Ohm's law and a Fisher 500 C

Mikeorg

Active Member
I've put the new bridge rectifier in and got the Phono stage running again, but now have to do the math as to the correct impedance V-I-R, I am working on understanding and learning Ohm's law~IS There and easier softer way?~If I'm to continue working on electrical components I am going to have to! And so here is the "question" and please take into account I am a hobbyist not a trained Electrician, How do I get these readings Voltage and current and when I can calculate the resistance needed,and where in the circuit will I place the resister? I do hope that these questions don't insult anyone but I am trying to learn this the best I can. Thanks...IMG_4129.JPG IMG_4083.JPG IMG_4082.JPG
 
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Are you trying to do something specific or are you just wanting to learn?
 
Are you trying to do something specific or are you just wanting to learn?
Well pin 4 on tube V17 of the phono section was at -22 after I have replaced the selenium rectifier"Before I got the tubes in that section to light up" and after I cleaned and tightened the pins the reading went up to -27 so I'm thinking maybe I need to put a resister somewhere in line to bring this reading down to -22 again?~Learning to do something specific...
 
rufleruf has the right idea to ask about line voltage, but you are 22% high between -22 to -27 which is more than can be attributed to line voltage alone. You could insert a high wattage (10W) 10 ohm resistor in the line coming off the top of the bridge and then measure the voltage across the 10ohm resistor. From that voltage measurement you can back-calculate the current draw and then figure out the real resistor you need.

1. To calculate current I = (V / R) where initially you are just using R = 10 ohms and you want to measure the DC voltage across the 10 ohm resistor with a multimeter set for DC volts.

2. Once you know "I" then you can determine the real resistor you need where the R = (V / I) and you would have measured "I" already and you know you want to drop 5V for the 'V' to get back down from -27 to -22.

3. Then you can figure out the power dissipation for your resistor which is Power = V x I (I'd multiply the resulting Power by 2 and find a resistor with at least that much power rating so you would have a safety factor.)

Also, if you end up mitigating some of your high line voltage with a variac or a bucking transformer then the voltage difference you have to correct for will be less than the 5V mentioned above.

Let us know if you use this suggestion. I'd be interested to hear what is the current draw in your testing.
 
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When I asked I was thinking about my variac - it goes to 12 in the Spinal Tap sense of things. At full on I get 145 volts ac at the plug. Could account for what you are seeing.

I had no idea my variac at 100% was so high until I put a meter on it.
 
When I asked I was thinking about my variac - it goes to 12 in the Spinal Tap sense of things. At full on I get 145 volts ac at the plug. Could account for what you are seeing.

I had no idea my variac at 100% was so high until I put a meter on it.

Many variacs have "overdrive" and 140 V is quite common. There are often multiple output taps that might allow you to limit it to 120 V.
 
I have a big line drawn on it at 120 now - it's an old one without a gauge - am tempted to install a gauge. If I am doing anything that affects internal voltages I just plug it into the wall so it will behave as expected in my house.
 
What's the OP's line voltage? Are all the voltages in the unit higher than spec or just this one spot?
 
InkedIMG_4147_LI.jpg I measured the line voltage from my home sockets and they run 119v I added a 18 ohm resister into the bias section that I took the selenium rectifier out of and added the full bridge rectifier into, the reading on V17 12AX7 is -25.7 on pin 4 and -12 on pin 5, the reading from the power tubes V8 through V11 is #8 -21.0 #9 -21.0 #10 -19.1 #11 -19.1...View attachment 1302824
 
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