Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: fpolak on January 23, 2021, 08:29:04 am
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I purchased Mercury magnetics FTDP/240-M, with 380v secondary output... an excesive B+ voltages for original clone
It says in description;
Fender Tweed Deluxe (10–15 watts)
Origin: The specs did not vary on virtually all original (1950–57) Deluxe PTs (with the exception of the mounting brackets). Our clone is from a 1954
When I connect to load, voltages goes to 420DCV. An excesive b+ compared with 370DCV of original schematic.
What can i do to drop voltages to correct values??? I'm between mosfet + zener diode from RG Keen or dropping voltage resistor
What do you recommend without affecting amp tone?
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The voltage quoted is probably within specs -- 5E3 schematics usually say +/- 20% on voltages.
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If 6V6s bias is too cold, voltage will be high.
I'll stay with that voltage ; 420Vdc
I, sometime use Zeners stack with success;
https://robrobinette.com/Generic_Tube_Amp_Mods.htm#B+1_Voltage
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I purchased Mercury magnetics FTDP/240-M, with 380v secondary output... an excesive B+ voltages for original clone
When I connect to load, voltages goes to 420DCV. An excesive b+ compared with 370DCV of original schematic.
380VAC secondary seems way too high for a 5E3. I used a 330VAC PT and got 340VDC for B+.
I've never seen a Fender schematic that showed voltages for a 5E3. Can you post an original schematic that shows 370VDC for B+?
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PT current is good to know, not only voltage.
More no load current may result in higher DC voltage for same secondary voltage.
AB6763 Deluxe Reverb schematic ( in front of me) show a 330 -0 330 PT for a 420 volts DC
How can you have only 340 VDC Sluckey ?
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AB6763 Deluxe Reverb schematic ( in front of me) show a 330 -0 330 PT for a 420 volts DC
Completely different amp to a 5E3
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There is quite a variance in B+ voltages , depending on the 5Y3 used . Recently I swapped different 5Y3's into a 5E3 clone and the B+ varied from 376 volts to 354 volts . All NOS tubes, RCA , GE , Rogers .
What is the voltage at the 6V6 cathodes ?
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380VAC secondary seems way too high for a 5E3. I used a 330VAC PT and got 340VDC for B+.
I've never seen a Fender schematic that showed voltages for a 5E3. Can you post an original schematic that shows 370VDC for B+?
Yes, 380VAC it too high for this amp, but it says original specs....
I read voltages from my attached picture
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I solved voltages problem. I connected 220VAC line to 240VAC primary cable (after 220v cable are connected). Now I have 384VDC output for B+, to much closer to 370.
Now I need to correct bias, because I change 250 ohms cathode resistor to 470 ohms/10w. What value do I need now to correct bias??
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AB6763 Deluxe Reverb schematic ( in front of me) show a 330 -0 330 PT for a 420 volts DC
How can you have only 340 VDC Sluckey ?
You're comparing apples and oranges. Fixed bias amp idles at much less current than cathode bias amp. Therefore, same PT will give higher B+ for fixed bias amp due to lighter load. Also, the DR uses a 5AR4 rectifier and 5E3 uses a 5Y3 rectifier. The 5Y3 has a higher voltage drop than the 5AR4.
Example... I used this same 330V PT in a Deluxe AB763 amp using 5AR4. My B+ was 428V. Bias was set for 24mA for each 6V6. When I put this PT in my 5E3 using 5Y3, my B+ dropped to 340V. Bias was set for 36mA for each 6V6.
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What value do I need now to correct bias??
measure the VDC at the top of whatever Cathode R you have in, then;
vdc/R = tube current.
Tube current * plate VDC ~~~~~~~= to tube power in Watts
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I solved voltages problem. I connected 220VAC line to 240VAC primary cable (after 220v cable are connected). Now I have 384VDC output for B+, to much closer to 370.
Now I need to correct bias, because I change 250 ohms cathode resistor to 470 ohms/10w. What value do I need now to correct bias??
First off, one little note... That Weber layout is NOT the original. If you want the same voltages that are listed on the Weber layout you will have much better luck by using the Weber W025130 PT (only 340VAC) and the Weber WY3GT solid state rectifier. My point is, 370V is not a magic number and should not be used as a standard for a 5E3 amp.
If you are not already using a NOS 5Y3, then get one. It will drop B+ more than a modern 5Y3.
I think your amp will be very happy with B+ at 384V. If you like the sound with the 470Ω cat resistor, I'd leave it alone. 250Ω is the Fender correct value and if you change back your B+ will decrease.
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First off, one little note... That Weber layout is NOT the original. If you want the same voltages that are listed on the Weber layout you will have much better luck by using the Weber W025130 PT (only 340VAC) and the Weber WY3GT solid state rectifier. My point is, 370V is not a magic number and should not be used as a standard for a 5E3 amp.
If you are not already using a NOS 5Y3, then get one. It will drop B+ more than a modern 5Y3.
I think your amp will be very happy with B+ at 384V. If you like the sound with the 470Ω cat resistor, I'd leave it alone. 250Ω is the Fender correct value and if you change back your B+ will decrease.
I'm using an old RCA USA 5Y3
So, 384 B+ is correct value for 5e3. If I want lower B+, can I change 470 to 250 cat resistor? So much closer to original, is it correct???
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What value do I need now to correct bias??
measure the VDC at the top of whatever Cathode R you have in, then;
vdc/R = tube current.
Tube current * plate VDC ~~~~~~~= to tube power in Watts
24 vdc/465 ohms = 0,0516129032258065
0,051 mA * 373 vdc = 19,25 w???
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If you are using 1 R for both tubes, then divide (current) by 2 1st - sorry for the oversight - i always use 1 R per tube
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The 19.25W figure you have is the power across the tubes plus the power across the cathode resistor, which is 1.2W. So, tube power is actually 18W. Here's a better way...
To determine power dissipated by the tube you must know the voltage ***ACROSS*** the tube and the current through the tube.
Here's how... Measure the plate voltage and the cathode voltage. Subtract the cathode voltage from the plate voltage. This is the voltage across the tube.
Calculate the current through the tube by dividing the cathode voltage by the value of the cathode resistor.
Finally, calculate the tube power dissipation by multiplying the voltage across the tube times the current through the tube.
Be aware that the power number you get is for two tubes and it is only accurate if the plate voltage for each tube is the same.
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The 19.25W figure you have is the power across the tubes plus the power across the cathode resistor, which is 1.2W. So, tube power is actually 18W. Here's a better way...
To determine power dissipated by the tube you must know the voltage ***ACROSS*** the tube and the current through the tube.
Here's how... Measure the plate voltage and the cathode voltage. Subtract the cathode voltage from the plate voltage. This is the voltage across the tube.
Calculate the current through the tube by dividing the cathode voltage by the value of the cathode resistor.
Finally, calculate the tube power dissipation by multiplying the voltage across the tube times the current through the tube.
Be aware that the power number you get is for two tubes and it is only accurate if the plate voltage for each tube is the same.
382 VDC cathode voltage - 373 VDC plate voltage = 9
382 VDC cathode voltage / 465 ohms cathode resistor = 0,821
9 * 0.821 = 7,389 Watts???
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382 VDC cathode voltage
Check that one again, should be <20vdc
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382 VDC cathode voltage
Check that one again, should be <20vdc
373 VDC plate voltage - 24,2 VDC cathode voltage = 348,8
24,2 VDC cathode voltage / 465 ohms cathode resistor = 0,052
348,8 * 0,052 = 18,15 Watts
9 watts per tube it's so cold. I'll change cathode resistor from 470 to original 250 ohms and check voltages again.
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AB6763 Deluxe Reverb schematic ( in front of me) show a 330 -0 330 PT for a 420 volts DC
Completely different amp to a 5E3
My answer is about Sluckey's transformer, same AC volt as a AB763 DR
I understand is apple and orange.
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I understand is apple and orange.
Does that mean you understand my answer to your question, "How can you have only 340 VDC Sluckey ?"
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9 watts per tube it's so cold. I'll change cathode resistor from 470 to original 250 ohms and check voltages again.
I changed cathode resistor. New values;
342 VDC plate voltage - 18,5 VDC cathode voltage = 323,5
18,5 VDC cathode voltage / 254 ohms cathode resistor = 0,072
323,5 * 0,072 = 23,56 Watts
11,78 Watts per tube.
Is it a good value for this amp? What's best values for this amp/tubes?
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I think your numbers are perfect. Very close to my numbers.
http://sluckeyamps.com/5e3/5e3.pdf
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I'm very confused... I calculated bias with robrobinette's web page but it dissipate 11,8 watts at 84.3%
Rob Robinette says class A/B max safe dissipation is about 70%, but in other part says up to 85% is safe.
What's maximum safe dissipation for 5E3?
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Rob Robinette says class A/B max safe dissipation is about 70%, but in other part says up to 85% is safe.
This statement applies to ***FIXED BIAS*** amps only.
Your 5E3 is CATHODE BIASED. Safe max. dissipation is 100% for any cathode biased amp.
How does your amp sound?
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Rob Robinette says class A/B max safe dissipation is about 70%, but in other part says up to 85% is safe.
This statement applies to ***FIXED BIAS*** amps only.
Your 5E3 is CATHODE BIASED. Safe max. dissipation is 100% for any cathode biased amp.
How does your amp sound?
Sounds great!! I concerned about max dissipation, but now it's clear.
Thank you!!!