Electronics

tgarrity · Joined: 05 Dec 2006 Posts: 2

Hi All,

I have a two-year degree in electronics that I haven't used in, oh about 30 years, and need some help here.

I have a little plug-in transformer (110V AC in, 9V DC, 300mA out) that was used for powering something. I want to use it to power my callerid box instead of burning through batteries all the time. The plug is right, but the voltage is wrong. I need 6V DC @ 50mA min., according to the callerid manual.

Well I thought this was simple enough, I just need a resistor in series with the output to drop 3V down to where I needed it. According to my math, 3V/.3 = 10 ohms. Well when I put the resistor in and measured it, I see 15 V. Obviously the output isn't correct, so back to the drawing board.

OK, so it's 15V. So I need to drop 9V not 3, so 9/.3 = 30 ohms. No good, still 15 V output. Hmmm, maybe I'm doing this wrong. Maybe it should be 50 mA instead? So 9V/.05 = 300 ohms. Yeah, that's gotta be it! Nope. Still 15 V output. Sheesh!

It's a full wave rectifier (4 diodes), and I checked them all, and they all appear to be good. Simple enough circuit--small transformer, 4 diodes, and a capacitor (filter?).

What am I doing wrong? Thanks!

Terry

neon · Joined: 25 Feb 2006 Posts: 581

well obviously you have an open circuit or you forget that caller ID usualy do not draw 50ma that is a lot for LCD display. 50 ma was maybe for charging the battery that you don't have any more . is it right that you 15v on the caller ID? you may have fried the box. you want 6 volts then use a 5.9v zener and and use zener load pluss load current to figure it the resistor[ ei.] load 50ma and zener 50ma then 15-6= 9v 9v/100ma=90 ohms in series from 15v you should be ok. need more help ask.

tgarrity · Joined: 05 Dec 2006 Posts: 2

Hi Neon, thanks for your reply. The 50 ma was the spec on the callerid box, I didn't measure it. Also, it uses alkaline batteries, which are not rechargeable, so I don't think that the 50ma is for charging.

The callerid box requires 6V, as I said. I have not connected the PS to the callerid yet, I just measured the output of the PS with no load, so I haven't fried it. Yet.

I thought about using a zener, but shouldn't I be able to achieve this with just a series resistor??

I also don't quite get why the output of the PS is 15V when it says 9V right on it. Would it be different under load?

Thanks,

Terry

neon · Joined: 25 Feb 2006 Posts: 581

a 9v PS even with no load shouln'd go to 15 v dc but it could go to 12.6 vdc if it still OK. IF this PS is a switching type then anything is possible. I must insist that you provide a 50ma load and a 100ma load and see what hapens. I think that your diodes inside you ps. are gone check the ac components on the 15 vdc with a 50ma load it should be in the mv it is after all it is a 10% load right.yes you may use a series resitor to get to 6 v provided that the load is constant in you case it is not therefore zener or regulator of some form is required. By the way these transformer of 6 volts cost a couple of dollars i sugest you buy the right transformer it would be cheaper then go trought this rigging and waste power as heaters.

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