0
\$\begingroup\$

I'm pretty new to all this. Please forgive me if this is an obvious question.

I am working with an ESP8266 module. They generally have a max operating current draw of somewhere around 250 mA. I have noticed that under certain conditions it can seize up, and starts drawing current at an unusually high rate.

I am operating this off a steady 3.3v DC supply, so I was thinking of using a resistor to limit the current going into the module when this happens, but would like for someone to double check my math here.

At 3.3v a 10 ohm resistor should cap me well within my safe range at 330 mA, correct? Going further with the math, 330 mA at 3.3v would be just over a watt of power. So would I need a resistor rated for over a Watt?

Improve this question
\$\endgroup\$
  • 2
    \$\begingroup\$ Why would you want to limit the current, the ESP will draw what it needs. It is a WiFi module, so the current it draws is mainly determind by the amount of network traffic it transmits, which is highly variable, hence a variable current is to be expected. Do you want to prevent it from transmitting?? If your car draws more gas when it is riding uphill at high speed, what do you do, pinch the gas tube??? \$\endgroup\$ – Wouter van Ooijen Sep 12 '15 at 21:16
  • \$\begingroup\$ Because it is going into an overcurrent situation at certain points. 330 mA should be more than enough to drive it, but until I can figure out what is causing this condition, I wanted to prevent it from burning itself up. \$\endgroup\$ – brad tee Sep 12 '15 at 21:32
  • \$\begingroup\$ Are you doing something funny with one or more of its IO pins? Are you sure of those current figures, how do you measure these? (A multimeter won't do, the current is highly fluctuating.) \$\endgroup\$ – Wouter van Ooijen Sep 13 '15 at 7:49
1
\$\begingroup\$

You have a 3.3V dc supply and you want to limit current taken by using a 10 ohm resistor. If the normal operating current is 250 mA then the volt drop across the 10 ohm resistor will be 2.5 volts taking the 3V3 supply down to 0.8 volts across the device.

It won't work like this of course because I suspect the minimum operating voltage will be a shade under 3V BUT I couldn't find that info so you need to work that out.

My advice is get a 6V wall wart and a 3V3 LDO regulator and feed the regulator with but have the 10 ohm in series feeding the regulator.

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

P=IV, so you need at least .33 Amps* 3.3 Volts, or 1.1 Watts. Best to overspec rather than underspec. I'd probably go for about 2Watts, if I were to use one resistor. You could also use five 50 ohm quarter watt resistors in parallel.

In any case, running a resistor to limit current on your power line is probably not a very good idea.

Improve this answer
\$\endgroup\$
  • \$\begingroup\$ Thanks for the quick answer! What would you suggest instead of the resistor? \$\endgroup\$ – brad tee Sep 12 '15 at 19:58
  • 1
    \$\begingroup\$ @bradtee I'd let the module draw what it wants to, unless there is some reason not to. Make sure you have a power supply capable of providing enough current. \$\endgroup\$ – Scott Seidman Sep 12 '15 at 20:11
  • 2
    \$\begingroup\$ A resettable polyfuse \$\endgroup\$ – Passerby Sep 12 '15 at 20:55
  • \$\begingroup\$ @Passerby might have the simplest answer as this would potentially resolve the seize up issue by performing a hardware reset. But if the module is this buggy I would be looking for a more reliable one. \$\endgroup\$ – BenG Sep 12 '15 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.