Uniden Scanner takes a AC 10 Volt power supply, can I substitute another type of Adapter without problems?

Question

I have a BC244LCT Uniden Bearcat Scanner w/o Power Adapter.
Manual states it only takes a:
10 VAC using AC-144U AC Adapter in a AC 10V power jack
I would like to know if I will or can use another voltage w.o problems to device?

Answer 1

You're going to have to get it as close to that value as possible. I wouldn't go over that value, but you could probably go under by a volt. It also has to be AC. Most adapters output DC so this is going to be a harder one to find. 9V AC adapters aren't that uncommon and MAY work for you. There's another figure that you really need to make a good judgement on an adapter and thats amperage, probably milliamps or mA in this case. I went and looked up the AC-144U on Google and found that it's 300mA. This figure can be higher, and should be at least slightly higher if you're going to try and go to 9V AC. Just don't go any lower than 300mA. It can be as high as you want. Radio Shack carries a 9VAC 800mA adapter. I'm not sure if the plug will fit, but it should electrically work. They have a return policy so try it out or better yet, bring the scanner with you and try it. The stock number is 273-1770 and it costs $20. However, I'd suggest getting the original. You can get it online for $15 here (and probably elsewhere) - http://www.hrosecure.com/firstclass/store09/batteriesandchargers.html

Answer 2

Uniden Ac Adapter

Answer 3

I wouldn't recommend it. Go a few volts higher and it'll probably work, but you will eventually fry the circuitry inside the scanner. Go any lower and the radio's performance will suffer. Your best options are either to a. contact uniden and try to get an exact replacement or b. take a trip to Radio Shack. They will most likely have an adapter that you can use with that radio without causing too many problems.

Answer 4

because of ac alerts ought to nicely be utilized in lengthy distance transmissions.. at the same time as DC can't.. The loss in ac ought to nicely be recovered by purely including a lead or lag circuit.. at the same time as loss in DC can't be recovered.. And as you've reported, the go portion of the cord will boost because the present transmitted will boost, DC isn't affordable as ac.. get at the same time: a particular cord can carry a maximum quantity of cutting-edge. Now, utilizing DC, this can finally end up utilizing constrained quantity of DC which losses can't be recovered. Now utilizing ac, a intense cutting-edge signal ought to nicely be switched over to a intense voltage-low cutting-edge signal which losses ought to nicely be recovered and this signal ought to nicely be switched over lower back to a low voltage-intense cutting-edge signal..