0
\$\begingroup\$

I have a couple of these http://www.toshiba.com/taec/components2/Datasheet_Sync//149/215.pdf I am using for level shifting 3.3 volts to 5v from a beaglebone to a stepper motor driver. The frequencies will range between 2.5kHz - 800 hZ. My questions are:

1) Do I need capacitors, and what would be good values?

2) The datasheet lists DC Vcc/GND current MAX as += 200mA, does this mean I need a current limiting resistor in series with my 5v line (Vcc) and Vcc pin? My TA says I don't, but I would like to double check.

3) For one of the IC's I want to tie the inputs to Vcc and use a Oen (G1/G2) to toggle them all at once. Do I need current limiters between Vcc and the input pins? Again my TA says I don't, but it would be nice to have a good explanation and another opinion.

The stepper driver draws between 8mA-16mA and has a 240 ohm resistor in series with it's inputs.

\$\endgroup\$

2 Answers 2

1
\$\begingroup\$

Do I need capacitors, and what would be good values?

Bypass capacitors between VCC and GND are always a good idea. For a low-frequency circuit like this, a combination of 1 uF and 10 uF in parallel is probably okay. If you're working on a breadboard, start with just 1 uF and add 10 uF if you see issues.

The datasheet lists DC Vcc/GND current MAX as += 200mA, does this mean I need a current limiting resistor in series with my 5v line (Vcc) and Vcc pin?

No. It means you need to limit the total load drawn by all the things connected to the outputs to less than 200 mA (and also follow the other limits like drawing no more than 50 mA from any individual output pin). Putting a limiting resistor in series with the VCC pin would reduce the provided VCC voltage and likely cause unexpected behavior.

For one of the IC's I want to tie the inputs to Vcc and use a Oen (G1/G2) to toggle them all at once. Do I need current limiters between Vcc and the input pins?

No, but it won't hurt anything. The "Input leakage current" spec in the DC Characteristics table tells you the input pins won't draw more than 1 uA when connected to VCC.

In real designs is typical to put a 5 - 10 kOhm resistor in this situation because it allows the pull-up to be overridden in testing and confirm the chip is mounted correctly and functioning correctly.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks, it's all much more clear now. I think you made a small mistake in your first paragraph with 1uA and 10uA instead of uF. *edit- looks like you caught it before my comment. \$\endgroup\$
    – user66817
    Commented Mar 7, 2015 at 16:33
  • \$\begingroup\$ I changed your unpredictable to unexpected behavior. You would find that it would be fully possible to model the chip behavior with a resistor in series with its VCC pin. As such the device behavior would be predictable but it would certainly not give the normally expected behavior. \$\endgroup\$ Commented Mar 7, 2015 at 18:00
2
\$\begingroup\$

Do I need capacitors, and what would be good values?

I can't guess where you think these capacitors might be necessary.

However, there should always be a bypass cap from power to ground at any power pin. 1 µF ceramic is a good value today. The loop between the power and ground pins thru the cap should be as small as reasonably possible.

need a current limiting resistor in series with my 5v line (Vcc) and Vcc pin?

No. Absolutely not. You don't want to make the impedance of the power supply as seen by the chip any higher than necessary. You want to NOT put any resistance between the 5 V rail and the power pin of your chip.

For one of the IC's I want to tie the inputs to Vcc and use a Oen (G1/G2) to toggle them all at once.

OK so far.

Do I need current limiters between Vcc and the input pins?

The inputs are meant to be driven at 5 V and 0 V. They will draw whatever small current they draw, but there is no need to limit that current somehow.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.