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?
Best Answer
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.