So that means there's a slight imbalance on the OT as well as between the tubes, seems your current setup is a bit better.
~Phil
Measuring for voltage drop (probe placement)
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Re: Measuring for voltage drop (probe placement)
tUber Nerd!
Re: Measuring for voltage drop (probe placement)
Well since you folks showed me what was wrong when I was trying to use the “shunt” method and I am very aware of the what lyies inside the box I tried it out using all safety precautions.
So using the Marstan shunt bias method below is what I came up with. And of coarse DO NOT TRY THIS METHOD IF YOU ARE NOT FAMILIAR WORKING WITH DEADLY HIGH VOLTAGES.
Using resistor method
V4 = 37.8ma
V5 = 30.2 ma
Marstan shunt method:
V4 = 38.5 ohms - 426.5vdc on pin 3 - (17.5/426.5 = .0410 = 41.0), 38.5*.0410 = 1.578v
V5 = 40 ohms - 425vdc on pin 3 - (17.5/425 = .0411 = 41.1), 40.0*.0411 = 1.66v
Reading pin 3 to CT
V4 = 1.297
V5 = 1.140
Set to below
V4 = 1.536v
V5 = 1.374v
So using the Marstan shunt bias method below is what I came up with. And of coarse DO NOT TRY THIS METHOD IF YOU ARE NOT FAMILIAR WORKING WITH DEADLY HIGH VOLTAGES.
Using resistor method
V4 = 37.8ma
V5 = 30.2 ma
Marstan shunt method:
V4 = 38.5 ohms - 426.5vdc on pin 3 - (17.5/426.5 = .0410 = 41.0), 38.5*.0410 = 1.578v
V5 = 40 ohms - 425vdc on pin 3 - (17.5/425 = .0411 = 41.1), 40.0*.0411 = 1.66v
Reading pin 3 to CT
V4 = 1.297
V5 = 1.140
Set to below
V4 = 1.536v
V5 = 1.374v
- pompeiisneaks
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Re: Measuring for voltage drop (probe placement)
Maybe I missed it, but you didn't finish the mA calculations from your shunt method:
V4: 1.536V / 38.5 ohms = 39.9mA
V5: 1.374V / 40 ohms = 34.4mA
To me that's quite odd, because the resistor method includes screen current which is purported to add around 5mA of current, but your resistor method is ~5mA less than the shunt method?
But the method I'd always been told was to just measure the voltage drop across the pin to CT which you have as:
V4: 1.297 / 38.5 ohms = 33.7mA
V5: 1.140 / 40 ohms = 28.5mA
which fits more with the 'minus screen current' idea. This marstan method, how is it more accurate if it's seeming to go against the other known data points? (I don't know it myself)
~Phil
V4: 1.536V / 38.5 ohms = 39.9mA
V5: 1.374V / 40 ohms = 34.4mA
To me that's quite odd, because the resistor method includes screen current which is purported to add around 5mA of current, but your resistor method is ~5mA less than the shunt method?
But the method I'd always been told was to just measure the voltage drop across the pin to CT which you have as:
V4: 1.297 / 38.5 ohms = 33.7mA
V5: 1.140 / 40 ohms = 28.5mA
which fits more with the 'minus screen current' idea. This marstan method, how is it more accurate if it's seeming to go against the other known data points? (I don't know it myself)
~Phil
tUber Nerd!
Re: Measuring for voltage drop (probe placement)
Phil,
Below is how in their document they come up with a calculated ma.
So here are my values and I italicized the section from the document for measurements that I used.
17.5/426.5 = .0410 = 41.0
3) Let's say that you want to bias the amp for 70% of the maximum plate dissipation of the power tube(KT66 and EL34's plate dissipation is 25 Watts maximum). Take 25W times 70% equals 17.5W. This is the plate dissipation that you will use for the remainder of the calculations.
4) Now we need to find the current draw needed to get 70% plate dissipation at 450VDC. Take 17.5W divided by 450VDC and you will get 0.0388A(38.8mA). This is the current draw per tube needed to get 70% plate dissipation with 450VDC on the plates.
5) Now we need to find the voltage drop across one half of the primary needed to give the 0.0388A of current. Take 0.0388A times 41 ohms equals 1.59VDC. This is the voltage you will be setting the bias pot to read while measuring across half of the primary of the output transformer. Check both halves of the transformer to make sure that both tubes are reading close to each other. Some transformers have a larger resistance difference between the two halves and will have to have the bias setup as a compromise(balance)between the two tubes. One tube will be set at a higher plate dissipation and the other will be set at a lower plate dissipation than the original calculated plate dissipation.
Just so everyone is clear, this is just a method someone told me about so wanted to try it. I had and continue to use resistor method. This was a test and only a test if it was a real life..........
James
Below is how in their document they come up with a calculated ma.
So here are my values and I italicized the section from the document for measurements that I used.
17.5/426.5 = .0410 = 41.0
3) Let's say that you want to bias the amp for 70% of the maximum plate dissipation of the power tube(KT66 and EL34's plate dissipation is 25 Watts maximum). Take 25W times 70% equals 17.5W. This is the plate dissipation that you will use for the remainder of the calculations.
4) Now we need to find the current draw needed to get 70% plate dissipation at 450VDC. Take 17.5W divided by 450VDC and you will get 0.0388A(38.8mA). This is the current draw per tube needed to get 70% plate dissipation with 450VDC on the plates.
5) Now we need to find the voltage drop across one half of the primary needed to give the 0.0388A of current. Take 0.0388A times 41 ohms equals 1.59VDC. This is the voltage you will be setting the bias pot to read while measuring across half of the primary of the output transformer. Check both halves of the transformer to make sure that both tubes are reading close to each other. Some transformers have a larger resistance difference between the two halves and will have to have the bias setup as a compromise(balance)between the two tubes. One tube will be set at a higher plate dissipation and the other will be set at a lower plate dissipation than the original calculated plate dissipation.
Just so everyone is clear, this is just a method someone told me about so wanted to try it. I had and continue to use resistor method. This was a test and only a test if it was a real life..........
James
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Re: Measuring for voltage drop (probe placement)
OH I see now, you're calculating the voltage across the transformer winding you need to hit your spot. Makes sense.Exojam wrote: ↑Fri Mar 08, 2019 7:28 pm Phil,
Below is how in their document they come up with a calculated ma.
So here are my values and I italicized the section from the document for measurements that I used.
17.5/426.5 = .0410 = 41.0
3) Let's say that you want to bias the amp for 70% of the maximum plate dissipation of the power tube(KT66 and EL34's plate dissipation is 25 Watts maximum). Take 25W times 70% equals 17.5W. This is the plate dissipation that you will use for the remainder of the calculations.
4) Now we need to find the current draw needed to get 70% plate dissipation at 450VDC. Take 17.5W divided by 450VDC and you will get 0.0388A(38.8mA). This is the current draw per tube needed to get 70% plate dissipation with 450VDC on the plates.
5) Now we need to find the voltage drop across one half of the primary needed to give the 0.0388A of current. Take 0.0388A times 41 ohms equals 1.59VDC. This is the voltage you will be setting the bias pot to read while measuring across half of the primary of the output transformer. Check both halves of the transformer to make sure that both tubes are reading close to each other. Some transformers have a larger resistance difference between the two halves and will have to have the bias setup as a compromise(balance)between the two tubes. One tube will be set at a higher plate dissipation and the other will be set at a lower plate dissipation than the original calculated plate dissipation.
Just so everyone is clear, this is just a method someone told me about so wanted to try it. I had and continue to use resistor method. This was a test and only a test if it was a real life..........
James
thanks,
~Phil
tUber Nerd!