Wow, first post here. I first posted this in another Zener thread. That was a hijack really. Here's what I'm up to. If you put a zener diode between the center tap of a HV winding and ground on a tube power transformer it will lower the PS voltage by an amount just shy of Vz. Problem is the diode needs to dissapate (say) 210MA at (say) 75 volts. 15.75 watts. The reason for even doing this is to recycle old amplifier transformers (I hate throwing anything away) and most common power tubes really do produce a different tone at different places on the curve. These are guitar amps by the way.

I designed this regulator a million years ago to regulate screen voltage on an AM transmitter modulator tube. Rs bias's the .5w zeners providing a reference voltage for the Darlington transistor that I was using. It worked great. But one thing you don't want in a guitar amp is regulated B+. Things need to sag when you push the amp. It's all part of that elusive "tone" thing. The other two diodes protect the transistor from excessive voltage at B+ reg and in case of zener failure.

Now here's my question. I don't understand Vebo. If you used this circuit as a zener current multiplyer instead of a series regulator could you ground the emitter, or would that let all the smoke out? What I mean here is to wire the transistor with the collector tapped off the HV supply just downstream of the rectifier, before the filter string, with the emitter going to ground. I'm trying to multiply the Pd of the cheap and readily available zeners, substituting a $2 transistor and .10 zener for an incredibly hard to find $25 zener diode. What I will probably do is string three 5 watt 25 volt diodes and jury rig a weird heat sink at the center tap and be done with it.

thanks in advance for replies
Dan H

 

If I understand you correctly, you want to add a Darlington transistor to the zener which is in series with the GND connection, so that you can use cheap zeners.
As you probably know, a Darlington is a circuit with 2 transistors, so I drew them this way to make it clear why R~1.5V/Iz.

 

Thanks Ron! I got about 20 of these transistors. They are NPN. This is a great help.

Q: You've used Ohms Law R=E/I except that you've used ~ instead of =. Why is that?

Dan

 

Referring to the circuit labeled "this apparently lets the smoke out...". That circuit is wrongly designed and cannot work in any way. The full rectified output voltage of the transformer goes directly to the capacitor and to C-E of the transistor. The transistor will never conduct anything as there is no B current. There is no current going through the zener either (except if the transistor burns and shorts C-B in which case the zener fries too). If the transistor burns out it just means the C-E voltage limit was exceeded. I am not sure what it is you are trying to do (my guess is a series regulator) but this is not it and will not work.

 

Thanks Ron! I got about 20 of these transistors. They are NPN. This is a great help.

Q: You've used Ohms Law R=E/I except that you've used ~ instead of =. Why is that?

Dan

"~" is a shorthand way of typing "≈" (approximately equals). To get "≈", I had to Google an extended ascii table, find "≈", and then type "(alt)247". If anyone knows a faster way to enter non-keyboard characters, please enlighten me.
I'm assuming you will use a heatsink on your Darlingtons. 16 watts is a lot of heat to get rid of.

 

Another way is to install a large, low ohm, resistor such as those used in automobile starter circuits. Go to your local auto parts store and ask for a Chrysler AL-795, and connect this between the CT and ground. It also helps protect the xfmr during short circuits (the resistor smokes instead of the xfmr).

 

Referring to the circuit labeled "this apparently lets the smoke out...". That circuit is wrongly designed and cannot work in any way. The full rectified output voltage of the transformer goes directly to the capacitor and to C-E of the transistor. The transistor will never conduct anything as there is no B current. There is no current going through the zener either (except if the transistor burns and shorts C-B in which case the zener fries too). If the transistor burns out it just means the C-E voltage limit was exceeded. I am not sure what it is you are trying to do (my guess is a series regulator) but this is not it and will not work.

Right! I designed it as a series regulator and was trying to use it in typical zener shunt type voltage dump. That's why I asked for help. I don't know s*it about transistors. I took the series regulator design directly out of a book.

 

Another way is to install a large, low ohm, resistor such as those used in automobile starter circuits. Go to your local auto parts store and ask for a Chrysler AL-795, and connect this between the CT and ground. It also helps protect the xfmr during short circuits (the resistor smokes instead of the xfmr).

Thanks yes, that is one way. Unfortunatly in Class AB1 or 2 where there are large swings in plate current, the voltage will flop around way too much and sound terrible. The big resistor works pretty well in Class A amps.

 

"~" is a shorthand way of typing "≈" (approximately equals). To get "≈", I had to Google an extended ascii table, find "≈", and then type "(alt)247". If anyone knows a faster way to enter non-keyboard characters, please enlighten me.
I'm assuming you will use a heatsink on your Darlingtons. 16 watts is a lot of heat to get rid of.

Ok thanks Ron. I'll probably use a little pot to find the sweet spot then solder in a fixed value. The 1.5 volt value I assume is the "more or less" standard zener bias point. And on the heat sink, oh yeah. Standard mica insulated 220 heat sink. When I used these in screen regulation I was dealing with (say) 5-10 ma. Not too hot. But this should light things up. I'm tempted to screw it right to the .090 aluminum chassis with some heat goop. I suppose I should look at the spec on the heat sink and see what it's rated for huh.

Dan H

 

Ok thanks Ron. I'll probably use a little pot to find the sweet spot then solder in a fixed value. The 1.5 volt value I assume is the "more or less" standard zener bias point.

I don't think you need to worry about a sweet spot. Just look up the zener's nominal operating current and calculate a resistor value. Zeners are low impedance, so if the current is not exactly at the calculated current, it won't matter. You can burn your zener up if you use a pot without a resistor in series with it.

 

Right! I designed it as a series regulator and was trying to use it in typical zener shunt type voltage dump. That's why I asked for help. I don't know s*it about transistors. I took the series regulator design directly out of a book.

I think you made a few mistakes when copying the design. That cannot work under any circumstances. Cut the connection between the collector and the capacitor and connect the emitter to the + of the capacitor instead. Now the transistor is in series. Next you need a resistor between the zener and the collector (i.e. the input voltage from the rectifiers). This will polarise the zener.

 

I think you made a few mistakes when copying the design. That cannot work under any circumstances. Cut the connection between the collector and the capacitor and connect the emitter to the + of the capacitor instead. Now the transistor is in series. Next you need a resistor between the zener and the collector (i.e. the input voltage from the rectifiers). This will polarise the zener.

What are we talking about here? The first scheme or the bastard second scheme where I just shorted the emitter to ground? The first sheme works. You put 325VDC in at the collector and you get 273VDC out the emitter regulated. That's with 91V .5 watt zeners. Rs bias's the zeners. I know the second scheme doesn't work. That's why I asked for help. I don't know what I'm doing with this sand state stuff. I'm a tube guy. All those arrows and stuff confuse me. Ask me why making the plate resistor bigger makes the stage gain higher while the plate voltage drops. On top of that I'm old. With a fragile ego. So include drawings please