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Old 23rd October 2006, 06:56 PM   #1
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Default How to calc Resistor Values to reduce Voltage?

B+ secondary reading is 257V. Using 6dj8 circuit, and this tube is rated .325 amp, and requires 200V...so I need to reduce 57V (257-200)

So Ohm's Law says V=IR or 57V / .325 = R = 175 Ohm.

And P = VI or 57V x .325 amp = 18.5 watts x 2 (to be safe) = 37 watt.

I must be doing this way wrong as another fella suggested a 25K ohm resistor rated at 2 watts.

Please set me straight.
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Old 23rd October 2006, 07:22 PM   #2
AndrewT is offline AndrewT  Scotland
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Hi,
what plate current will your tube operate at?
Now calculate what resistor you need at the quiescent plate current.

Now what range of plate currents flow when the tube is amplifying a signal.
What happens to the plate voltage as the plate current varies.

Is this good for an audio amplifier?
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Old 23rd October 2006, 07:29 PM   #3
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Andrew is right. Your formula is correct, but it is not the RATED plate current that you need to plug in but the ACTUAL plate current in the circuit, of course. Makes sense, no? Same for the anode voltage. What is the ACTUAL voltage in your circuit? That determines (with the supply voltage) the number of volts you need to waste, and that gives you the R and the P.

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Old 23rd October 2006, 07:37 PM   #4
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The voltage reading is 257 volts. 200 volts is the spec for the 6dj8 output stage I assembled.

I'm a beginner. I can measure voltage and ohms. But I haven't measured current yet. I'll have to read up on how to do this with my multimeter.
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Old 24th October 2006, 06:31 AM   #5
jarthel is offline jarthel  Australia
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Quote:
Originally posted by riotubes
The voltage reading is 257 volts. 200 volts is the spec for the 6dj8 output stage I assembled.

I'm a beginner. I can measure voltage and ohms. But I haven't measured current yet. I'll have to read up on how to do this with my multimeter.

to measure current, insert a 5W ceramic 10ohm resistor between B+ and plate. measure voltage. use ohm's law to compute for current.
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Old 28th October 2006, 04:54 AM   #6
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Default measuring voltage "across" a cap?

Thanks Jarthel. I'll pick up a resistor tomorrow and try.

When someone says "measure the voltage across the cap" that is soldered to a circuit board, and the leads are not accessible (except by turning the board upside down) does this mean that you can place the DVM leads: (a) on the solder joints of the cap's leads on the underside of the board or (b) just on the cap itself respecting the positive and neutral orientation of the cap?
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Old 31st October 2006, 01:58 AM   #7
jarthel is offline jarthel  Australia
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Default Re: measuring voltage "across" a cap?

Quote:
Originally posted by riotubes

When someone says "measure the voltage across the cap" that is soldered to a circuit board, and the leads are not accessible (except by turning the board upside down) does this mean that you can place the DVM leads: (a) on the solder joints of the cap's leads on the underside of the board or (b) just on the cap itself respecting the positive and neutral orientation of the cap?
options a and b seems the same to me. where else would you measure a voltage on point that is connected directly to the leads of the caps. so just turn the pcb upside-down and measure the voltage on the solder joints.

as for polarity if it's an elcap and you get negative voltage, just remove and the negative symbol and you get the proper voltage.
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