Dzturbd, here is the pinout of the power adapter. To test the battery, use a voltmeter to locate the positive and negative battery terminals and establish the polarity. I want to test the voltage at the DC Power Jack to see if the power jack is bad (or not receiving power). So, I was going to plug the AC power adapter into the jack and put the multimeter on the DC Jack to see if it reads 1.9 volts, but I don't know what DC Jack pin is the positive and which is the negative and I'm afraid I'll short the.
Hey everyone I'm working on a project where I want to be able to power a thin mini-itx motherboard two different ways. Here is what I'm trying to do:19V battery will be connected to a relay which is connected to the DC input of the motherboard. The port for the power adapter will also be connected through a relay to the DC-IN of the motherboard and to the charging port of the battery. When the adapter is present the adapter relay is closed and the battery relay is opened.
When no adapter is present it closes the battery relay and opens the adapter relay. The goal being that the motherboard always has power and I can charge the battery without discharging the battery at the same time since the power adapter can handle both powering the motherboard and charging the battery.Since the battery voltage will vary based on its charge i'm concerned that there may be short periods of time where a. This circuit may do at least part of what you want. V1 is the adapter and V2 is the battery.D1 and D2 prevent current from flowing back to either source. When S1 is closed (before the relay opens), the diodes D1 and D2 feed along the higher of the two sources to the output. When the relay opens, the battery (V2) is disconnected so even if it is higher than the adapter it will not be drained. It does not matter how long the relay takes, since the diodes will allow one or the other source to power the board during the transition.There's nothing here that would allow the battery to be charged- not enough information, so this is a yet incomplete answer.– Schematic created using.
$begingroup$ Hey thanks. I'm going to to grab that CircuitLab program that it is saying you used. I considered using paint, but I figured that would be worse than none.
What is the purpose of D3 in your diagram? For charging I just need 19V into the battery through a different port on the physical battery. I'm thinking just use that off of the switch so when the switch closes it initiates the charging of the battery through the adapter. Short circuiting of the battery should be prevented through D1.
Is that logical? $endgroup$–Feb 10 '14 at 0:27. If you use a changeover relay there will be no chance that the two power sources can become connected to each other BUT there will be a small time period when 'switching' that no supply is connected.
This is going to be in the order of about 10 milli-seconds so maybe, if you can calculate the current needed by the motherboard you can use a capacitor to 'hold-up' the voltage for that short period in time.Alternatively there should be no problem putting diodes across each contact so while the relay contact is in motion, the larger of the two supplies will 'fill the gap': -Once the relay contact has moved to its resting position there will be no voltage drop across the diode because it is shorted by the relay contact.
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