hello

I have a prc 320, military HF radio. it takes 175mA on RX and 1.2A on TX. I have a 6v 12Ah SLA rechargeable battery, a dc-dc step up booster
http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&item=130926346159&ssPageName=ADME:X:RTQ:GB:1123

I need 24-28v to power the radio. I can get just over 30v from the step up booster. the input current is 3A maximum. if I used the 12Ah battery would that damage the booster. would I require a current limiting device before the booster. I know some may suggest a protective device for good measure but the load would only draw 1.2A maximum on TX. so, would I require some limiting device. im still a little unsure of the operation of a booster circuit. it says 3A maximum and I know that whatever the load demands is what it would get. but im worried that the device would just pull in whatever is available without regard to the output. probably doesn't make sense. the maximum output current of the booster is 2A without a heat sink. I think maybe that the 1.2A is a little too close, could it wear out the booster over time(things like heat build up?)
any help would be great
thanks
simon

 

It would be ok except the regulator is to small. The 3 Amps refers to the input current. To get 1.2 amps out at 24 volts you would need (at 100% efficiency) 4.8 amps at 6 volts.

 

No. The absolute maximum current that boost circuit could provide (with the added heat sink) at 28v would be 0.71 Amperes - and that's if your ambient temperature is 25°C.

Why don't you just use two 12v 12ah SLA batteries instead? You'd get more than 4x the run time than you would with a single 6v 12ah battery, and wouldn't have to deal with a boost circuit.

 

hello

thanks for both replies. please could you explain the formulas used for working those values out, im new to this and don't know how you did it.

what I wanted to do is use a 6v battery that I could fit inside the shell of an original battery for the radio. I have 4 batteries which is enough for the radio but they are bulky and I have to charge them individually and that takes time. I would prefer one battery.

thanks
simon

 

A couple of ways to the same conclusion.

Sgt used the maximum power spec of the supply with a heatsink (20 watts). The formula for power is P=IE or I=P/E. so 20 watts/28 volts=.71 amps max.
Being lazy I just looked at the conversion from 6 volts to 24 volts. In this case you trade current for voltage so if you want 1.2 amps out you need 4 times that much in or 4.8 amps plus some loss in the conversion. Since the supply is only rated at 3 amps max it won't work.