Transformer current

[dr.power]

Hi guys,

I have got an step down transformer for my LM3886 power amplifier it is a dual transformer having a center tap, rated as 20V, So after rectifying it gives me +-56V. The seller told me that the said transformer is a 6A one, but I am suspect if he was right or not, so How can I measure the this factor by myself please?
The transformer gets hot when I connect it to my said LM3886 power amp. the power amplifier I have designed has a load of 4 ohms.

thanks a lot

 

[colin55]

Look at the size and weight of the transformer and compare the VA ratings in a catalogue.

 

[alec_t]

2 x 20V windings will give you +28V and -28V rails (not +-56V)? If all the power were dissipated by the load the peak load current (with the amplifier clipping) would be close to 2 x 28/4 = 14A. There are additional power losses in the amp, so you could be asking the tranny to supply >> 14A. No wonder it gets hot if it's rated at 6A (is that peak or rms, btw ?).

How can I measure the this factor by myself please?

With considerable difficulty! You would need to ramp up the winding current to detect the onset of core saturation, while measuring winding temperature to ensure over-heating (however you define that) doesn't occur.

 

[ronsimpson]

I agree with Alec_t.

For those that don't know where the numbers came from:
The transformer has a windings with a center cap.
The windings can be labeled 40VCT or dual 20V. That is RMS voltage not the voltage you get when rectified and filtered.
20V X 1.414 = 28 volts. 40V X 1.414 = 56 volts. The diodes peak rectify and you get the peak voltage.

Transformers are rated in RMS volts just like the power line is 110 or 220 VAC (RMS).
Most transformers are rated for RMS current. If you put a power resistor across the transformer it would deliver 6A of load current. The voltage and current will be sign wave. If you use diodes and rectify and filter before using the power the transformer will not deliver the 6A of current. The current is narrow pulses of current at the peak of the sign wave. The peak current might be 20 amps while the average is only 4A.

A 6A RMS rated transformer can only output about 4 amps after the rectification and filtering.

 

[Reloadron]

I like Colin's method. As in weigh it.

Additionally this is a good read on the subject. The available current and voltage downstream of rectification really depends on the filtering used. Also, as mentioned in the link, the diode drop is not figured in.

Also Ron, is it that:

The diodes peak rectify and you get the peak voltage.

or is it only that with a cap out there for filtering the cap charges to the peak value (1.414 * RMS value less diode drop)? Sans a cap I don't think we would see anything above the rectified RMS value less diode drop so the 1.414 would not apply.

Ron

 

[ronsimpson]

It looks to me that DrPower is measuring the output with out a load so it is 1.414, less the diode drop.

I went back and looked at post #1. He said he gets +/- 56 volts. We (I) read that as 56 volts, which is what you would get from a 40 volts (CT) transformer with full wave rectification and little load. (maybe 55 or 54 because of the diodes) and (56 volts if the line is 5 volts high)

If he is really getting +56 and -56 then it is not at 20 volt transformer.

Dr.Power needs to give us more information. Is it really +/-56 volts?
Transformer: 20 volts dual CT, or 40 volts CT, or 40 CT+40 CT, ????
Why get transformers from e-bay?

 

[audioguru]

The datasheet for his audio amplifier IC says that its output with a plus and minus 28V supply is 68W into 4 ohms. Its heating is 47W so a stereo amplifier takes 230W when playing at full blast. The 40V center-tapped current is 230W/40V= 5.75A.

But nobody plays music or speech continuously at full blast so a 4A transformer will be fine. The transformer will get pretty hot but it should not melt.

 

[Roff]

<snip> The current is not narrow pulses of current at the peak of the sign wave.<snip>

This is an unfortunate typo, as it makes the meaning exactly the opposite of what was intended.
I'm pretty sure he meant

The current is now narrow pulses of current at the peak of the sine wave.

 

[ronsimpson]

Thank you Ron. I will go back and edit that.