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 Post subject: Zener diode as voltage drop to replace field coil...help?
PostPosted: Jan Thu 17, 2019 1:08 am 
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Joined: Jan Thu 11, 2007 2:23 am
Posts: 579
Location: Maumee, OH
Hi all, I recently picked up a Leslie 21H amplifier on eBay initially with the intention of using it in a Leslie 45. The 45 doesn't use a field coil speaker like the 21H does, so I needed to find a way to replace the field coil resistance. Initially, I used a 50W 2.7k resistor, but I found that it got REALLY hot in operation; like, uncomfortably hot.

Convinced that there was a better solution, I continued searching the internet, and turned up this old topic where a technician named Harvey Olsen suggested using a 39V 5W zener diode in place of the field coil. Here's the relevant passage:

Quote:
The original speaker field coil also acts as a voltage
regulator of sorts by loading down the power supply. Without the coil in the
circuit, the output screen voltage tends to climb. This causes the output stage to
draw more plate current than it should.

In both 6V6 and 6L6 amps the output screen grids should be 39 volts DC
*under* the plates. The easiest way to accomplish this today is to wire a 5 watt,
39 volt zener in series with the 400 ohm power resistor in the B+ circuit. The
zener is installed after the resistor but before the DC takeoff for output
screens. The cathode (banded) end of the zener diode goes toward the screens.


So, I bought the parts and tried it out. My first attempt was an epic failure because I totally misinterpreted his instructions. After replacing the .25 amp fuse and installing a fresh 39V zener diode the *right* way, I tried it out again. I could see the screen voltage rising to a good level as it was warming up, but I also saw the magic smoke coming out of the zener diode, so I quickly powered down, and now I'm here.

Here's the 21H schematic for reference:
Image

As I read it, the zener diode should basically be connected where the field coil was installed before, with the band pointing toward the line labeled "295V" on the schematic, and the other end connected to the line labeled "19V". Apparently, this is either not correct, or this technique just doesn't work for some reason.

Any tips or thoughts? I feel like I'm missing something, especially since a 50W resistor was barely able to handle that job before. I don't think the current to the screens is that high, so in that way I would think that a 5W zener would be OK, but I'm a little fuzzy on how the zener is actually working here; the theory is that the zener drops 39V from the 335V supply to produce approximately 296V, but every reference I can find on zener circuits would imply that this would be trying to supply 39V to the screens and drop 296V as I connected it.

Thanks!


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 1:38 am 
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Joined: Jan Thu 01, 1970 1:00 am
Posts: 5158
Location: Rochester NY USA
The 400 Ohm resistor and the field coil form a voltage divider, reducing the screen voltage. Putting the Zener in series with the 400 Ohm resistor will drop the voltage without the field coil connected. I'd use a 5W Zener. The Zener does NOT connect in place of the field coil. That WILL smoke!


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 1:44 am 
Member

Joined: Jan Thu 11, 2007 2:23 am
Posts: 579
Location: Maumee, OH
Ok, so that was how I originally read it, but I second guessed myself. So the “19V” side of the field coil is left disconnected, and the screens connect to the banded end of the zener, the other end of which is connected to the resistor. Right?

(I can make a drawing shortly if that description didn’t help.)


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 4:55 am 
Member

Joined: Jan Thu 01, 1970 1:00 am
Posts: 5158
Location: Rochester NY USA
The Zener works in reverse conduction, so the banded end would connect to the positive supply. It can connect between the + supply and fuse, fuse and resistor, resistor and screens, whatever works best for layout. Series components can go in any order. And when you're done, screen voltage should be about 39V less than plate voltage.


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 5:19 pm 
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Joined: Jan Thu 11, 2007 2:23 am
Posts: 579
Location: Maumee, OH
OK, so the BANDED end of the zener points TOWARD the resistor?

However, flipping the zener diode would make a lot more sense. I really need to figure this out soon anyway, I'm running out of .25 amp fuses! :lol:

EDIT: Tried it with the band TOWARD the resistor, I now get a ~40V difference between B+ and screen voltage! Hooray! Thanks for the help!


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 9:13 pm 
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Joined: Jan Thu 01, 1970 1:00 am
Posts: 4199
Location: Boston, MA USA
That's a lot of current to drop 40V through a 400 ohm resistor. Better check to make sure you are not exceeding the screen grid dissipation spec.

-David


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 Post subject: Re: Zener diode as voltage drop to replace field coil...help
PostPosted: Jan Thu 17, 2019 9:28 pm 
Member

Joined: Jan Thu 11, 2007 2:23 am
Posts: 579
Location: Maumee, OH
dberman51 wrote:
That's a lot of current to drop 40V through a 400 ohm resistor. Better check to make sure you are not exceeding the screen grid dissipation spec.

-David


Interesting point. "Design Center Value" for grid #2 max power dissipation is 2 watts on a 6V6GT. Using V=IR with V=40 and R=400 I get I=100 milliamps; reworking all that, that gives me 4 watts. However isn't that shared between all 4 6V6 tubes?


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