Chip amp power supply- a beginners guide

[Tripmaster]

Originally Posted by pacificblue

The relay coil is rated with 0,9 W at 12 V, hence draws 75 mA. 240 V x 75 mA = 18 W plus the consumption of the rest of the control circuit. It seems the manufacturer relies on the 1µ5 capacitor to dissipate most of the waste heat. Pretty scary in view of everybody striving for energy efficiency.

Richard, would you publish the manufacturer's name, so that everybody can avoid buying from him?

Hi David

I'm not sure its fair to give the name of the vendor as they maybe unaware of this. I wouldn't want it to affect the sales of their other products. The vendors been quite helpful so far.

The manufacture is LITE Audio

Regards

[Tripmaster]

Originally Posted by audio1st

Hi Richard, thanks for the Christmas brain teaser I have no idea what the answer is
I tried the voltdrop part of the cct and voltage was reduced to 6V ac, lightbulb tester glowed (60W bulb) and the cct drew 20VA.

Thanks for trying Barry

[pacificblue]

Originally Posted by Bill_P

No, the reactive feed is non-dissipative. See EDN PDF for more information.

I can't find any proof of your assumption in that link.

Theory in the form of Kirchhoff's Current Law predicts a power consumption of > 18 W. audio1st measured a consumption of 20 VA in practice. The relay consumes only 0,9 W, which leaves most of the power to be dissipated. The 1µ5 capacitor is the only device, where that can happen due to its function in the circuit (voltage divider) and due to its physical size.

[Bill_P]

Quoted from the article:

Direct rectification normally
gives a voltage approaching
the peak value of
the mains voltage, but placing
an impedance in series
with the ac input reduces
the dc output voltage. If the
impedance is resistive, low efficiency results. However, if the
impedance is reactive, the circuit is essentially lossless,
except for diode losses and a small power loss in the parasitic
series resistance.

I don't know how much more plain it could be...

[audio1st]

Originally Posted by pacificblue

I can't find any proof of your assumption in that link.

Theory in the form of Kirchhoff's Current Law predicts a power consumption of > 18 W. audio1st measured a consumption of 20 VA in practice. The relay consumes only 0,9 W, which leaves most of the power to be dissipated. The 1µ5 capacitor is the only device, where that can happen due to its function in the circuit (voltage divider) and due to its physical size.

Hi David, my test was carried out with a 47r load resistor because I did not no the value of the resistor that parallels the coil resistance.

Barry.

[pacificblue]

"Essentially lossless" is not a plain explanation. It is a statement that becomes absurd, when you read later on in the text

the power factor is approximately 12/(230X1.14)=0.045.

That is pretty similar to the statements of CFL aka "energy-saving lamp" manufacturers that tell you their lamps only consume 20 W, but 'forget' to tell you that their power factor of ~0,5 makes it necessary to produce 40 VA of apparent power in a power plant to make the lamps measure as consuming 20 W of real power.

audio1st measured a consumption of 20 VA to activate a 0,9 W relay coil. The difference must go somewhere. That is plain to me.

[pacificblue]

Originally Posted by audio1st

Hi David, my test was carried out with a 47r load resistor because I did not no the value of the resistor that parallels the coil resistance.

A 12 V 0,9 W coil corresponds to 160 Ohms. The difference is not too big. When you simulate it in LTSpice, you get a total power consumption of 18 W, just as Kirchhoff's law predicts.

David

[audio1st]

Tested with 160 ohm load, figures are, 20VA, 2W and .08 power factor.
DC voltage across load now 15V, was 5V with 47ohm load.
The resistor I thought was parallel to the coil is only connected before the relay is powered, I guess 150ohms, this keeps the load and voltage constant.

Barry.

[Puffin]

So, in laymans terms are these soft start modules any good?

[AndrewT]

Hi,
most mains loads are inductive and that lowers the power/load factor.
If we take current through a series connected capacitor to our circuit (soft start control), do we raise the power/load factor of our total household demand?