The switching boost regulator is a good idea for maintaining the operating supply for the repeater, but unless that regulator also has low-voltage shutdown provisions, to protect the battery from damage due to being drawn down too deeply, then a seperate cutoff is needed. Depending on the holding voltage of the relay to do this is not a good idea - a relay that will pull in at 12 VDC might not drop out until the supplied voltage is less than 8 volts - and this is well below the 10V or so level that you ought to consider the "never cross this boundry" level for the discharge level of the battery. Further, the relay coil will continue to draw the battery down, as long as it's connected. The simplest way I've seen for implementing a more positive means of assuring disconnection is to use a transistor, a zener diode, and a bias current limiting resistor. The circuit is designed such that the bias to turn on the transistor comes through the series connection of the resistor and zener diode from the opposite polarity from that which the switching transistor emitter is connected. The zener diode is selected for its knee voltage to be 0.7 VDC less than the desired cutoff voltage. When the voltage across the zener falls below its knee voltage minus the base/emitter drop of the switching transistor, the transistor will stop conducting, opening the current path through the relay coil. Actually, it'll probably happen a little above this voltage, as the transistor base is robbed of enough current to keep the transistor conducting enough to keep the relay pulled in. Use a transistor that has enough power dissipation that when it's not saturated, i.e. as the battery is getting low, the transistor won't be cooked by acting as a voltage divider between the battery and the relay coil. The transistor can be either PNP or NPN. If you use NPN, the emitter will be connected to the (-) rail, the resistor-zener pair, and one end of the relay coil to the (+) rail. The other end of the relay coil will be connected to the transtors's collector. Reverse this if you're using a PNP transistor. A power MOSFET could be used for this, but the transistion voltage for gate conduction is usually less well nailed down. Hope that helps, and 73 Bob, KD7NM
_____ From: [email protected] [mailto:[EMAIL PROTECTED] On Behalf Of David Struebel Sent: Monday, October 20, 2008 11:37 AM To: [email protected] Subject: Re: [Repeater-Builder] Shutting Down Battery Back up You might also want to consider the approach shown in the article on p 76 in the November 2008 issue of QST where a boost regulator has been placed in line...This will allow the equipment to still see 13.5 to 13.8 volts even where the battery has discharged below the 12 volt level. 73 Dave WB2FTX ----- Original Message ----- From: ka9qjg <mailto:[EMAIL PROTECTED]> To: [email protected] Sent: Sunday, October 19, 2008 10:29 PM Subject: [Repeater-Builder] Shutting Down Battery Back up I have a Astron 60 Amp Power supply with the battery back up option it works Great I have it Fused. However in the Testing I noticed that on Battery Back up I only loose about 10 Watts that is fine, But as the Battery drains down the Repeater gets Distorted and of Course this is normal too Because of the input Voltage getting Lower , Now the question and I have not seen this talked about I would assume all I would need is a Normally closed Relay and as the Voltage dropped below a Certain Level it would open and just break the connection to the Battery back up , Is this the way to do it Thanks Don KA9QJG _____ No virus found in this incoming message. Checked by AVG - http://www.avg.com Version: 8.0.173 / Virus Database: 270.8.1/1734 - Release Date: 10/20/2008 7:25 AM
