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I have an amp that requires 12-20v dc power. I'm looking into a solution for a portable battery pack that I can charge while running the amp as well with discharge and overcharge protection.

This is what im thinking:

18v Power Supply --> Diode --> Resistor --> 12v Battery --> Device

Will this charge the battery and power my device while it is plugged in, and then when it is unplugged use the battery as its power source?

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  • \$\begingroup\$ We have a "base" question about wall warts, but I'm not sure if we have one about batteries. Maybe we should. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 15 '14 at 6:07
  • \$\begingroup\$ I know about similar solutions for small lithium batteries, but I guess there are for 12 V batteries as well. \$\endgroup\$ – clabacchio Dec 15 '14 at 8:49
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This will work for you:

enter image description here

CHG is the battery charger, BAT is the 12V battery, LOAD is the amp, and K1 is a Double-Pole Double-Throw relay, with K1A being one set of contacts and K1B being the other.

"NC" refers to the normally closed fixed contacts, "NO" to the normally open fixed contacts, and "C" to the movable common contacts.

In operation, what happens is that the common contact (the slanty line) is spring-loaded and, when there's no power to the coil, the spring's tension forces the common contact up against the normally closed contact, creating an electrical connection between them. When power is applied to the coil, however, the magnetic field generated by the coil attracts the common contact away from the normally closed contact, breaking that connection and forcing the common contact up against the normally-open contact, which creates an electrical connection between the normally-open and the common contacts.

The action is similar to that of a double-pole double-throw toggle switch except that the switching is done electrically instead of manually.

With the 18V supply disconnected from the circuit, the relay will be de-energized and the connection between the charger and the battery will be broken, but the connection between the battery and the load (the amp) will be made through K1B-NC and C, allowing the amp to run off the battery.

With the 18V supply connected to the circuit, the relay will be energized and the battery will be disconnected from the load and connected to the charger through K1A-NO and C, allowing the battery to be charged.

In addition, the load will now be disconnected from the battery and connected to the 18V supply through K1B-C and NO, allowing the amp to be run from the 18V supply while the battery is charging.

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  • \$\begingroup\$ i'm new to circuit diagrams. I tried looking online at cheat sheets for your symbols but i only got a few. any chance you could explain it a little more? thanks \$\endgroup\$ – user2570937 Dec 15 '14 at 23:58
  • \$\begingroup\$ Sure; I'll be happy to edit my answer to include a circuit description, but I won't be able to finish it until tomorrow. \$\endgroup\$ – EM Fields Dec 16 '14 at 0:42
  • \$\begingroup\$ thanks for the info. I got one of these: radioshack.com/12vdc-10a-dpdt-plug-in-relay/… is k1A and k1B the com pins? \$\endgroup\$ – user2570937 Dec 19 '14 at 7:08
  • \$\begingroup\$ Thanks for the accept. :-) No, K1A and K1B are the two sets of SPDT contacts in the relay. "NO" are the normally open contacts, "NC" are the normally closed contacts, and "C" are the common contacts. This should help to clear up any confusion. \$\endgroup\$ – EM Fields Dec 19 '14 at 7:52
  • \$\begingroup\$ it says link not found \$\endgroup\$ – user2570937 Dec 19 '14 at 21:52
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If the charging current is less than the active current into the load then no, your battery will not be charged. However, the rate of discharge will be reduced. You need to ensure that the charging current is greater than the load current.

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  • \$\begingroup\$ ok, im using 3.5a charger and my device only pulls around 2 so I should be fine for that. Do I need more than a 12v charger as well? My device is 12v. and is the diode needed? \$\endgroup\$ – user2570937 Dec 15 '14 at 9:55
  • \$\begingroup\$ The diode probably isn't needed if it's a dc output from your charger. If the charger is intended for a 12 volt battery then no more voltage should be needed but you might find it actually produces a little more than 12V. \$\endgroup\$ – Andy aka Dec 15 '14 at 11:16

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