I've got Sparkfun's FT245RL Usb to Fifo Breakout Board. I need 5V output from data bit ports(D0-D7). While I'm working with it, I've found out I need to unsolder SJ1 jumper between Vcc and Vccio then connect Vusb to Vccio in order to get 5V output. I did it but unfortunately I cannot get 5V output from that ports. Did I miss something? Did I do it wrong? Or do I need to supply external power?

  • \$\begingroup\$ What you describe should work, so when you say you can't get 5V from the outputs, what do you mean? What voltage level do you get? \$\endgroup\$ – alexan_e Apr 6 '14 at 14:00
  • \$\begingroup\$ For example when I set D0 port to HIGH, I get 2.8V from D0. \$\endgroup\$ – st. Apr 6 '14 at 19:02
  • \$\begingroup\$ What level do you get with Vccio=3.3V? \$\endgroup\$ – alexan_e Apr 6 '14 at 19:43
  • \$\begingroup\$ I've made some adjustments and I'm getting ~4.85V output voltage from pins when they're HIGH and unloaded. You can check out the answer below. \$\endgroup\$ – st. Apr 7 '14 at 9:31

Contrary to what you wrote, SJ1 is not between VCCIO and VCC. It selects whether the FT245 chip gets IO-side driver power from the built in 3.3V regulator (default) or from the target device via JP3 pin VCCIO.

So you must either leave VCCIO connected to the 3.3V regulator via jumper SJ1 (which results in max 3.3V output swing), or cut SJ1 and connect VCCIO to some other supply, perhaps from the external target system, or from USB. You have done the latter, so that's promising.

So two further points: 1. Check that 5V is actually getting to the VCCIO pin using a voltmeter. If it is, then...

  1. Note that the output pins have only limited ability to deliver current, and the more current needed by the external load, the lower will be their max voltage in the high state. Spec sheet says Voh typical 4.1V and min 3.2V with a load of only 2mA.

Only if the outputs are unloaded will you see close to 5V. That's fine if your external system has high-impedance inputs, as, for example, Arduino inputs would be. But if you are testing with an external system connected but powered down, then those external connections will present a significant load due to their input protection diodes.

Likewise if you are using these outputs to drive LEDs or even transistors (with base resistors) you will see lower output voltages.

So... test with external system disconnected.

Schematic: https://www.sparkfun.com/datasheets/BreakoutBoards/FT245RL-Breakout-Schematic-v2.pdf Data sheet: http://www.ftdichip.com/Support/Documents/DataSheets/ICs/DS_FT245R.pdf

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  • \$\begingroup\$ OK I've made some adjustments and I'm getting ~4.85V output voltage from pins when they're HIGH and unloaded. I checked with a voltmeter. That's really cool. But here's the deal, I want to activate relay switches using these pins. However, they're not working. I'm using 05VDC relays, and they're working while they're connected to VUSB pin and also V3.3 pin. I don't understand why. Maybe current is not enough for relays? What should I do to make it work? \$\endgroup\$ – st. Apr 7 '14 at 8:54
  • \$\begingroup\$ Post the relay spec if you like. You are lucky if you can drive the very smallest relays with an output like this. Instead you need to use the logic output to control a BJT transistor or FET, which can in turn switch the relay coil. Regardless of how you do it, you will need protection diodes to avoid the coil creating a high voltage ("reverse EMF") pulse that will destroy the component controlling it. Google images for "arduino controlling a relay" for examples. Typical: electronics.stackexchange.com/questions/33312/… \$\endgroup\$ – gwideman Apr 7 '14 at 9:23
  • \$\begingroup\$ Here's the relay's specs megasan.com/service/pdfhandler.ashx?fileid=3398 \$\endgroup\$ – st. Apr 7 '14 at 9:29
  • \$\begingroup\$ Coil data shows requirement of 71mA at 5V. Using a BJT, saturated (fully on) to switch that requires BJT to dissipate about 0.3V * 70mA = 20mW, so a fairly small transistor will do, like a common 2N2222, or 2N3904. To make sure it's fully on, we need say IB ~= 70mA/50 = 1.3mA. Assuming Vout from FT245 of say 3.7V, and VB of 0.7, R = V/I = 3/1.3m ~= 1.8k to 2.5k. Be sure to include the protection diode. 1N4001 or similar will do. \$\endgroup\$ – gwideman Apr 7 '14 at 9:43
  • \$\begingroup\$ That said, if you need the relay to switch only a small amount of current compared to the 10A this one is good for, you can use a smaller relay and not need to waste so much current driving the coil. This would be a consideration if you're running the controller off battery or USB, where total power supply current is limited. \$\endgroup\$ – gwideman Apr 7 '14 at 9:46

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