Help! DC voltage is less than 1.4 times. Why???

[Mamun ur rashid]

Hi,

I'm using chinese 5.1 home theater (microlab FC861). Few days ago I opened it to see internal circuits although my amp is fully working. there is a transformer which is rated output 16-0-16 volt AC 2.2 Amps. Input 220-240 volt.

When the amp is idle the ac volt is slightly lower 15-0-15 volt ac (when mains is 220). Bit the DC volt is only about 18-19 volt. It uses full wave dc by using 2 rectifier. If we calculate output should be 15 x 1.4 = 21 volt. but it is only 19 volt or less. I checked the rectifiers, they both are ok and forward voltage drop is about 300-400 milivolt (on multimeter). I checked the DC filter capacitors they both ok 4700 uf each.

Whats the problem? why dc voltage is less than 21? please help!

BTW amp using 2 STA540 IC. Thanks

 

[JonSnell Electronic]

It sounds like you have a half wave bridge rectifier using just two diodes. A full wave bridge rectifier gives 1.4 times the average AC voltage as DC, less the voltage drop of the diodes. A half wave bridge gives slightly less voltage but more current.

 

[DF96]

You will not necessarily get a meaningful DC voltage measurement across the diodes, so ignore that. Expect power diodes to drop anything from 0.5V to 1V each.

You probably have a full wave rectifier. A half wave rectifier would give more ripple, less voltage and less current and cause extra transformer heating.

The 1.4x figure you calculate is peak DC voltage; this will droop between charging pulses so DC meter will probably give the average.

There is nothing surprising about your figures.

 

[nigelwright7557]

If its a bridge rectifier you will lose a couple of volts across the diodes.

 

[Mark Tillotson]

Hi,
I checked the rectifiers, they both are ok and forward voltage drop is about 300-400 milivolt (on multimeter).

Then they are schottky diodes, which could be a problem with reverse leakage and self-heating if they are not on a heat sink. Probably not at these voltages, but reverse leakage in schottkys is an exponential function of device temperature...

Any rectified transformer output voltage will droop below the simple theoretical voltage due to the high current pulses involved.

These cause voltage losses due to IR drop across transformer windings. The peak currents will cause higher voltage drops on the diodes than might be naively assumed too, and the internal resistance of the filter caps will also cause IR losses - basically there are a lot of small losses to add up, and a few volts are easily lost.

The charging pulses can easily be 10x the load current.

 

[kodabmx]

See the following handy guide from our friends at Hammond... https://www.hammfg.com/files/products/700/hammond-recitifier-guide.pdf

Also, the voltage drop will be higher with more current flowing through the diodes...

 

[Gnobuddy]

...reverse leakage in schottkys is an exponential function of device temperature...

That is also true of conventional PN junction diodes.

The rule of thumb I learned was that the reverse leakage current of a junction diode doubles for every 10 °C rise in temperature.

If I did my logarithms right, that corresponds to the exponential equation Is = Io exp(0.069315 * T), where Is is the reverse leakage (saturation) current, Io is a constant, and T is the temperature in Celsius.

The exponential temperature dependence of the leakage current was a big problem in the era of germanium semiconductor devices. Silicon lowered the (Io) term in that equation so much that we can often get away without even thinking about reverse leakage current.


-Gnobuddy

 

[PRR]

STA540 is rated 22V operating (an up-rated car-radio chip). The rectifier is 2-diode on center-tap winding. The wall voltage is at the low end of what the name-plate asks (16V rated, 15V observed). If the DC voltage on this wall outlet were much higher, then on a 245V outlet it would be too high for good chip life.

I agree that you can't simply read diode forward voltage with the big current spikes. 0.6V at medium current is "typical" for common SI diodes, but a sim of a similar plan shows 0.986V in the narrow current spikes.

I admit that I can not duplicate your DC result, unless I sim an incredibly cheap transformer, but I think it is working as intended.

 

[Mamun ur rashid]

Then they are schottky diodes, which could be a problem with reverse leakage and self-heating if they are not on a heat sink. Probably not at these voltages, but reverse leakage in schottkys is an exponential function of device temperature...

Any rectified transformer output voltage will droop below the simple theoretical voltage due to the high current pulses involved.

These cause voltage losses due to IR drop across transformer windings. The peak currents will cause higher voltage drops on the diodes than might be naively assumed too, and the internal resistance of the filter caps will also cause IR losses - basically there are a lot of small losses to add up, and a few volts are easily lost.

The charging pulses can easily be 10x the load current.

No diodes are not in heatsink. Should I replace these diodes? Thanks

 

[Mamun ur rashid]

Thanks everyone for your helps