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The nRF24L01 operates at a voltage range of 1.9 – 3.6V.

I have a power supply of 6V. I need to regulate to 3.3V. I used an LD33V (3.3V regulator), but I need other solution, so I decided to use a 3.3V Zener (half watt.)

I got garbage values on the serial monitor.

I believe that is due to low power at the Zener output.

Can someone suggest a good power supply method to power nRF24LO1 to regulate 6V to 3.3V?

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    \$\begingroup\$ Your first solution is the obvious one, so what is the issue with using a LDO regulator? \$\endgroup\$ – Kartman Feb 23 at 8:05
  • \$\begingroup\$ thank you for answering my question. My system is battery powered and it need to function atleast 15 days. I'm using ultrasonic sensor and atmega 328P. if i use LDO regulator to power nRF24L01 it dissipate power which decrease battery life. \$\endgroup\$ – Manu Devappa Feb 23 at 9:01
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    \$\begingroup\$ But the zener approach is wasting even more power! The only solution to safe power compared to your LDO is a switched regulator. \$\endgroup\$ – jusaca Feb 23 at 9:43
  • \$\begingroup\$ @ManuDevappa: Thanks for asking this question. You got me to thinking about one of my own projects. The suggestion I made to solve your problem turned out to be the solution to a problem that I had. I've just ordered some of the Adafruit LM3671 modules to run some ESP32S modules from solar charged lithium polymer cells. \$\endgroup\$ – JRE Feb 23 at 20:52
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There are two ways that you can use a Zener in your circuit. Neither is all that good.

First, there's the standard "shunt regulator" circuit.

It looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

You have to size R1 to provide the full current to the nRF24L01 at 3.3V. That's about 14 milliamperes.

That also means that your circuit will always consume 14 mA - even when the nRF24L01 is sleeping.

If you can assume that the 6V will never go above 6.6V, then you could do the following:

schematic

simulate this circuit

That will consume less than 14mA all the time, but I'm not sure what value of R1 you would use. The value depends on the current/voltage curve of the diode and how much current it takes for the Zener to hit 3.3V

Neither circuit will regulate all that precisely, and neither will do well when the load (nRF24L01) draws current in burts - which the nRF24L01 will do.

You can add a couple of capacitors in parallel to the nRF24L01 to handle the bursts, but that won't do anything for the poor regulation or for the fact that both of those regulators are wasteful. Both waste power all the time - even when the nRF24L01 isn't doing anything.

For efficiency, you need a buck converter with a low quiescent current.

Something like the LTC3388 would do the job. It consumes less than 1 microampere of current, and only operates when the output capacitor voltage drops - the buck converter will be automatically turned off most of the time when the nRF24L01 is inactive.

That's kind of an expensive chip (and you'd have to build a PCB for it,) though maybe you need that extremely low quiescent current for your task.

Alternatively, many not so specialized buck converters are (compared to your Zener circuit) low power and efficient. Adafruit makes an LM3671 based 3.3V output buck regulator module with a quiescent current of 16 microamperes. That's still worlds better than your Zener regulator or the LD33V you were originally using.

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  • \$\begingroup\$ Thank you so much. it helps me a lot. \$\endgroup\$ – Manu Devappa Feb 25 at 6:53
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If already an LDO consumes too much power for your liking, you could consider the following:

from your 6V, use Diodes to generate a drop of ~3 Volt. After the diodes, place a 1 MOhm resistor and a cap to ground. The resistor is there to limit voltage excursions beyond 3.6 V

This will "regulate" worse than an LDO, but stay within range of your IC. Quiescient current can be tuned very low.

Also, this is only applicable for very low current loads, as efficiency cannot be better than ~50% in this case. For more current hungry ICs, you quickly come to a range, where switched regulators offer the best efficiency despite their quiscient current.

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  • \$\begingroup\$ thank you for your suggestions \$\endgroup\$ – Manu Devappa Feb 25 at 6:56
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Whatever you put in series with your circuit (LDO, resistor, diodes etc) to "burn off" the extra 2.7 V will do just that: waste energy. The Zener is even worse because it also draws considerable current in parallel with your circuit.

Despite the fact that switched mode power supplies (buck converters in your case) are obviously more efficient because they don't get rid of the extra voltage by converting it into heat, you should consider this option with care. These active circuits also need power to work, their inductors are not ideal, and the switching process itself wastes energy.

Without knowing the power consumption of your device you will not be able to determine for sure if using a more efficient regulator (under high loads) will be better overall.

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Unfortunately, any circuit that regulates the output voltage by producing a voltage drop will dissipate a power equal to \$P_{Loss} = V_{Drop} · I_{Load}\$ .

In your specific case you have a \$V_{drop} = 2.7 V\$ so, for a current of 12.3mA (\$I_{VDD_2M}\$ from nRF240L01 datasheet) your minimum wasted power it will be of 33.21mW.
To that value, you have to add the additional loss for the device polarization.
For LDO regulators, for example, you will have a ground current (difference between input current and output current) of some μA, while in the zener circuit you have the zener polarization current that it's of order of mA.
Another problem of the Zener solution is that if you design the series resistor value for a certain current, if the load draws less current you will dissipate more power on the zener.

In conclusion, if your goal is to save power, the Zener circuit is a bad choice. LDO regulator are the best choice for non-switching solutions.

For your application, I recommend a charge pump IC like this.
It provides a regulated voltage of 3.3V with an input voltage up to 6.5V, a quiescente current less than 40μA and the efficiency can be up to 90% depending on the current.

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  • \$\begingroup\$ Thank you so much. \$\endgroup\$ – Manu Devappa Feb 25 at 6:55

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