• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Do diodes in the power section produce heat?

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So if I use power diodes to rectify the high voltage wires coming off the power transformer, will the diodes produce heat? I was planning on just soldering the diodes directly to the high voltage wires and heat shrinking them individually but then realized if they gave off any heat at all it'd be like a little sauna in there and possibly overheat the diodes. This is basically the idea.

H.V. wire 1 - (butt connection ) - Diode - (butt connection) - Diode - B+
H.V. wire 2 - (butt connection ) - Diode - (butt connection) - Diode - B+

So basically the B+ would be where they connect and then each wire through diodes would have heat shrink over them. Hopefully that makes sense, any thoughts??? Thanks!
 
The diodes will run hotter if you insulate them.

But as they get hotter, the Vdrop (Vf) goes down. So the increase in heat is not quite as bad as first seems. Have a look at the datasheet.

Diodes conduct a lot of their heat to the lead outs and then to the connections.
The body+leads+connections work as a heat dissipating system.
 
The flying wire technique is fine. If it's a HV section for a tube amp, diode dissipation is about 70mW worst-case for each tube supplied by this diode. As Merlinb suggests, 300mW or below will be fine. During turn-on is when the diodes are really tested until the caps are charged.
 
In defense of the old fashioned terminal strip method of mounting components, the one thing you have going for it is that the wires and component-leads are firmly bound to a mechanically stable strip. The second thing is, that if you're experiencing problems and suspect a component, its leads aren't completely covered in heat-shrink. And, if you decide a component is bad, it is easy to clip it out or desolder it, replace it, and get back working again quickly. Heat shrink? Not so much so.

GoatGuy
 
…During turn-on is when the diodes are really tested until the caps are charged.

Yes: specifically they have to dissipate whatever energy is held by capacitor at the least, divided by charging time. This is one of the good reasons to also include a series small value resistor… as it takes the load, and slows down (stretches out) that power-dissipation interval.

GoatGuy
 
It would be common for the B+ and gnd junctions of that diode bridge to be soldered to terminal strip tabs, or directly to the main filter capacitor terminals. In that common scenario, the capacitor terminals become a major heatsink to the diodes, as the lead length from diode junctions to capacitor terminals may be quite short.

If possible, it is best to keep the transformer leadout wires twisted and short and not near sensitive circuitry. Each lead would solder to a junction of two diode leads (unless you use a bridge package) - I would tend to insulate or protect those two junctions, as they pretty much have the highest idle working voltage in the amp.
 
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