Sunn Solarus solid state power supply

[navdave]

Hello all I'm repairing a early Sunn Solarus that had a burnt power transformer and I'm having problems matching the voltage of the previous power transformers solid state supply. The replacement has no bias tap so I used a 2 watt 100k resistor to tap off the HT winding for the bias and solid state supply duties. Voltage was only 4volts I need 22 so I wired a 100k pot in series with the 10k resistor in the first voltage divider to try bring up the voltage to no avail. Any suggestions?

Oops the 2w 100k resistor is missing in the schem I drew up but its there..

[eleventeen]

I want to be sure I understand where you put the 100K series resistor (connected directly to one of the HV leads off the PT?) but two things:


That's a pretty big value, most of the time when I am stealing a bias feed from a HV winding I am in in the 15K - 22K - 33K - 47K range for the first R the HV winding sees....then going to a pot...with yet another resistor to ground. Perhaps the whole thing (as I see it) is in the 100K range....all three resistances in series. IMO your 100K might be too big a value, particularly if powering both things.


I would tend to use completely separate feeds from the HV to feed to low-voltage supply and the bias supply, I think. If you feed them through the same series resistor, that first R has to be a lot smaller. No question the SS preamp uses very little power, but the bias should use almost *none*. You could have a situation where one application (the preamp) uses 50 times as much power as the bias...to me, I would want better isolation between those two elements. That's my gut feeling about it.

[sluckey]

I would tend to use completely separate feeds from the HV to feed to low-voltage supply and the bias supply

I agree. Use the 100K to feed into the bias diode. Use another smaller resistor to feed into the other diode.

[navdave]

I had a 100k pot in series with the 10k resistor on the solid state power supply right after the 680r resistor to try to bring the solid state voltage up from 4v to 22v. A single 100k resistor is tapped into one lug of the HT winding feeding both the bias and solid state. The bias works fine. Guess I can tap the other leg of the HT winding with a 56k resistor to feed the solid state side.

I want to be sure I understand where you put the 100K series resistor (connected directly to one of the HV leads off the PT?) but two things:


That's a pretty big value, most of the time when I am stealing a bias feed from a HV winding I am in in the 15K - 22K - 33K - 47K range for the first R the HV winding sees....then going to a pot...with yet another resistor to ground. Perhaps the whole thing (as I see it) is in the 100K range....all three resistances in series. IMO your 100K might be too big a value, particularly if powering both things.


I would tend to use completely separate feeds from the HV to feed to low-voltage supply and the bias supply, I think. If you feed them through the same series resistor, that first R has to be a lot smaller. No question the SS preamp uses very little power, but the bias should use almost *none*. You could have a situation where one application (the preamp) uses 50 times as much power as the bias...to me, I would want better isolation between those two elements. That's my gut feeling about it.

[navdave]

I'll give that a try and report back.

I would tend to use completely separate feeds from the HV to feed to low-voltage supply and the bias supply

I agree. Use the 100K to feed into the bias diode. Use another smaller resistor to feed into the other diode.

[eleventeen]

I had a 100k pot in series with the 10k resistor on the solid state power supply right after the 680r resistor



This is somewhat of a clue: In general terms, when a part or a voltage level is ten times another one and we mush them together; we usually think of the thing that is ten times (or...in the case of current draw....1/10th as large) that of another part as "swamping" the other one. What you have here is 100K : 10K : [less than] 1K. These are all 10:1 relationships. They are out of scale with each other. The item that is going to dominate the action of your voltage divider is the 100K pot (assuming, perhaps incorrectly, that it is near full-rotation and full resistance.) The 680 ohm is in effect a mere rounding error from the standpoint of the 100K pot. And a pretty small one at that, less than 1%! You could pay big money for a 100K ohm 1% resistor and measure it and find it to be 100,680 ohms or 99,320 ohms (or even looser) and have no cause for complaint! Thus, left over for the 680 ohm resistor to "deal with" after the 100K is done with whatever volts you throw at them is less than 1%  of what is feeding the 100K!


So, on the pulled-out-of-the-blue assumption that your PT is 375-0-375 and the reading on a Solarus schematic that the tremolo uses 24 mils (.024 amp) Because we can see, on the the filter for the SS section two voltage readings, 22.5 volts, a 330 ohm resistor, then 16.5 volts. So 330 ohms drops 8 volts. 8 / 330 = .024.


We would expect to see .7 times 375 volts or 262 volts on the output of the rectifier diode. We want to get that 262 down to 22.5 volts, we need a resistor that is (262-22.5) / .024 = 9979 ohms. So you want about 10K as a dropping resistor--whether it's a single resistor or a pot and a fixed R--something in that range. This is why your 100K pot leaves you with next to nothing. It's eating too many volts. Those resistors or the R/pot combo have to be of decent size, too...because power-wise, you are talking about (.024) * (.024) * 10K = 5.76 watts. 

[navdave]

Ok I went as low as 10k tapped in to the high voltage supply and that's doing the trick but there is an awful lot of heat dropping across this 5 watt resistor. The question of reliability comes to mind... How long will a 5 watt 10k resistor last holding back 350 volts ac?

[eleventeen]

It's dissipating 5.76 watts. A 5 watt resistor is not a good reliability/longevity choice, as you imply. You're gonna have to either go to one bigger wattage 10K resistor; 10 watts being an obvious choice; or series connect several lower ohms-valued or parallel connect some higher-value ones to hit your 10K target. Real estate consumed, number of tie points needed, and mechanical mounting (if any) become considerations. Or a voltage regulator tube or two, LOL! (Just kidding)

[navdave]

Cheers 🍻! And thank you for all the help I'm thinking two 5 watt 20k's in parallel as space is tight in this amp.

It's dissipating 5.76 watts. A 5 watt resistor is not a good reliability/longevity choice, as you imply. You're gonna have to either go to one bigger wattage 10K resistor; 10 watts being an obvious choice; or series connect several lower ohms-valued or parallel connect some higher-value ones to hit your 10K target. Real estate consumed, number of tie points needed, and mechanical mounting (if any) become considerations. Or a voltage regulator tube or two, LOL! (Just kidding)

[ernie_jr]

ND,
what I have done on my SUNN rebuilds for the bias is get 12.6 volt transformer from radio shack. Connect the 12.6 V lead to the 6.3 V tap on your P/T, you will now get 60 volts off the small P/T. One lead to ground, one to the bias diode so it is feeding the same voltage the original SUNN bias tap had. Nice part is, you dont have to play with different resistor values, just hook it up to the stock SUNN bias circuit.
Good luck,
Ernie

[6G6]

I like Ernie's idea.

[sluckey]

Me too.

[navdave]

That is a great idea but would it leave enough power to run the solid state side? Without the proper voltage the photo cell in the tremolo circuit won't turn on and no preamp signal will pass through it.

ND,
what I have done on my SUNN rebuilds for the bias is get 12.6 volt transformer from radio shack. Connect the 12.6 V lead to the 6.3 V tap on your P/T, you will now get 60 volts off the small P/T. One lead to ground, one to the bias diode so it is feeding the same voltage the original SUNN bias tap had. Nice part is, you dont have to play with different resistor values, just hook it up to the stock SUNN bias circuit.
Good luck,
Ernie

[eleventeen]

At the full voltage proposed, 60 volts (which you are going to whack back to 22.5 or so) you would talking about 60 volts * 24 mils


Or 60 * .024 = 1.44 watts.


Imagine a teensy fil transformer that can only power a single 12AX7 = 6.3 volts @ 300 mils. That's 1.89 watts. I don't think there actually is such a thing, smallest I have seen is 2 amps, or about 7 times that. So I can't see there being anything near a power problem with the above-described scheme.

[navdave]

Then off to radio shack I go thats way more reliable than the two resistors I'm using.

At the full voltage proposed, 60 volts (which you are going to whack back to 22.5 or so) you would talking about 60 volts * 24 mils


Or 60 * .024 = 1.44 watts.


Imagine a teensy fil transformer that can only power a single 12AX7 = 6.3 volts @ 300 mils. That's 1.89 watts. I don't think there actually is such a thing, smallest I have seen is 2 amps, or about 7 times that. So I can't see there being anything near a power problem with the above-described scheme.