes that can be operated at either 6.3V or 12.6V, or in the CT irrelevant in this case.

Thanks!
 
is it simply Vf * current consumed by the load for each of the four diodes?
Not precisely, but maybe I am nitpicking here. Power dissipated is the integral of the voltage across the diode and the current through it. If V and I were constant then yes, the integral is just V * I. But the voltage across the diode varies non-linearly with current, and in the case of a real load, ie with a filter capacitor, the current comes in spikes, and there is no really simple formula. You must come up with a formula for diode voltage vs current, then calculate ( based on load and on filter capacitor ) when conduction starts and stops and set the limits of the integral accordingly. Then do the integral, if possible.

But, having said that, Vf * I dc may be a good enough approximation.

The good news is that I dont think there is a real practical need to calculate diode power dissipation, just get ones rated for the correct inverse voltage and average current, and they should be OK. Others may comment, but I think that is a safe practical approach.

As for the heater center tap, as you say it comes into play with tubes with 12.6 (series) or 6.3 (parallel) connections. Or as acenter tap for connecting resistors or pots to ground for hum reduction.
 
I would agree with Robert on diode dissipation. Only to add: Yes, it is the same/diode for the two circuits you gave.

Regarding 12/6V heaters, 12V is (or are? - or perhaps were!) mainly there for automobile use - 12V battery. In audio amps one would normally (especially where the heater supply is AC) use the 6V arrangement to keep 50/60 Hz induction at bay. But in pre-amps where some prefer to play safe and use dc on heaters, it is slightly easier to connect the lot in series on 12V, because 150mA heater current is more easily smoothed than 300mA. (As an aside, you may have read the dangers of using raw un-/poorly smoothed so-called dc-on-heaters, often introducing more interference from the ripple than the original 6V ac would have done). But these days with sophisticated i.c. regulators available the lower current option is a minor advantage.
 

luvdunhill

Member
2006-07-09 6:59 pm
thanks guys!

the reason I'm interesting in calculating this, is I am using DC heaters and am expecting that I will need to heat sink the diodes. Thus I need to figure out how much heat there will be, in addition to the requirements of the regulator. If there is no reason to heat sink the diodes and I am wrong in assuming so, then this would be nice to know :)

so, given I have both switchable 12.6v DC and 6.3v DC output, where does the heater CT come into play? Perhaps I am just a bit dense... I figured just use pin 2 and 7 and it would magically work with both voltages... seems I'm figured incorrectly :) Just a guess, for 12.6v just disregard the CT, and for 6.3v tie pins 2 and 7 together and use the CT as ground?
 

woody

Member
Paid Member
2002-01-15 12:57 am
Tyrone Ga. U.S.A.
Sure hard to say without knowing what current you wil be drawing, the diode rating and case style you are using.
Say you need .3 to .6 A then 4 10A diodes in your bridge
would be fine without a sink unless mounted right next to the
tube. But I am thinking about building something needing 10A
these same diodes would work for me but would need a big sink.
 

luvdunhill

Member
2006-07-09 6:59 pm
Johan Potgieter said:
Re your last sentence: Correct. But keep in mind what I said about smoothing. Considering that, and depending on the heater power supply circuit, you might have a smoother voltage at 150mA than 300mA, thus wiring for 12V.

ok, well, it's working at 6.3v pulling 3.4A of heater current through 2 tubes. Nothing is really getting as hot as I thought. Ripple measures to be 2.2mV. I'll measure and see what the regulator is dropping. I'll probably try the 12.6v tomorrow and measure everything again and report back :)
 
Continued:

Oops, just before I create havoc:

Just as I pressed the "submit" button, I noticed that you mention pins 2 and 7. For what tube is that? I am used to miniature tubes (12AT7 et al), where the 12V heater is between pins 4 and 5, with center tap at pin 9 (thus 4 and 5 tied together and pin 9 the other connection for 6V). Pins 2 and 7 are the usual heater connections for octal tubes, also 12V types (e.g. 12SL7), but I am not aware of any such having a center tap.

Can we just clear this up before I cause you to mess up something. What am I missing?