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I have made a portable device charger with a 5 volt voltage regulator(7805-TO220 package),a 9 volt battery and a few capacitors.The problem is,when I connect the charger to a certain device(whose original charger outputs 5 volts and 2A) and the device starts charging, after a few minutes,the image starts flashing and then it shuts down,then the start-up screen appears,then the screen turns black and so on a few times,then it shuts down completely and remains in this state.I connected the circuit to another device whose original charger outputs 5,1 volts and 700 mA and all works just fine.

schematic

simulate this circuit – Schematic created using CircuitLab

Why does the first device behave this way?
I don'think it's about the voltage or current.
The 9 volt battery is a Varta High Energy model-6LP3146. EDIT:I noticed that when charging the second device,its battery drops after a while,whether discharging on its own or being discharged by my charger(I added a diode between the output pin of the regulator and the positive terminal of the device,so it's unlikely).

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  • \$\begingroup\$ Provide details of the 9V battery. \$\endgroup\$ – Leon Heller Aug 28 '15 at 12:42
  • \$\begingroup\$ Do you have enough cooling for the linear regulator (7805)? It may be hot and turn of. \$\endgroup\$ – Botnic Aug 28 '15 at 13:47
  • \$\begingroup\$ No,I checked it every time.And it also has a heatsink. \$\endgroup\$ – Daniel Tork Aug 28 '15 at 14:28
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You don't mention what case style of 7805 you have. If i was guessing you are using a TO220. That case can only support up to 1A output current and even then requires substantial heat sinking to do so. Some 7805s had thermal protection and would shut down if overheating. Check the voltage to the device when it shuts down. Perhaps the voltage drops out.

Also you cryptically say that it's a 5v device. If i also had to guess it's a USB device. Some USB devices require that an identified is detected on the data lines to announce that the device is connected to a charger. For example my HTC phone charger shorts the data lines together. Also my Galaxy tab doesn't like my HTC charger because Samsung uses a different scheme to identify the charger.

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  • \$\begingroup\$ Its original charger is not USB. \$\endgroup\$ – Daniel Tork Aug 28 '15 at 13:23
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They typical applications on page 18 and 19 of the datasheet show an input capacitor of 0.33uF and an output capacitor of 0.1uF. Your input is 300x larger and your output is 100x larger than the datasheet shows. I'm not sure if this is causing any issues.

This regulator also can't handle 2A output current, so it might be browning out. Also, a 9V battery doesn't have too much capacity to start with, so it could also be dead. Bear in mind that terminal voltage under no-load conditions isn't the best way to measure capacity.

You should check terminal voltage under load to see if your battery is supplying the 7.5V minimum required for a 5V output - I would be this is why your device keeps shutting down.

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    \$\begingroup\$ FYI - This figure on the battery capacity page I linked shows that, at a discharge rate of 500mA, you'll hit the 7.5V cutoff voltage after you've discharged about 0.02Ah of capacity. At 0.5A, that means you hit the cutoff voltage in (0.02Ah/0.5A) = 0.04 hours, or about 2.4 minutes. It looks like, by their curves, no 9V battery could supply adequate supply voltage to keep the voltage regulator on after about 150 seconds of charging. \$\endgroup\$ – Chuck Aug 28 '15 at 13:07
  • \$\begingroup\$ So you are saying that I'm consuming the battery for nothing? \$\endgroup\$ – Daniel Tork Aug 28 '15 at 13:16
  • \$\begingroup\$ I'm not quite sure what you're asking, but the way that you have setup your system I don't think you're getting much use from the battery. \$\endgroup\$ – Chuck Aug 28 '15 at 13:17
  • \$\begingroup\$ Can you suggest what should I change to make it feasible? \$\endgroup\$ – Daniel Tork Aug 28 '15 at 13:21
  • \$\begingroup\$ Use a different battery or double up on your 9V batteries. The regulator handles up to 35V in, so you can put two 9V batteries in series and never have to worry about falling below the cutoff voltage until both batteries are dead. \$\endgroup\$ – Chuck Aug 28 '15 at 14:03

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