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I was wondering about this part of a circuit diagram. I believe the component marked as JP12 is a DC barrel connector, which I thought had three pins. One to detect input, one with V+ and one with ground.

However I thought this device was also receiving 5.0V from USB. The diagram shows pin 1 of the DC connector directly connected to the VBUS line, does this mean the DC connection is also powering the USB device. If so, what is pin 2 being used for.

Also, I have no idea which of the three pins in JP12 is supposed to be ground. Am I mistaken, and this DC connector does not supply ground?

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    \$\begingroup\$ Btw Usb supplies 4.75 to 5.25 if within standard tolerance. \$\endgroup\$ – Passerby May 26 '15 at 23:05
  • \$\begingroup\$ I see, I suppose what I was thinking was that the DC supply was sending the 5V to the VBUS pin. It is important to note that there is variation though. \$\endgroup\$ – Harley Armstrong May 26 '15 at 23:26
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CN2 is probably the barrel connector you are looking for. It is labelled as a "DC_2.0mm" which would be a 2mm diameter DC-jack, which can be known as barrel connectors. CN2 is connected to the circuit GND and the VCC net.

JP12 looks like a jumber to allow you to set the power source which U2 regulator is fed from to generate 3.3V. If pins 1 and 2 are shorted together U2 is fed from the 5.0V net, where as if pins 2 and 3 are shorted together U2 is fed from the VCC net. The jumper will usually either be a set of 2.54mm posts which you can short together using a special shunt piece that fits on top like this:

jumper pins

Or surface-mount pads close together that you can short with a solder blob or a surface-mount resistor like this:

SM jumper

So in this circuit, the USB connector CN1 always provides 5V when connected, but the current available will be limited by the USB host (usually to 500mA). The designer has provided an optional way of supplying your own supply (which could be greater than 5V as long as U2 does not overheat) if the power drawn by the 3.3V circuit is too great for USB. You can choose where to supply your circuit from by shorting the appropriate pins of jumper JP12.

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JP12 is a three-pin male header. A jumper connected to two adjacent pins is used to select which of 5.0V and VCC will be used as the supply for the 3.3V regulator.

The barrel jack is CN2, on the bottom left of the diagram.

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USB provides up to 0.5A, so if you need more, you put external power supply.

Usually external side of the plug is minus. Use DVM to check it.

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  • \$\begingroup\$ Right, but they appear to be stepping down the input voltage to 3.3V for one pin, then the other one is connected to +5V, I just don't follow, does this mean there are 2 positive pins? \$\endgroup\$ – Harley Armstrong May 26 '15 at 18:38
  • \$\begingroup\$ 3.3v is for internal use, i don't see it on the connector. \$\endgroup\$ – Gregory Kornblum May 26 '15 at 18:42
  • \$\begingroup\$ By the way, there is that jumper to select 5v source. \$\endgroup\$ – Gregory Kornblum May 26 '15 at 18:44
  • \$\begingroup\$ Yes, I think I misunderstood, The JP12 is not a power connector, it is a switch. \$\endgroup\$ – Harley Armstrong May 26 '15 at 18:47
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JP12 is a three-position post header, used as an SPDT switch to select power from either USB or CN2 (presumably a coaxial power connector).

You would use a shorting plug on JP12 to connect between pins 1 and 2 to use USB power, or between 2 and 3 to use power from CN2.

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