I'm trying to brush a little bit on my toroid transformer knowledge here. I have a new class A amplifier up and running.
It's powered by a 500VA transformer with 36VAC secondaries rated for 7A each. I tried to measured secondary current under load but
didn't have much luck. Secondary DC voltage is 47 volts loaded. Secondary DC current is 2A per side. How close am I to the theoretical
recommended VA limit of my transformer. Any easy formulas for future calculations?
Regards,
Dan
It's powered by a 500VA transformer with 36VAC secondaries rated for 7A each. I tried to measured secondary current under load but
didn't have much luck. Secondary DC voltage is 47 volts loaded. Secondary DC current is 2A per side. How close am I to the theoretical
recommended VA limit of my transformer. Any easy formulas for future calculations?
Regards,
Dan
Why can't you measure secondary current. How do you do it and what goes wrong.
You write "with 36VAC secondaries" (plural). Secondary DC voltage (singular) is 47VDC. Secondary DC current is 2A per side. Is it a symmetrical +/- 47 power supply? Please explain the circuit.
If the secondary DC current is 2A per side assuming you have +/- 47V, you draw approximately 200W total. This is way below the 500VA rating of the transformer. The power factor will be less than one, but with such a huge margin it does not really matter. The average current thru the transformer is relevant so don't break your head thinking about current waveforms etc. Watts are Watts.
In Europe we have nice energy meters to measure consumption of appliances and devices. Do those exist in USA? If so, you can use one. They will indicate power and power factor. So (power / power factor) = VA.
You write "with 36VAC secondaries" (plural). Secondary DC voltage (singular) is 47VDC. Secondary DC current is 2A per side. Is it a symmetrical +/- 47 power supply? Please explain the circuit.
If the secondary DC current is 2A per side assuming you have +/- 47V, you draw approximately 200W total. This is way below the 500VA rating of the transformer. The power factor will be less than one, but with such a huge margin it does not really matter. The average current thru the transformer is relevant so don't break your head thinking about current waveforms etc. Watts are Watts.
In Europe we have nice energy meters to measure consumption of appliances and devices. Do those exist in USA? If so, you can use one. They will indicate power and power factor. So (power / power factor) = VA.
Explain the circuit. 500VA transformer with dual secondaries rated at 36VAC and 7Amps AC each. Each secondary feeds a bridge rectifier and a CLC rectification
circuit. 47000uF/2.5mH/47000uF. Loaded voltage is 47VDC. Loaded DC (per channel) is 2 amps DC.
I tried to do an ac current measurement with a meter between the ac secondary and the bridge and it didn't work. More reading required or a different setup.
Regards,
Dan
circuit. 47000uF/2.5mH/47000uF. Loaded voltage is 47VDC. Loaded DC (per channel) is 2 amps DC.
I tried to do an ac current measurement with a meter between the ac secondary and the bridge and it didn't work. More reading required or a different setup.
Regards,
Dan
As @techtool says, it is probably the waveform of the current. However "didn't work" is not very clear. What did you see, was it something else than expected, was it unstable?
Never mind, unfortunately 98% of all electric engineers still think that a Digital Multimeter is useful because it provides 4 or 5 easy to read digits. And therefor is more more accurate and easier than an analog meter. For almost all audio related measurements an analog meter is better. It is a bit harder to read, but it is more accurate than the flashing digits of a DMM despite of the perceived inaccuracy of an analog scale.
I know, the next 17 posts will point out that DMM of brand X can measure frequencies up to Y, brand P has an update rate of Q times/second. And brand A can measure distorted waveforms of type B. They are all correct. It is just that you cannot measure what you wanted to measure.
Sorry for the rant. Your amplifier is class A, so power consumption is reasonably constant.With 200W you are fine.
PS. Ooops you are Canadian. Apologies for asking about "am energy meter available in the USA"
Dual secondaries so 2 x 36V x 7A = makes 504 VA. After rectifying and buffering you ruffly get 36V X root 2. ~ 50V DC each secondary dependent type of rectifier, reservoir capacitors and load. And since energy never comes free, the useful current after rectifying will sink from 7A to 5A.