I'm developing a small micro-ohm meter that I will be powering from USB, due to the ubiquity of USB power bricks and computer power.

I've run into a few questions:

  1. What do I do with the data lines of the USB? I'm not using them, I'm just interested in USB power, should I tie them to ground, to each other, to +5V?

  2. Do I need to switch from "low power mode" to "high power mode"? This seems to indicate that you need to ask for high power using something called "enumeration", but the same article also talks about a 2.25W max... which I know is no longer true.

  3. Max capacitance for the USB 2.0 specification according to the above is 10µF. I've got significantly more than that. This leads me to believe that I should be using something like the NCP380, MIC2545, or TPS2112. These are high side switches that can limit inrush current, allowing me to have more capacitance after the switch. Is this still necessary? Or has the capacitance limit increased since the above linked document was written?

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    \$\begingroup\$ For this to be answerable you need to decide if your goal is to meet USB specifications, or if you are content with something that merely happens to work in the vast majority of cases. \$\endgroup\$ Aug 18, 2016 at 16:45
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    \$\begingroup\$ @ChrisStratton I don't know how stringent USB specifications are, and how difficult they will be to obey. Until I know that, I don't know if I want to worry about obeying them. Can you help with that? \$\endgroup\$ Aug 18, 2016 at 16:48
  • \$\begingroup\$ You have basically stated the older specifications, if you want to look at new standards you need to read those documents or summaries thereof. But consider even if the new versions turned out to make things easier, you would not comply with the specifications that older hosts were possibly built to. So, it comes back to the original key question - do you want to comply with the specs or not? You might survey what is on the market before you make a hasty answer to that. \$\endgroup\$ Aug 18, 2016 at 16:51
  • \$\begingroup\$ @ChrisStratton This is a device for me only, not a market. If I have to limit myself to newer power supplies than I am ok with that. It looks like the Battery Charging revision 1.2 is the standard that I would be interested in obeying, rather than the USB 2.0 standard. \$\endgroup\$ Aug 18, 2016 at 16:55
  • \$\begingroup\$ Obviously, you make your own choices. But you might consider that you live in a world where USB ports are molested by motorized dog figurines... \$\endgroup\$ Aug 18, 2016 at 16:58

3 Answers 3


For your particular purpose, you don't need to do anything, just connect GND (black wire) and VBUS (red wire) to your design.

(1) leave D+/D- unconnected;

(2) don't worry about any "low" or "high" power modes. These modes are for host to supervise and police overall power that host believes is available to supply. It never worked correctly. If you don't pull any data wires up, the host will not even know if your cable is even plugged in.

(3) Do not worry much about 10uF limit. In worst case you might get "port overcurrent" message on a host, but it is rare. Or you might disrupt other USB devices that might be working in adjacent ports. But there are plenty of DC-DC converters that use input caps of 10uF or less, so it is advisable to use a better DC-DC converter in your design. If you are not using any DC-DC de-coupling from VBUS supply line, it is a very bad idea for a sensitive analog design.

Regarding the BC1.2, forget this completely. All you need to care is not to exceed 500mA DC limit.

Clarification for (2): The OP is not building a USB device. The 100mA is a requirement for USB devices, not hosts. The OP wants to use the port capability of a USB host. Every host port must have an ability to supply 500mA, regardless if there are "low" or "high" powered devices. So 500mA is always fine. If the port will be behind a bus-powered hub, there could be a different story.

  • \$\begingroup\$ I need more than 100mA of current... I still don't need "high power" mode? Will I run into problems with some chargers not giving me enough power? \$\endgroup\$ Aug 18, 2016 at 18:21
  • \$\begingroup\$ No, every charger I have ever seen would give you much more than 110mA. You will be just fine. As I said, "power modes" are for host bookkeeping, a formality, while power ports never have this kind of control, and predominantly boldly supply at least 500mA, and usually way-way above that. \$\endgroup\$ Aug 18, 2016 at 19:03

USB supplies come in a few flavours and standards.

Basically low power (portable hosts), negotiable power (most smart hosts and professional mains adapters) and brute power (DIY and many car type adapters).

If you do not plan to negotiate for extra power you can ignore the data pins altogether. Some devices use a 5th pin to do other clever tricks sometimes in barely standard ways.

There were 3 common ways of signalling a desire for more current from a host when I was looking at this a while back. The old way, the new way and the Apple ways. If you look at IC datasheets that support these standards you will see what your device would be required to do to trigger the power boost if this is what you need.

If you have a post regulator with generous capacitance (before and or after) you should be able to use any type of supply and use what it can give.

EDIT: Most of the brute force chargers supply full power all the time and while they are very practical they may not be as safe with short circuits as an intelligent supply. The low power hosts are usually incapable of providing more than the USB minimum by design to conserve their own battery life.



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    \$\begingroup\$ "Ignore the power pins altogether". I just want to make sure you mean the signal pins... :) \$\endgroup\$ Aug 18, 2016 at 17:38
  • \$\begingroup\$ Right you are, typo that I can fix. \$\endgroup\$
    – KalleMP
    Aug 18, 2016 at 17:45

USB port provides power by default - this is necessary for the device enumeration.

  1. Thus, you could simply draw the power from the USB power pins. If your multimeter will not use USB communication, then you probably do not need to do anything with data line. The standard unit load available is 100mA (for USB2). No device is permitted to take more than this before it has been configured by the host (according to the standard).
  2. In case your power requirements are higher than the default (100mA USB2 and 150mAUSB3), you would need to enumerate the device indicating higher power needs. This requires the whole USB enumeration protocol, where your device will be communicating using data lines.
  3. The capacitance problem is related to hot-pluging your device while other devices are connected to the same hub. Thus, high capacitance device may case a voltage drop/ reset of other devices. If this is not an issue, you should be fine.

So, if your device just uses a little power and does not require communication with the host - I would not worry about any of your questions.

  • \$\begingroup\$ I need a bit more than 100mA (around 110 mA or so), which places me above the low power mode. And what about the capacitance limit, is that not something I should worry about? \$\endgroup\$ Aug 18, 2016 at 18:28
  • \$\begingroup\$ @Naz: USB ports "are not configured" to 100/150mA limit. It is a host system bookkeeping limit. The limit might be enforced only on some modern over-sophisticated single-port low-powered tablets/phones, but even this is very, very unlikely. Any normal USB port is capable to supply at least 500mA. Of course, if you have your host loaded with a bunch of USB hard drives, USB can simply die. No big deal, just exercise reason and don't overload USB host. \$\endgroup\$ Aug 18, 2016 at 19:15
  • \$\begingroup\$ @AndrewSpott: what is your VBUS capacitance? \$\endgroup\$ Aug 18, 2016 at 19:17
  • \$\begingroup\$ @Ali Chen: At the moment, 25µF. However, I was looking to decouple the power supply with a low pass filter at ~1kHz, which would give me close to 100µF. \$\endgroup\$ Aug 18, 2016 at 19:18
  • \$\begingroup\$ @AliChen Just updated the "configuration" expression - you are right. However, I do not know what happens when you exceed the limits. If no "current measurement logic" is present, then you can probably (ab)use all possible power without "telling" it to the host. \$\endgroup\$
    – Nazar
    Aug 18, 2016 at 19:31

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