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This question already has an answer here:

I am trying to create a power supply circuit for the ESP8266, using a voltage divider and a cellphone charger rated at 5V, 1A as power supply.

I have resistors connected like this:

enter image description here

The resistors are rated at 1 - 6 W. Using an online calculator for power dissipation i get 900W per resistor for the ESP8266 max current of 300mA. I am using 3 10K resistors with two in series for the 20K part.

I guess this is the correct reasoning.

Could someone explain if I'm doing something wrong and if this will work?

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marked as duplicate by PeterJ, Community Feb 7 '16 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ Look at Peter Bennett's answer here : electronics.stackexchange.com/questions/111646/… \$\endgroup\$ – Marla Feb 6 '16 at 17:12
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    \$\begingroup\$ A resistor voltage divider is not suitable for use as a power supply. Never. Not ever. \$\endgroup\$ – brhans Feb 6 '16 at 17:12
  • \$\begingroup\$ Yes but I'm asking because i don't have a regulator with me. What will happen in this circuit? Will the resistors melt? \$\endgroup\$ – JJ_Jason Feb 6 '16 at 17:17
  • \$\begingroup\$ The output voltage of the divider will vary depending on the current drawn from the output. Your ESP8266 will effectively be a variable resistor in parallel with the 20 K resistor. I believe the ESP8266 can draw 250 mA or so when transmitting - that would theoreticaly produce 2500 volts across the 10K resistor!! \$\endgroup\$ – Peter Bennett Feb 6 '16 at 17:22
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    \$\begingroup\$ We should have a giant "no, bad dog" sign for any voltage divider power supply question. \$\endgroup\$ – Passerby Feb 6 '16 at 17:55
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When unloaded the R divider will generate 3.3 V, but when you try to consume current (e.g. the ESP8266), the voltage will fall -- in this case to unusable levels.

To the ESP8266, this circuit looks like a 3.3 V source (which it wants), but in series with a resistance of 10k//20k = 6.66k ohm. Thus if it tries to consume current, the voltage will fall, and it won't function.

Nothing will melt.

You could probably use 2-3 diodes in series to drop the 5 V to 3.3 V (actually 5-2*0.7 = 3.4), but this isn't very stable (likely the 5 V is variable and may be as high as 5.5 V which would damage the ESP8266.

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