09-12-2011, 10:37 AM
Ah HA! I wish I had found this before I recommended an 86 power transformer to you...
I owe you an apology, Robert. At the time I recommended the 86 transformer, my mind was on wrapping up college instead of being fully focused on your transformer problem.
I thought I had a Model 86 & 82 service manual, but if I do, I can't find it right now. So until this morning, I was unable to find the specifications for the 86 power transformer. I have it now...thanks to an ancient Philco parts catalog from 1933 or so.
Specifications:
Model 511 (Part No. 3073)
High voltage: 375 VCT (187.5-0-187.5), 37 mA; 1.5 V @ 4.3 A (26 tubes); 2.47 V @ 1.87 A (27); 4.85 VCT @ 0.35 A (71A); 4.8 V @ 2 A (80).
Model 86 (Part No. 3271)
High voltage: 575 VCT (287.5-0-287.5), 80 mA; 1.5 V @ 4.3 A (26 tubes); 2.47 V @ 1.87 A (27); 5 VCT @ 0.6 A (71A); 4.8 V @ 2 A (80).
So you need to drop 100 volts in your B+ line.
The easy fix would be to add a resistor in line with the B+ source to drop the voltage to a level acceptable to the 511. It's still a lot cheaper than having Heyboer custom wind a 511 transformer. Yeah, I know I called it a "kludge" but sometimes you gotta do what ya gotta do...
The Utah will work with the 511 if it has an audio output transformer to match the 511 high impedance output to the Utah's voice coil. I used to have one of those units, finally gave it away at an IHRS swap last Fall. But as far as dropping the B+, the Utah won't really help you, since its electrodynamic speaker has a built-in power supply.
Just add a resistor to the B+ source of the 511. We know the current draw of the B+ in the 511 - 37 mA. We know we need to drop 100 volts. So, applying Ohm's law - R=E/I - we get R = 100 divided by 0.037 which equals 2702.7 ohms. Let's use 2700 as a nice round figure. Now, we need to determine the dissipation in watts. For this, P=I squared R, so P = 0.037 x 0.037 x 2700 = 3.7 watts. Let's double that for safety, which comes out to a little over 7 watts. Therefore, you need a 2700 ohm resistor rated at 10 watts to safely drop the B+ to a level usable by the 511.
You will need to break the connection between terminal 2 of the filter capacitor bank and terminal 2 of the large power resistor, and insert the new 2700 ohm resistor in this line, as shown here:
[Image: http://i326.photobucket.com/albums/k420/...mod_sm.jpg]
Click the image, or click here for a larger version of this partial schematic.
Again, I apologize for the mixup.
I owe you an apology, Robert. At the time I recommended the 86 transformer, my mind was on wrapping up college instead of being fully focused on your transformer problem.
I thought I had a Model 86 & 82 service manual, but if I do, I can't find it right now. So until this morning, I was unable to find the specifications for the 86 power transformer. I have it now...thanks to an ancient Philco parts catalog from 1933 or so.
Specifications:
Model 511 (Part No. 3073)
High voltage: 375 VCT (187.5-0-187.5), 37 mA; 1.5 V @ 4.3 A (26 tubes); 2.47 V @ 1.87 A (27); 4.85 VCT @ 0.35 A (71A); 4.8 V @ 2 A (80).
Model 86 (Part No. 3271)
High voltage: 575 VCT (287.5-0-287.5), 80 mA; 1.5 V @ 4.3 A (26 tubes); 2.47 V @ 1.87 A (27); 5 VCT @ 0.6 A (71A); 4.8 V @ 2 A (80).
So you need to drop 100 volts in your B+ line.
The easy fix would be to add a resistor in line with the B+ source to drop the voltage to a level acceptable to the 511. It's still a lot cheaper than having Heyboer custom wind a 511 transformer. Yeah, I know I called it a "kludge" but sometimes you gotta do what ya gotta do...
The Utah will work with the 511 if it has an audio output transformer to match the 511 high impedance output to the Utah's voice coil. I used to have one of those units, finally gave it away at an IHRS swap last Fall. But as far as dropping the B+, the Utah won't really help you, since its electrodynamic speaker has a built-in power supply.
Just add a resistor to the B+ source of the 511. We know the current draw of the B+ in the 511 - 37 mA. We know we need to drop 100 volts. So, applying Ohm's law - R=E/I - we get R = 100 divided by 0.037 which equals 2702.7 ohms. Let's use 2700 as a nice round figure. Now, we need to determine the dissipation in watts. For this, P=I squared R, so P = 0.037 x 0.037 x 2700 = 3.7 watts. Let's double that for safety, which comes out to a little over 7 watts. Therefore, you need a 2700 ohm resistor rated at 10 watts to safely drop the B+ to a level usable by the 511.
You will need to break the connection between terminal 2 of the filter capacitor bank and terminal 2 of the large power resistor, and insert the new 2700 ohm resistor in this line, as shown here:
[Image: http://i326.photobucket.com/albums/k420/...mod_sm.jpg]
Click the image, or click here for a larger version of this partial schematic.
Again, I apologize for the mixup.
--
Ron Ramirez
Ferdinand IN
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