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I have a RF module (nRF24L01) and Li-ion battery with 3.7V nominal voltage, so when it is fully charged it can be a bit higher than 3.7V (I guess).

RF module can work from 1.9V to 3.6V and draws <13mA.
I know about LDO regulators, but I don't have such, rather don't want to buy it and I want to keep design simple.

So, becasue of 1.9-3.6V and low power consumption can I use simple, regular diode, like 1n4007, on Vcc line, to drop voltage?
Typical voltage drop is 0.7V. I know it varies with current etc. but I only need to lower it a bit and can do it up to 1.9V. So?

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  • \$\begingroup\$ yes you can as long the diode you have chosen will be properly biased at the required current. Just one stage (to get it to about 3V) will be enough, such that not too much power will be wasted on the diodes. \$\endgroup\$ – Eugene Sh. Nov 26 '15 at 19:36
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So a Li-ion battery typically charges to 4.2v. You would have to look at the Vf vs I curve for the specific diode to see if it dropes the voltage enough. Based on the current you mentioned, with only 1 diode in series it would be very close to the upper limit, so you would really need more like 2 diodes. That being said, LDO regulators are cheep for only 100mA, it really would be a better solution. Also keep in mind if you are charging the Li-ion battery in the circuit, the charge voltages could be higher than 4.2V depending on the specific charger. A similar solution that could be better if you only want a single component would be to put a zener diode in series with the power, reverse biased so it dropes the zener voltage. This would be more accurate and constant then a diode.

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  • \$\begingroup\$ Ok, maybe I will use LDO regulator, but there's one case I'm not sure about. My whole circuit operates from 3V-5V. So voltage on the battery can be as low as 3V. I wonder what happens when there is lower voltage than LDO reg requires. I will get for example 3.3V reg. and will power it with 3.3V. So will it stop working (not nice) or just produce lower voltage, like 2.8V (which is ok)? \$\endgroup\$ – zupazt3 Nov 26 '15 at 19:54
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    \$\begingroup\$ The LDO will just basically pass through the input voltage minus the voltage drop of the regulator. For small modern LDOs, you can get voltage drops of sub 100mVs. So say it is a 3.3v LDO with a 100mV drop, with 3v in the output would be 2.9v. \$\endgroup\$ – MadHatter Nov 26 '15 at 20:06
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Use a LDO regulator like the LP2980: -

enter image description here

There'll likely be a few more like this just use your google fu (3V 50mA LDO regulator) and hunt around for a few minutes and check the data sheets for this sort of graph that tells you how benignly they fall out of regulation at input voltages below 3V: -

enter image description here

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