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#1
10-02-2008, 05:01 PM
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Silicon for selenium rectifiers: resistor size?
Hello TV experts,
When replacing the original selenium rectifiers in a voltage doubler circuit, using modern silicon units, how do you estimate the required series resistors? The TV under consideration is a 21-inch, 90-degree unit. Thanks in advance for any practical advice. 
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#3
10-02-2008, 05:55 PM
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Hello andy,
Thanks for the reply, here is my attempt at calculating the resistors. However, I found that the resistors would dissipate almost as much as the rest of the electronics, so I thought there was an error. What do you think? First get the total power consumption from the data plate. Let's call this P_tot. The heaters are drawing 600 mA at 117 V, that is 70 watts. Define P_DC = P_tot - 70. P_DC is essentially generated by the 264 V supply, so the power supply current is essentially I_DC = P_DC/264. Now you want resistors that will drop the voltage from 475 to 264 while passing I_DC, that is 2R = (475 - 264)/I_DC, or R = 105.5/I_DC. Numerical example: assume total power is 140 W. Then P_DC = 70 W, I_DC = 70/264 = 0.265 A and R = 398 ohms or roughly 390 ohms. Each resistor will dissipate 398*0.265*0.265, that is 28 W, that is a total of 56 W. P.S. 475 is the voltage measured at the B+ bus without resistors, while 264 is the nominal value from the schematic. Last edited by electroking; 10-02-2008 at 05:56 PM. Reason: added note
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#4
10-02-2008, 07:11 PM
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I beleive the heaters in the set are not drawing any current from the point I measured the 475 volts. Nor the B+
The heaters are powered right from the ac from the wall. This resistor would be placed past this point if I'm correct. So I would not include the heater current in your equations.
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#5
10-02-2008, 07:18 PM
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One complication is that the current thru the rectifier and the resistor isn't a constant DC, but a large pulse of current happening at the top of the incoming AC waveform. This pulse of current is essentially the first filter cap being topped off by the incoming top of the AC waveform, to replenish the voltage charge that was drained off the B+ by the rest of the set since the last replenishment The duty cycle of this pulse is related to the amount of ripple the B+ will have. Not a linear relationship, but still related. A rough approximation: If the duty cycle of the recharge pulse is 5%, that would make for 20 times the normal B+ load current this pulse will peak at. 100ma of B+ load current would make a 2A max pulse recharge current happen thru the rectifier. (You'll find a spec in rectifier tube data sheets "peak current", that's what this spec is for, and why it's possible to kill a rectifier tube with caps that are too big, as bigger caps will make for shorter duty cycles but larger current peaks). The resistor in series with the new silicon diode will only see this recharge current pulse, so when you want to figure out the required resistance you'd need to know that peak current. Problem is that you probably won't know it well enough to make a good calculation, so be prepared to adjust the resistance once you have it in the set. And why you end up with a much lower resistance than what you though you'd need...
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#6
10-02-2008, 08:02 PM
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Quote:
a way of determining the power used by the DC supply, that's why I suggested subtracting the heater power from the total power consumed by the TV set. As a rough guess, I would pick two 100-ohm, 10-watt resistors and mount one in series with each diode.
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#7
10-02-2008, 08:06 PM
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Alright sounds like it's worth a shot.
I'll have news at least by tomorrow. Thanks guys
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#8
10-02-2008, 09:35 PM
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Alright I have jury rigged up some resistors.(BIg ones)
And the voltage is down to respectable 244 volts. I am still trying to get it right in spec. But it brought the width of the screen back to a good spot. But the horiz lines are still vertically streched out near the top. Other than that I'd say things are coming very well
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#10
10-03-2008, 12:42 AM
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The tv had a load. But I was just thinking it could have been my tester messing with me.
My newer one is giving me the lower readings, which I'm guessing are right. I'm gonna retest with lower values.
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#11
10-03-2008, 01:03 AM
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Quote:
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#12
10-03-2008, 01:28 AM
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Yes, I was telling myself, a doubler should not get higher than twice the peak value
of line voltage, that is 2*120*1.41=338. Good night.
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#13
10-03-2008, 12:48 PM
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Yes,
It was definatley the batteries in my tester. Sorry guys. I just never even thought about it. All I could think was, "That Is Wrong.....Very Wrong..." And I just didn't wanna see those numbers again. And since the numbers are so much lower than I thought. My resistors are waay to big. But I have an idea. A 76 dodge ignition system resistor. It's sitting in the garage has 5 and 1 ohms of resistance with probably 20 watts of dissipation. And we don't have the truck anymore so it's the perfect candidate. I'll be back, with proper readings.
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#14
10-03-2008, 01:08 PM
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Quote:
one in series with each diode. Good luck.
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#15
10-03-2008, 01:41 PM
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Quote:
The new reading I got with two 5 ohm resistors in series with both diodes was MUCH closer. 142-144+ and 274-276+ volts. I'm very close Actually if I can find another resistor from that truck I'm gonna replace the other 5 ohm resistor with it. Then set both to 6 ohms. That should get the voltage right in spec. And these giant resisitors look very reliable. Just like that old truck.....
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