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I need to make this Arduino Nano-like board with one single connector: a mini USB type B which will of course be connected to the PC for programming and interface. Additionally, as it will be used away from the PC, I need to be able to connect a 6V+ battery pack using the same connector. There's no way I can fit a second connector for this.

Usually the FT232 (or similar) won't take more than 5.5V, so I've thought of placing an LDO between the USB-VCC input and the rest of the circuit. The LDO would drop about 0.3V but solve my overvoltage problem. It could take 9V easily.

My question is: will the voltage drop cause problems with the USB communication (since the FT232 will get power from a slightly lower voltage source)?

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  • \$\begingroup\$ Not enough information to answer the question. Basically you need to list all the parts that use power directly from USB and see what their minimum operating voltage is. Then consider worst case USB voltage (4.5). Subtract a little bit for dropout of regulator (check specific regulator datasheet) and then see if the lowest possible voltage is acceptable for all the parts. I have seen many processors which do not need or want 5V from USB. But for the USB to work correctly, there is usually an input that needs to go high when USB VBUS is attached. \$\endgroup\$
    – user57037
    Aug 17, 2016 at 15:26
  • \$\begingroup\$ @mkeith I'm more concerned about the PC side of things and the effect in the differential signals. \$\endgroup\$ Aug 17, 2016 at 16:02
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    \$\begingroup\$ You mean you are worried that without 5V on the microprocessor board, it will not generate correct USB signals for the host/PC? Most USB phy's I have seen use 3.3V not 5. You need to figure out what is required by YOUR hardware. \$\endgroup\$
    – user57037
    Aug 17, 2016 at 16:07
  • \$\begingroup\$ @mkeith That's exactly what I was worrying about. There's no problem, then. Do you want to expand it as an answer for me to accept it? \$\endgroup\$ Aug 17, 2016 at 16:09
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    \$\begingroup\$ Consider making the 5V regulator a part of the battery pack. Or, maybe the regulator could be a detachable module that sits between the 6V battery pack and your device's USB jack. \$\endgroup\$ Aug 17, 2016 at 16:31

3 Answers 3

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If you are using a FT232R chip, this should work, provided you have a minimal voltage of 4.0V at the output of your LDO with all conditions. Other FT232 chips might have similar properties, check the datasheets.

The datasheet of FT232R states for VCC on page 11: +3.3V to +5.25V supply to the device core. However, in Note 1 (page 12), they precise that to use the internal oscillator (I suppose you want to save the space for a crystal) you need a minimum VCC of 4.0V.

Thus, the manufacturer ensures that the FT232R will reliably work, without an external crystal, with a VCC between 4.0V and 5.25V. This of course includes the USB communication with the USB host, which is of course an essential capability of the chip.

If you look at Chapter 6.2 of the datasheet (self-powered configuration), you can see that the self-powered circuit can be supplied with any voltage between 4V and 5.25V and still reliably communicate with the host. There's no reason this would not remain true if the power is derived from the USB supply voltage instead.

Although not mentioned in your question, another point you need to consider is the power supply voltage of the ATMega chip (I guess you will be using a ATMega168 or ATMega328, if you want to be compatible with Arduino Nano). Everything will be fine with your 6V external power supply as you will end up with 5.0V power. However, if you power from USB, the LDO will introduce a drop on the VBUS voltage, which is 5.0V nominal. If you use a 0.3V drop LDO, you will have 4.7V on the microcontroller, which is still fine. However, if VBUS goes below 4.80V, your VCC will drop under 4.50V, which is the minimum required voltage to operate at full speed (20 MHz), and the maximum frequency decreases with the voltage (see Chapter 33.3 of the datasheet for the ATMega 168P) In this case, you will have to operate at less than 20 MHz. If you use the 16 MHz typical of Arduino boards, you should always be in the safe area.

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  • \$\begingroup\$ I'm more concerned about the PC side of things and the effect in the differential signals. \$\endgroup\$ Aug 17, 2016 at 16:01
  • \$\begingroup\$ Can you follow mkeith's suggestion in the question comments to expand your answer? \$\endgroup\$ Aug 17, 2016 at 16:20
  • \$\begingroup\$ Sure, I added more details about the communication with the PC and the operating frequency of the microcontroller. If this does not match your hardware, please comment or edit your question to give more information about what µC you are actually using. \$\endgroup\$
    – Ale
    Aug 17, 2016 at 17:10
  • \$\begingroup\$ Doesn't the Arduino run at 16 mHz by default? \$\endgroup\$
    – Passerby
    Aug 17, 2016 at 17:15
  • \$\begingroup\$ The Arduino yes, as I mentioned in the answer. However, as the OP mentions a "Arduino Nano-like board" with no more details, it might be a custom development, and we can make no assumption about its clock frequency. 16 MHz seems probable, but it is not certain, so a comment on this seemed to be pertinent. \$\endgroup\$
    – Ale
    Aug 17, 2016 at 17:37
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Usb Power is typically 5V +- 5%, or 4.75 to 5.25 volts to be with spec. USB data is 3.3V. The FTDI has an on board regulator to drop it down, as does the Arduino board. As the FTDI and ATMEGA both have very wide latitude in voltage range, dropping it to 4.7 (6%) wouldn't be an issue, as the FTDI uses a good LDO internally. Many usb devices simply use a LDO and run off 3.3V instead.

That said, the solution is simpler. Add the LDO to your battery pack before the usb cable you will be cutting up.

Alternatively, use a USB power bank. 5V out, rechargeable, come in a variety of shapes, chargeable by any usb power supply or computer, etc. (Mind you, some have a minimum current draw that a simple Arduino wouldn't draw, so you need to shop around)

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  • \$\begingroup\$ That's a good idea! \$\endgroup\$ Aug 17, 2016 at 19:07
  • \$\begingroup\$ +1 for the USB power bank, which is an off-the-shelf solution to the problem. If you can find one that can simultaneously be charging itself while providing power downstream, you can leave the board more or less permanently connected to the power bank (which becomes its "UPS"). \$\endgroup\$ Aug 17, 2016 at 20:09
  • \$\begingroup\$ Please be aware that some "USB power banks" may stop supplying power when current draw falls below certain level. SO the low-power instrument connected to this power bank might stop working. \$\endgroup\$ Aug 18, 2016 at 20:50
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According to FT232 specifications, VCC pin can hold +6V. So you do not need to do anything, except maybe you need to limit the VCCIO pin to a somewhat lower level than 6V.

If you decide to use a LDO between VBUS and FT232, you can safely set the LDO output to +4V, which should be perfectly fine, all in accord with specifications. Or even to +3.3V if you are using external crystal for FT232. In both cases USB D+/D- signaling should meet USB signal specifications.

Addendum: to be on a safe side, an inexpensive in-series diode can be used to meet both ends of the particular voltage conditions.

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  • \$\begingroup\$ The 6.0V is a absolute maximum rating, you are not supposed to operate the device continuously at that voltage. Besides that, it is 6.0V for the FT232R, and 5.8V for the FT232H. \$\endgroup\$
    – Ale
    Aug 18, 2016 at 13:52
  • \$\begingroup\$ First, I am sure we are not talking abut production volume of 10,000 unit per week, this seems to be a small DIY project. Max rating usually have sizeable margins. Second, the 5.8V limit for FT232H chip is defined for "high-impedance" inputs. Max rating for VCC are not listed for the HS version of FTDI bridge. \$\endgroup\$ Aug 18, 2016 at 17:05
  • \$\begingroup\$ Sure, the OP looks like a DIY project, but other people will read the answers in the future and might erroneously deduce that the absolute maximum rating is a voltage you can safely use continuously. Your addition of a in-series diode seem a nice idea to meet the specification without permanently staying on the max rating. \$\endgroup\$
    – Ale
    Aug 19, 2016 at 8:42

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