Hi All,
My HV supply is a 5AR4 followed by a 50uF-10H-660uF CLC filtered supply. What power factor can I reasonably expect from such a supply? I know pure capacitive supplies are about 0.5 and LC filters close to 1, but not sure what it would be for a CLC filter.
The reason I ask is I am pulling about 150mA from a 200mA rated transformer secondary and it is getting very hot. I was hoping to pull even more current with some larger tubes but perhaps I'm already overdoing it.
Appreciate any thoughts!
Greg
My HV supply is a 5AR4 followed by a 50uF-10H-660uF CLC filtered supply. What power factor can I reasonably expect from such a supply? I know pure capacitive supplies are about 0.5 and LC filters close to 1, but not sure what it would be for a CLC filter.
The reason I ask is I am pulling about 150mA from a 200mA rated transformer secondary and it is getting very hot. I was hoping to pull even more current with some larger tubes but perhaps I'm already overdoing it.
Appreciate any thoughts!
Greg
Power factor is not constant and so easy to estimate exactly , it varies a lot with load and components . Buy a meter for power factor .
But if it runs very hot I doubt you can do something to change the situation , even knowing the power factor . It is what it is , replace it with a bigger one .
But if it runs very hot I doubt you can do something to change the situation , even knowing the power factor . It is what it is , replace it with a bigger one .
This is what Lundahl have on the power transformer datasheeet.
Output current from rectifier: 63% of above with condensor input rectifier, 95% of above with choke input rectifier.
Output current from rectifier: 63% of above with condensor input rectifier, 95% of above with choke input rectifier.
Center-tapped winding should be able to supply a DC load about the same as its AC rating; they're normally rated for DC load with cap input, since that's the most likely use. I simulated it in PSU Designer, assuming 100 Ohm center-tapped winding and got 200 mA in secondary with 150 mA load. More accurate with actual transformer measurements, though.
Kill-a-Watt measures RMS current, actual watts, VA, power factor for $28, there are knock-offs for less.
Kill-a-Watt measures RMS current, actual watts, VA, power factor for $28, there are knock-offs for less.
Thanks for the replies everyone and taking the time to sim it Tom.
Looks like I'm pretty much on the limit. I definitely won't be putting in the KT88's and have learnt a good lesson for the next amp. As it is, I get a couple of hours use and can keep my hand on it so I think I will live with it for now.
Looks like I'm pretty much on the limit. I definitely won't be putting in the KT88's and have learnt a good lesson for the next amp. As it is, I get a couple of hours use and can keep my hand on it so I think I will live with it for now.
Typically, a B+ filter that consists of a 5uF cap, then a 5H choke, followed by a 100uF cap will have a considerably lower power factor,
versus a 50uF cap, then a 20H choke, followed by a 55uF cap (same total filter capacitance in each filter).
All Generalizations have exceptions; including this Generalization.
versus a 50uF cap, then a 20H choke, followed by a 55uF cap (same total filter capacitance in each filter).
All Generalizations have exceptions; including this Generalization.
Thanks for the advice all. I simmed my transformer and power supply in psud2 and got 280mA RMS through the transformer, so not looking good - quite a bit above the 200mA spec.
On the plus side, I had previously written to Hammond to ask about what constitutes a normal operating temp and while they didn't provide a specific number, they did say that the transformers will run quite hot at rated power.
Looks like I have the excuse I need to build another amp haha.
On the plus side, I had previously written to Hammond to ask about what constitutes a normal operating temp and while they didn't provide a specific number, they did say that the transformers will run quite hot at rated power.
Looks like I have the excuse I need to build another amp haha.
If you can keep your hand on the PT after it's been running for 2 hours, that's not really hot. I would consider that to be relatively cool, actually. I have vintage gear that runs much hotter than that and the PTs have been functioning just fine for over 60 years.
If the PT is legitimately rated at 200mA and you're only drawing 150mA then you've got nothing to worry about.
merlin el mago,
Take a 350V-0-350V secondary, 2 silicon diodes, a 7H choke, 2 each 100uF capacitors, and a 5000 Ohm power resistor.
For example:
1. Wire it for full wave (center tap to ground [cap -]) diode anodes from 350v and 350V ends; 100uF capacitor input filter: - to ground, and + to the diodes cathodes; then 7H choke to the second 100uF cap; and to the 5000 Ohm load.
You will get a little less than 500VDC, and a little less than 100mA.
The Volt x Amps will be far more than the Watts.
Perhaps you will measure the primary load as ~ 50 Watts, but measure the primary load as ~ 90 Volt x Amps; on a KiloWatt meter.
Suppose the numbers actually measure as 50 Watts and 90 Volt x Amps, the power factor is 0.556.
2. Wire it for full wave (center tap to ground [cap -]) diode anodes from 350v and 350V ends; 7H choke input filter to the diodes cathodes; then 2 x 100uF cap, cap - to ground, choke output to caps +; and the 5000 Ohm load across the parallel 100uF caps.
You will get a little less than 315VDC, and a little less than 63mA.
The Volt x Amps will be almost equal to the watts.
Perhaps you will measure the primary load as ~ 20.5 Watts, but measure the primary load as ~ 23 Volt x Amps; on a KiloWatt meter.
Suppose the numbers actually measure as 20.5 Watts and 23 Volt x Amps, the power factor is 0.891.
3. But to be fair, we need ~ 500VDC, and ~ 100mA load, so here goes . . .
Use 555.6V-0-555.6V primary, and 7H choke input filter, and the rest of the circuit the same as the 350-0-350 choke input filter circuit, you will get a little less than 500V, and a little less than 100mA.
The primary Watts will be ~ 50 Watts, and ~ 56.1 Volt Amps; on a KiloWatt meter.
The load on the rectifier diodes: either a capacitor input filter, or a choke input filter . . .
As you can see, the type of input filter definitely will influence the power factor that the power mains sees.
Cap input filters have DCV out of about 1.4 x the rms input voltage. You will not get this much voltage, if the input cap is small in relation to the final DC load current on the B+.
Choke input filters have DCV out of about 0.9 x the rms input voltage.
Notes:
For a choke input filter, the inductance you need is at least 350/DC mA load current, that number in Henrys.
350/100mA = 3.5 Henry. 350/mA load is how you calculate critical inductance (for 60Hz power mains; for 50Hz power mains you multiply the Henrys by 60/50, which is 1.2). So if you calculated 3.5 Henry for 60Hz . . .
For 50Hz power mains you need 1.2 x 3.5H = 4.2 Henry
For a cap input filter, you need a fair sized input capacitance.
But a (small) 0.5uF input cap, 7H choke, and 200uF second capacitor following the choke output; is not really a cap input filter, Instead it is between a cap input filter definition, and a choke input filter definition.
Now, add in one or more secondaries, and load them with AC powered tube filaments (they present a resistive load), and that will affect the power factor (ratio of Watts to Volt x Amps) the load that the power mains 'sees' from the primary.
Take a 350V-0-350V secondary, 2 silicon diodes, a 7H choke, 2 each 100uF capacitors, and a 5000 Ohm power resistor.
For example:
1. Wire it for full wave (center tap to ground [cap -]) diode anodes from 350v and 350V ends; 100uF capacitor input filter: - to ground, and + to the diodes cathodes; then 7H choke to the second 100uF cap; and to the 5000 Ohm load.
You will get a little less than 500VDC, and a little less than 100mA.
The Volt x Amps will be far more than the Watts.
Perhaps you will measure the primary load as ~ 50 Watts, but measure the primary load as ~ 90 Volt x Amps; on a KiloWatt meter.
Suppose the numbers actually measure as 50 Watts and 90 Volt x Amps, the power factor is 0.556.
2. Wire it for full wave (center tap to ground [cap -]) diode anodes from 350v and 350V ends; 7H choke input filter to the diodes cathodes; then 2 x 100uF cap, cap - to ground, choke output to caps +; and the 5000 Ohm load across the parallel 100uF caps.
You will get a little less than 315VDC, and a little less than 63mA.
The Volt x Amps will be almost equal to the watts.
Perhaps you will measure the primary load as ~ 20.5 Watts, but measure the primary load as ~ 23 Volt x Amps; on a KiloWatt meter.
Suppose the numbers actually measure as 20.5 Watts and 23 Volt x Amps, the power factor is 0.891.
3. But to be fair, we need ~ 500VDC, and ~ 100mA load, so here goes . . .
Use 555.6V-0-555.6V primary, and 7H choke input filter, and the rest of the circuit the same as the 350-0-350 choke input filter circuit, you will get a little less than 500V, and a little less than 100mA.
The primary Watts will be ~ 50 Watts, and ~ 56.1 Volt Amps; on a KiloWatt meter.
The load on the rectifier diodes: either a capacitor input filter, or a choke input filter . . .
As you can see, the type of input filter definitely will influence the power factor that the power mains sees.
Cap input filters have DCV out of about 1.4 x the rms input voltage. You will not get this much voltage, if the input cap is small in relation to the final DC load current on the B+.
Choke input filters have DCV out of about 0.9 x the rms input voltage.
Notes:
For a choke input filter, the inductance you need is at least 350/DC mA load current, that number in Henrys.
350/100mA = 3.5 Henry. 350/mA load is how you calculate critical inductance (for 60Hz power mains; for 50Hz power mains you multiply the Henrys by 60/50, which is 1.2). So if you calculated 3.5 Henry for 60Hz . . .
For 50Hz power mains you need 1.2 x 3.5H = 4.2 Henry
For a cap input filter, you need a fair sized input capacitance.
But a (small) 0.5uF input cap, 7H choke, and 200uF second capacitor following the choke output; is not really a cap input filter, Instead it is between a cap input filter definition, and a choke input filter definition.
Now, add in one or more secondaries, and load them with AC powered tube filaments (they present a resistive load), and that will affect the power factor (ratio of Watts to Volt x Amps) the load that the power mains 'sees' from the primary.
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Thanks for the replies everyone, this has been very educational. I'm not too concerned with the current situation but my plans to run some bigger tubes are suspended. I think I will get a power meter just for curiosity's sake. Next time I will run a bigger xformer and/or choke input supply which has a PF close to 1.