Selenium rectifiers

[Findm-Keepm]

Originally Posted by Electronic M

Option B would be to take 7 modern diodes and string them in series as 7*0.7 = 4.9 which should be close enough to 5 to not matter much, and possibly cheaper than a dropping resistor.

He's got two selenium rectifiers in a voltage doubler configuration - best not to use series diodes in his application but they are fine in a straight half-wave configuration.

Of note, I had fun experimenting with an old GE Selenium today - put 40V from one of my Lambda supplies on the anode, and hung a 120 ohm 10 watt resistor off the cathode to ground - got 353 mA through the selenium, and had a volt and a half of loss through the selenium. My guess is they get leaky, and drop less across them as they age. I've got a coffee can full of Sarkes (ST logo) seleniums in the shop at home, and will try one or two later tonight. I think my Sorensen DCR40-40 is up to the task - just gotta get the right load without exceeding the limits of the selenium rectifier, as my shop at home is quite small and I don't want/need a stinkfest again. I had a Silvertone portable give up it's flyback in there last May, and it still smells of burnt flyback.

Cheers,

[Kamakiri]

Brian, if you'd be so kind, please review with me where I'm at, and the next step in the process.....when doing something that I've never done, I like to be very clear on what I'm doing.

Here's where I'm at right now.....

I have in my possession, the following parts (in quantities more than I need I assume):

(4) 1N5408 diodes, rated at 3 amps.

(8) 39 ohm 5 watt resistors

As shown, I've removed the selenium rectifiers, and installed two terminal strips, marked with the respective polarities. The underneath shot is just showing my work.

The selenium rectifiers that were in the set had a part number of 13-84630-2, and were rated 450 mA. Replaced at one time, I assume, with a heavier duty part.

[Bill Cahill]

Now, why do I have a feeling you will burn up that resistor????

[Kamakiri]

Originally Posted by Bill Cahill

Now, why do I have a feeling you will burn up that resistor????

Bill, I've got a lot of hours in working on this set and I'm trying to do something that I've never done before.

That said, mind elaborating?

[Electronic M]

Originally Posted by Bill Cahill

Now, why do I have a feeling you will burn up that resistor????

Those metal cased types don't burn they explode....Don't ask me how I know.

[Kamakiri]

Took me about a half hours' worth of racking my brain to trace everything out.....turns out that on this, M2 is on the left, and M1 is on the right.....but I guess it depends on how you're looking at the chassis.

Just one last question. I want to put the dropping resistor in between the filter choke and the resistor junction, and not between the diodes and the choke? If that's the case, I might want to mount that remotely with another Dale resistor when it's finished, as that's a REALLY tight junction.

[Findm-Keepm]

Originally Posted by Kamakiri

Took me about a half hours' worth of racking my brain to trace everything out.....turns out that on this, M2 is on the left, and M1 is on the right.....but I guess it depends on how you're looking at the chassis.

Just one last question. I want to put the dropping resistor in between the filter choke and the resistor junction, and not between the diodes and the choke? If that's the case, I might want to mount that remotely with another Dale resistor when it's finished, as that's a REALLY tight junction.

Before or after the choke is fine, it's all one series DC path. C2A and C2B, along with the choke form a Pi Filter, and additional DC resistance shouldn't be a problem. Technically, you could even raise the value of R91, the fusistor, but best to leave that alone.

Cheers,

[Kamakiri]

Got 235V on the B+, so by the Sams it's under by 20V. I only allowed it to run for about a minute, assuming that would be enough time to get the set up to operating voltage.

[Kamakiri]

Okay, now this is interesting.

Powered the set up again with another 39 ohm resistor in parallel, voltage is jumping between the 240 and maybe 260 range. Full raster, everything's looking good, but I think I'm smelling something.

Resistor is starting to burn. Shut down the set.

[Findm-Keepm]

added Series resistor burning? Is the voltage stable before the resistor?, at the cathode of M2? Powering up via variac, or cold start from wall?

B+ might be okay without the added series resistor - jumper it with a wire and bring it up on a variac. Monitor B+ at 120V, and post results - that'll give us a starting point from here.

Cheers,

[Findm-Keepm]

At the stated current in the sams, the resistor should drop about 12 or 13 volts, so if you are 20 volts low, I'd lose the resistor:

E=I*R
E=320ma x 39
E= 12.48V, and you are 20 low, so something is odd.

[Bill Cahill]

I don't agree. You must have an extra resistor to handle the surge. Further, I've told Tim this before, and, I'll state it again.
A tv with a 5 watt resistor? Balderdash!!!
I've see ten watt resistors in simple table radios. I've never seen a five watt loading resistor in a tv/ You need at least twenty watts.
Bill Cahill

[Kamakiri]

Just to clarify, I did a soft start on a variac, but not too slowly....I took about 20 seconds to bring it up to voltage. Initially, the reading was 305 volts, and dropped to 235 on the first try, with the one resistor in place. The voltage reading was tested at the B+ point that was on the schematic on the previous page in this thread.

I then shut it down, put in a second resistor in place, then again with a quick variac power up, the reading briefly stabilized at around 253 volts. I noticed that I had a full raster, snow, and static in the audio....exactly the way it should be. Observed it only for a few seconds, then went back to the meter and noticed that it started jumping all over the board, ranging from 240-260 volts. When I caught a light odor of something, I shut the set down immediately.

This is a picture of the resistor, along with brand new ones from comparison. Obviously I caught it in the very beginning stage. The picture doesn't show it, but there is a very light browning on one spot.

I clipped the lead of the resistor to check it afterwards. It's reading at 39.5 ohms.

[bandersen]

Right and P is also I2 X R. So if B+ is drawing 320mA, then you'll have (0.32 * 0.32) * 39 = 4 watts out of that resistor so a 5W should be OK, but maybe 7W for a little overhead. Or use a smaller 5W resistor like 33 or 27 ohms.

[Findm-Keepm]

Originally Posted by bandersen

Right and P is also I2 X R. So if B+ is drawing 320mA, then you'll have (0.32 * 0.32) * 39 = 4 watts out of that resistor so a 5W should be OK, but maybe 7W for a little overhead. Or use a smaller 5W resistor like 33 or 27 ohms.

He's 20V low, and the 39 ohms is dropping that. My original calculations called for a ~13 ohm resistor. (see post 27, page 2) Take three 39 ohm resistors, parallel them, and place them in the circuit. That'll give you the proper (lower) voltage drop, which should give you the proper B+, and plenty of overhead on the wattage. .33A X 13 = 4.29W, and you've got (3) 5 watters in parallel.