2007-09-20 19:38:29 · 6 answers · asked by Chuck S 1 in Science & Mathematics ➔ Engineering
You will have to get rid of 1.v of the volts or you may damage the device. A series resistor and zener diode would do the trick for very little money.
2007-09-20 19:43:35 · answer #1 · answered by ZORCH 6 · 0⤊ 1⤋
the direct answer is no.
however a 9v supply may be modified to provide the correct current. you didn't state if the device drew a constant 500 ma or whether it might vary based on operating conditions. If constant, the resistor will work. If not, the voltage drop across the resistor will change and nearly all the 9 volts will appear at the load when the current draw is low.
Technically the zener solution will not be satifactory because the zener will either drop out of regulation when the device draws the specified current, or the 9 volt supply must source somewhat more than 500 ma so some current can be diverted to the zener.
My suggestion is to use a 12 volt supply and follow that with a 7808 regulator chip. That will regulate the output over the entire range of load draw from zero to the full 500 ma.
2007-09-22 16:41:05 · answer #2 · answered by lare 7 · 0⤊ 0⤋
"can i" well, maybe, possibly the device will have a built in regulator that can handle the difference, but I would definitely not recommend it, as it could possibly result in destruction of the device.
However if the "transformer" (power supply) has a DC output, there are a couple of ways you could reduce the voltage to match, the resistor and zener diode previously recommended would be a pretty good solution if you can get a 7.5v zener. Another option, not as accurate, but possibly close enough, would be to put a couple of silicon rectifier diodes in series with the load, each conventional silicon diode will drop about 0.7v, so 2 in series will reduce 9v to about 7.6v, pretty close to what you need. On the other hand, if the transformer output is AC, I'd suggest getting a transformer that has a 7.5v output.
2007-09-21 03:35:02 · answer #3 · answered by tinkertailorcandlestickmaker 7 · 2⤊ 0⤋
no...it will overload the device....try to find a closer voltage difference if not....add resistors to reduce the voltage current
2007-09-21 02:54:33 · answer #4 · answered by dranoel78 1 · 0⤊ 1⤋
For some applications you might get away with it but if you were using MOSFET or CMOS semiconductors, you'd blow them.
2007-09-21 03:29:04 · answer #5 · answered by Anonymous · 0⤊ 0⤋
no
2007-09-21 05:57:43 · answer #6 · answered by ghost_rider 1 · 0⤊ 0⤋