proteus999

I am wondering if I can use a 9v dc power supply (or battery) on a piece of equiptment that requires a 9v Ac wall wart? Won't the internal rectifier pass the 9v DC input voltage thru?
Thanks
Dave

audioguru

9VAC has a peak voltage of 12.7V which is reduced to about 11.2V by the full wave rectifier bridge. A little 9V alkaline battery's voltage quickly drops to 7.2V. Then the full wave rectifier bridge will reduce it to 5.7V which is much too low.
A 9VDC power supply will have its voltage reduced to 7.5VDC which is also much less than 11.2VDC.

Yry it. It might work.

__________________
Uncle $crooge

Sceadwian

Peak voltage shouldn't matter so much, it's RMS voltage that's important. I did a simple simulation of a 9VDC source and a 9VAC source (peak voltage of 12.8V) and passed them through a full wave bridge rectifier made out of 1N4148 diodes and I got 7.57 volts RMS on the DC source and 8.49 volts RMS on the AC source after rectification using a 10u filter capacitor and a 500ohm load just to simulate a basic filter capacitor and load. A 1u filter cap drops the AC lines RMS voltage down to 7.73 so the size of the filter cap is relatively important.

__________________

audioguru

A little 1uF or 10uF capacitor is not a filter for a mains powered rectifier. Try 1000uF or more.
The 1N4148 is a low current signal diode, not a rectifier because its voltage drop is too high (typically more than 1.4V at only 800mA) when it tries to conduct a few amps to briefly charge the filter capacitor. Try a 1N4002. Its typical voltage drop is less than 1.3V at 40A!

__________________
Uncle $crooge

Sceadwian

With the 500ohm load even the signal diode model should approximately mimic a better rectifier diode under higher load. I just used the 4148 because it was the first diode on the model list, the load is only 20ma's RMS. Just saying he's going to have one hell of a time figuring the difference between a 9VDC and 9VAC voltage source to his device. Likely because the only thing I can think of that commonly uses an AC wallwart are cheap cordless power tools, they could care less what voltage they get really, just as long as the new supply isn't capable of providing too much additional current. AC or DC wallwart's are typically current limited to the inductive coupling/capacity of the initial transformer stage, which is going to be the same from one device to the other. The bridge rectifier just happens to be on the device in his case which saves a few cents per device for the maker of the device as it's probably almost incidental to add a bridge rectifier to the device, a secondary power supply manufacturer is going to add a markup to the rectified output as opposed to an AC one, just because they can.

__________________

kchriste

Another reason to make a wallwart with an AC output is to maximize the VA you can get out of a certain size wart. Without the bridge, cap, and wiring there is more room for a beefy transformer.
Part of the problem stems from the fact that, in the capacitive input filter supply, the conduction angle of the rectifiers is really small resulting in a peak rectifier current 3-5 times the average value. This will increase the rectifier forward voltage drop much more than you would initially expect.
10uF, as Audio mentioned, is WAY too small of a value even for a 20ma load. Here is a handy little formula for approximating ripple in a capacitive input power supply:
Vrip = I/FC
Where Vrip is the ripple voltage in volts. I is the current drawn from the supply in amps. F is the frequency AFTER the rectifiers. C is the filter capacitor value in farads. Of course, ripple can't exceed the peak input voltage, but if the formula returns such a value; your input capacitor is WAY too small.

__________________
--- The days of the digital watch are numbered. ---

audioguru

I have used this graph of the amount of ripple voltage. I added the formula by K Christe.

Attached Images

Full-wave rectifier ripple voltage.PNG (33.2 KB, 8 views)

__________________
Uncle $crooge