Francois G,
As you know, (I say again for others who do not know) the critical inductance calculation is obtained from knowing the load current; the mains power frequency; and the 2x full wave rectification frequency is most often used - so it is assumed to be that; and not 1/2 wave rectification.
The problem that comes with the calculation is what is the minimum current draw of the amplifier.
If there are 4 output tubes drawing 35mA each, that is 140mA.
If they each draw 42.5mA, that is 170mA.
You used the correct factor of 420 for 50Hz (100Hz full wave). For 60Hz power (120Hz full wave) the factor is 350.
And 2.5H is correct for 170mA.
I do like to use a choke that is larger than the critical inductance number, especially if the quality of the choke is unknown. 5H gives a very nice margin.
If we use true choke input for the power supply, then we will get 0.9 x the Vrms from the power transformer secondary (but of course we have to account for the additional loss due to primary DCR, secondary DCR, rectifier voltage drop, and choke DCR.
The voltage drop of a tube rectifier is less for choke input filter than it is for capacitor input filter (average current, versus very large current transients), and the same goes for all the other DCRs, more voltage drop with capacitor input, versus voltage drops with choke input.
But the cap input starts with 1.4 x the Vrms of the secondary; that means we have to use a higher voltage secondary to get the same B+ voltage as the capacotor input filter (1.4 x Vrms capacitor; 0.9 x Vrms choke input).
What runs cooler?
For a given B+ voltage, and given B+ load current . . .
The heat of the power transformer primary, secondary; choke, and rectifier tube are far less with choke input (lower, average current).
The heat of the power transformer primary, secondary; (choke after the input capacitor), and rectifier tube are far more with capacitor input (larger peak current).
Heat is I squared x R. ( I = current)
The integral of I squared x R for the capacitor input filter, is more than the integral for I squared R of the choke input filter.
Then there is the 'pseudo choke input filter' that uses a small capacitor in front of the choke.
Consider the peak current versus average current of the following:
47uF capacitor at 100Hz is 34 Ohms of capacitive reactance (capacitor input filter) Peak Current
2uF capacitor at 100Hz is 796 Ohms of capacitive reactance (the small capacitor before the choke on a pseudo choke input filter) Peak Current
5H choke at 100Hz is 3,140 Ohms of inductive reactance (choke input filter) Average Current
Which of those 3 reactances is harder for the power transformer DCRs, and the vacuum tube to put current into?
Just saying
As you know, (I say again for others who do not know) the critical inductance calculation is obtained from knowing the load current; the mains power frequency; and the 2x full wave rectification frequency is most often used - so it is assumed to be that; and not 1/2 wave rectification.
The problem that comes with the calculation is what is the minimum current draw of the amplifier.
If there are 4 output tubes drawing 35mA each, that is 140mA.
If they each draw 42.5mA, that is 170mA.
You used the correct factor of 420 for 50Hz (100Hz full wave). For 60Hz power (120Hz full wave) the factor is 350.
And 2.5H is correct for 170mA.
I do like to use a choke that is larger than the critical inductance number, especially if the quality of the choke is unknown. 5H gives a very nice margin.
If we use true choke input for the power supply, then we will get 0.9 x the Vrms from the power transformer secondary (but of course we have to account for the additional loss due to primary DCR, secondary DCR, rectifier voltage drop, and choke DCR.
The voltage drop of a tube rectifier is less for choke input filter than it is for capacitor input filter (average current, versus very large current transients), and the same goes for all the other DCRs, more voltage drop with capacitor input, versus voltage drops with choke input.
But the cap input starts with 1.4 x the Vrms of the secondary; that means we have to use a higher voltage secondary to get the same B+ voltage as the capacotor input filter (1.4 x Vrms capacitor; 0.9 x Vrms choke input).
What runs cooler?
For a given B+ voltage, and given B+ load current . . .
The heat of the power transformer primary, secondary; choke, and rectifier tube are far less with choke input (lower, average current).
The heat of the power transformer primary, secondary; (choke after the input capacitor), and rectifier tube are far more with capacitor input (larger peak current).
Heat is I squared x R. ( I = current)
The integral of I squared x R for the capacitor input filter, is more than the integral for I squared R of the choke input filter.
Then there is the 'pseudo choke input filter' that uses a small capacitor in front of the choke.
Consider the peak current versus average current of the following:
47uF capacitor at 100Hz is 34 Ohms of capacitive reactance (capacitor input filter) Peak Current
2uF capacitor at 100Hz is 796 Ohms of capacitive reactance (the small capacitor before the choke on a pseudo choke input filter) Peak Current
5H choke at 100Hz is 3,140 Ohms of inductive reactance (choke input filter) Average Current
Which of those 3 reactances is harder for the power transformer DCRs, and the vacuum tube to put current into?
Just saying
Correction to my post # 456:
Wrong:
I said:
"Example:
A pair of 100 Ohm 3 Watt resistors, one in each plate lead . . .
Or, a single 100 Ohm 10 Watt resistor in the cathode lead.
The power transformer and the rectifier tube will not notice the difference."
. . . My Bad, a typing/brain error.
Correction:
I meant to say:
"Example:
A pair of 50 Ohm 3 Watt resistors, one in each plate lead . . .
Or, a single 100 Ohm 10 Watt resistor in the cathode lead.
The power transformer and the rectifier tube will not notice the difference."
. . . That is true, accurate, and correct
No wonder nobody 'liked' that post.
And, generally the single 10 Watt resistor has more surface area than two 3 Watt resistors, so the single resistor runs cooler
(the same total power is dissipated in both cases).
Wrong:
I said:
"Example:
A pair of 100 Ohm 3 Watt resistors, one in each plate lead . . .
Or, a single 100 Ohm 10 Watt resistor in the cathode lead.
The power transformer and the rectifier tube will not notice the difference."
. . . My Bad, a typing/brain error.
Correction:
I meant to say:
"Example:
A pair of 50 Ohm 3 Watt resistors, one in each plate lead . . .
Or, a single 100 Ohm 10 Watt resistor in the cathode lead.
The power transformer and the rectifier tube will not notice the difference."
. . . That is true, accurate, and correct
No wonder nobody 'liked' that post.
And, generally the single 10 Watt resistor has more surface area than two 3 Watt resistors, so the single resistor runs cooler
(the same total power is dissipated in both cases).
I've already ordered a range of spares including a CL-80....
CL-90 for 240VAC? Damn.
Good explanation. Thanks.
Hehe. I was hoping I will learn something new in an explanation of
Hehe. I was hoping I will learn something new in an explanation of
More seriously, the Amperex data sheet shows specifically that a resistor in each diode lead is required to make up the “minimum resistance”. The only reason I could think of is that there is slightly better load sharing between the two diodes, compared to one resistor in the centertap. Or is there?
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Don’t worry. CL-80 has less initial resistance than CL-90 (47 vs 120) but your CL-80 will be fine. Not critical.
I was pondering why @calpe was actually measuring a B+ of 369Vdc, while the simulation above, with the addition of only the parallel pair of 100 Ohm resistors, shows a reduced B+ of 324Vdc. As was pointed out before there could be differences between what PSUDII predicts and the actual measurements, but this difference seemed more than I expected.
I’m afraid there is another mistake in this PSUDII simulation that went unnoticed. The transformer secondary voltage used in the simulation was 300Vac, but we know @calpe’s transformer gives 310Vac. Contrary to what the simulation says I expect the pair of 100 Ohm resistors will most likely not lower the B+ enough. Additional work might be needed to drop it further. The simplest is probably as 6A3sUMMER suggested - put more resistance between the centertap in the secondary and ground. Initially the 150 Ohm, 5 watt R1 could be tried in this location to determine a better final value, without much rework. In calculating the dissipation in R1 I noticed that its dissipation is rather high, so we may need a bigger resistor here for the final one.
Sorry we did not notice the possible error in the simulation before you ordered the additional resistors.
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Just done a quick test with a 150 Ohm (3W) from secondary winding centre tap in series to the 0v input (ground) on the Tubelab PCB - T1-RED-YEL 02.
A.C. input measures 615VAC
B+ = 314V D.C.
Do i need to increase slightly the resistor value to obtain the B+ to be within 320-340V D.C. ?
I would appreciate a brief explanation of how adding this resistor effects the lowering of the voltage?
A.C. input measures 615VAC
B+ = 314V D.C.
Do i need to increase slightly the resistor value to obtain the B+ to be within 320-340V D.C. ?
I would appreciate a brief explanation of how adding this resistor effects the lowering of the voltage?
Did you measure B+ = 314Vdc with all tubes functioning in steady-state and speakers (or load resistor) connected? Eager to see the cathode voltages.
You have not installed the two 100 Ohm resistors when you made this measurement, correct? Hold off on the 100 ohm if the SPP seems to be working fine now. I suggest that you wrap things up and play/listen to the amp for several days before trying to raise and further tweak the B+. With the B+ around 320 V and 300 ohm cathode resistors it should work safely and sound good already.
The CT to ground resistor (R1 in a different location but serving the original purpose) will need to be of lower value (drop less voltage across the resistor) if you want to increase B+ voltage. Before ordering the “final” resistor you need to calculate the current through R1 (I=V/R), and calculate heat dissipation.
According to my estimate it is dissipating more that I would like with a resistor rated for 5 watts. (However, I have not yet heard of anyone having a failed R1 on SPP, so it is probably OK)
You have not installed the two 100 Ohm resistors when you made this measurement, correct? Hold off on the 100 ohm if the SPP seems to be working fine now. I suggest that you wrap things up and play/listen to the amp for several days before trying to raise and further tweak the B+. With the B+ around 320 V and 300 ohm cathode resistors it should work safely and sound good already.
The CT to ground resistor (R1 in a different location but serving the original purpose) will need to be of lower value (drop less voltage across the resistor) if you want to increase B+ voltage. Before ordering the “final” resistor you need to calculate the current through R1 (I=V/R), and calculate heat dissipation.
According to my estimate it is dissipating more that I would like with a resistor rated for 5 watts. (However, I have not yet heard of anyone having a failed R1 on SPP, so it is probably OK)
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Yep, 314V D.C. was measured, with load resistors connected!
Only thing i've changed temporarily is to add that 150 Ohm 3 Watt from the secondary winding tap to ground.
Eventually, perhaps a slight drop of the resistor to get to 320V.... if i have one.
I'll begin putting the PCB back on the top chassis plate too.
Only thing i've changed temporarily is to add that 150 Ohm 3 Watt from the secondary winding tap to ground.
Eventually, perhaps a slight drop of the resistor to get to 320V.... if i have one.
I'll begin putting the PCB back on the top chassis plate too.
The final resistance of R1 (currently 150 Ohm, 5 watt in the new location) can be easily lowered by adding a properly sized resistor in parallel to it. That way the dissipation in the current R1 will be reduced too.
I think it is OK to put the SPP all together now, and enjoy it, and later add a parallel resistor to raise the B+, if you locate R1 in an accessible place.
I think it is OK to put the SPP all together now, and enjoy it, and later add a parallel resistor to raise the B+, if you locate R1 in an accessible place.
Can't you just plug your loudspeakers in now and see how that is? You still might have to re-jig the OPT connections if the UL taps are reversed to the primary winding. Now's the time for fettling.
And if you do decide to add another resistor in parallel to the 150R one, the wattage can be less if the value is greater. For instance a 1K resistor makes the parallel resistance 130R, but if 20V is dropped across the pair, then the current in the 1K resistor is just 20/1000 = 0.02mA. Power = V x I, 0.2W (whereas the other one dissipates 2.7W).
And if you do decide to add another resistor in parallel to the 150R one, the wattage can be less if the value is greater. For instance a 1K resistor makes the parallel resistance 130R, but if 20V is dropped across the pair, then the current in the 1K resistor is just 20/1000 = 0.02mA. Power = V x I, 0.2W (whereas the other one dissipates 2.7W).
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I mis-calculated again.
refer to my Post # 462. Error, Error, Error.
I have to stop making late night posts.
Suppose the need is for 50 Ohm 3 Watt resistors in series with each plate lead (two resistors).
Then, the alternative is a single 50 Ohm 10 Watt resistor in series with the filament to B+ filter lead, and no resistor in the plate leads.
Now, suppose the secondary DCR from plate to center tap lead is 100 Ohms (each secondary to center tap).
So,
When one plate conducts, we have 100 Ohms DCR, and 50 Ohms resistor to the plate (150 Ohms total). Then we have the plate to filament voltage drop (perhaps 50V for a 5Y3 at a given current).
For a capacitor input filter, the peak current might easily be as much as 5 times the DC current (100mA DC load, but peak plate current is 500mA).
So the 5Y3 voltage drop is much more than 50V.
Or . . .
When one plate conducts, we have 100 Ohms DCR, then we have the plate to filament voltage drop (perhaps 50V for a 5Y3 at a given current).
Then we have the 50 Ohm 10 Watt resistor after the filament. 100 Ohms DCR and 50 Ohms resistor to the filament (150 Ohms total). That is the same thing as when the two 50 Ohm resistors are instead in the plate leads.
For a capacitor input filter, the peak current might easily be as much as 5 times the DC current (100mA DC load, but peak plate current is 500mA).
So the 5Y3 voltage drop is much more than 50V.
Then there is the case of using a higher voltage secondary, no series resistor in the plate leads, no series resistor in the filament to B+ (Choke input) filter, and . . .
Yes, a Choke input filter.
Now, the voltage drop in the rectifier will be about 50V, unlike the much larger voltage drop when using a capacitor input filter.
refer to my Post # 462. Error, Error, Error.
I have to stop making late night posts.
Suppose the need is for 50 Ohm 3 Watt resistors in series with each plate lead (two resistors).
Then, the alternative is a single 50 Ohm 10 Watt resistor in series with the filament to B+ filter lead, and no resistor in the plate leads.
Now, suppose the secondary DCR from plate to center tap lead is 100 Ohms (each secondary to center tap).
So,
When one plate conducts, we have 100 Ohms DCR, and 50 Ohms resistor to the plate (150 Ohms total). Then we have the plate to filament voltage drop (perhaps 50V for a 5Y3 at a given current).
For a capacitor input filter, the peak current might easily be as much as 5 times the DC current (100mA DC load, but peak plate current is 500mA).
So the 5Y3 voltage drop is much more than 50V.
Or . . .
When one plate conducts, we have 100 Ohms DCR, then we have the plate to filament voltage drop (perhaps 50V for a 5Y3 at a given current).
Then we have the 50 Ohm 10 Watt resistor after the filament. 100 Ohms DCR and 50 Ohms resistor to the filament (150 Ohms total). That is the same thing as when the two 50 Ohm resistors are instead in the plate leads.
For a capacitor input filter, the peak current might easily be as much as 5 times the DC current (100mA DC load, but peak plate current is 500mA).
So the 5Y3 voltage drop is much more than 50V.
Then there is the case of using a higher voltage secondary, no series resistor in the plate leads, no series resistor in the filament to B+ (Choke input) filter, and . . .
Yes, a Choke input filter.
Now, the voltage drop in the rectifier will be about 50V, unlike the much larger voltage drop when using a capacitor input filter.
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Just an update...
B+ is 314 ˜ 316v
Cathodes: -
V102 = 10.4v, V101 = 10.8v
V202 = 10.5v, V201 = 10.8v
Can someone advise what will the Valve dissipation be?
I'm going to put it back together, check mounting position of R1 150R (or slightly lower) 5W.
Ideally, i'll have two resistors to share the warm dissipation and place them away from the PCB.
Having that heat between the 5AR4 and V100 (ECC 81) is not a good idea.
B+ is 314 ˜ 316v
Cathodes: -
V102 = 10.4v, V101 = 10.8v
V202 = 10.5v, V201 = 10.8v
Can someone advise what will the Valve dissipation be?
I'm going to put it back together, check mounting position of R1 150R (or slightly lower) 5W.
Ideally, i'll have two resistors to share the warm dissipation and place them away from the PCB.
Having that heat between the 5AR4 and V100 (ECC 81) is not a good idea.
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10.8 V / 300 Ohm = 0.036 A
(316 V - 10.8 V) * 0.036 A = 10.9872 W
take a look here:
https://www.diyaudio.com/community/threads/tubelab-spp-first-timer-build.388172/post-7283589
Nearly 500 posts, this has been a long journey. I get the impression that you are very risk averse, and maybe you had not considered some of the leaps of faith that would be necessary to produce a vintage amplifier solution using high voltages, and solving the interworking issues of parts that had to be selected from different sources.
What is the verdict so far? Has this been a good journey or a bad one?
What is the verdict so far? Has this been a good journey or a bad one?
What the hell, lets aim for the magic 500!!!
Matters that fall under the risk factor are the costs being more than i thought and i didn't want a burn up.
It's better to be safe than sorry and with a hole in your pocket.
And yes, constructing a Valve amp is a different world.
Unlike the Elektor Q-Watt amp i built a few years ago and still running daily.
Verdict? It has been a pleasure making it and more especially from all you guys with your support.
Just need to aim for the 320Volts and mount R1 120/130/140/150 Ohms 5 Watt in a safe place.
Thanks