Quote:
Originally Posted by Cobb
Many have tld me the voltage wouldnt be there. I am guessing with a cheaper non sign wave inverter it will.  Its only the sign wave ones that wont have the constant 120 volts I need. |
This isn't correct or at least overly pessimistic. Some inverters, maybe the one you have, produce a square wave output. That kind of inverter is so electrically noisy it really can't be used for anything but light bulbs. Most cheap inverters generate a 2 step wave that has about the same rms power as a sine wave. With a square wave inverter the rectified voltage will be the claimed rms voltage, if the inverter output is "120 vac rms" it will rectify to 120 vdc. But with a true sine wave inverter rated at 120 vac rms the rectified voltage will be about 170 vdc. But this will depend on load and the inverter peak load capability. When you rectify house current, you can assume the peak load capability is infinite. Assume you are using house current, you rectify it and use the dc to charge a capacitor, for smoothing, the capacitor may be discharging to a load at near constant current, but the cap can only charge when the rectified ac is above the capacitor's dc voltage. This occurs during a short time at the sine wave peak. So all the recharging current must flow during that brief period - and so if the constant load was 10 amps and the amount of time that the input voltage was above the capacitor voltage is about 1/10 of the time, then the charging current would be 100 amps for those short pulses. The rectified voltage will be about 1.4 times the rms ac voltage because the power company can deliver almost any current you want.
The problem here is that your inverter can't supply 10 times the rated rms current. But it might be able to supply twice its rms rating. So the voltage dips will be lower and the average dc will also be lower than if you had house current. In some psuedo-sine wave inverters the rectified DC is no higher than the rms ac output, just as if it were a square wave. If the inverter is undersized it won't drive a rectified dc load at all.
Also diode voltage drop is going to be insignificant for your use. There are 2 components to diode voltage drop, one is resistive the other has a negative thermal curve. So the higher the current, the hotter it gets and the lower the junction voltage, which is just more than offset by the resistive part. So a cold silicon diode may have a forward voltage of 0.65 v at some small forward current, and hot 0.4 v, at the same low current. But the resistive part could be 0.025 ohms so at 10 amps the hot diode drops 0.65 volts.
-mort