[solved] Nano with RFID-MC522 not always reading

Hello everyone,

I try to implement some sensors, motors which are controlled via an Arduino Nano. Depending on the state of the program a power relay will also be switched. Since the relay needs more power than the Nano can provide it needs its source elsewhere.

Now to my crazy idea. I have a "device" in close proximity to my planned PCB which can provide 24V/DC@3A. Is it possible to power my Nano over a "Three-step-voltage divider"?

Since I do not want to burn unnecessary energy over the Vin pin I limit the voltage to 8V (7-12V from the datasheet) and 5V for the power relay. I attached a schematic.

Is it possible to "leech" of this device without a voltage regulator?

No!! Resistive voltage dividers only work if the current going through the resistors is very much higher than any tiny currents taken out at the junctions. You can't POWER anything with one.

A 5V DC-DC converter powering both the Nano and the relay is what you need.

Steve

I had the feeling that this would be too easy. Why would there be a need for DC/DC converters if everything could be accomplished with a simple voltage divider.

I looked at some converters from Traco which can produce 9V (since the Nano needs 7-12V). I also would use a 5V-Relay with a Voltage regulator.

When I looked at the prices, I think it would be cheaper to buy a cheap AC/DC (9V) converter rather than build a whole power circuit.

Your Nano is 5V, relay is 5V, sensors are 5V, why would you buy a 9V power supply? :slight_smile:

As I already said you want a 5V DC-DC converter or power supply. You can power the Nano through its 5V pin and that's a lot more efficient than using a higher voltage to Vin or the barrel connector and then letting the regulator on the board waste power taking it down to 5V.

Steve

v3xX:
I looked at some converters from Traco which can produce 9V (since the Nano needs 7-12V).

The Nano most certainly does not require 7-12V. This is a gross confusion unfortunately introduced by the misleading descriptions on the Arduino site.

It operates on 5 Volts, so you need a 5 V regulated supply for it, connected to the "5V" pin and ground.

There is some voltageregulator on the Arduino, but I can not read the markings. Usually the regulators need a little more voltage than the output voltage in order to regulate correctly.

Does anybody has the correct datasheet of the regulator of an original Nano?

I looked under a microscope and found a marking on a different Nano.

Ams1117-5.0 https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.advanced-monolithic.com/pdf/ds1117.pdf&ved=2ahUKEwivt9HNwZ_qAhUCIcUKHaXyD3QQFjAAegQIARAB&usg=AOvVaw32b7yUSaNtlsyuBUGG7NQr

Everything between 5V- 17V should be fine (1.5V<Uin-Uout<12V)

Here, from the schematic.

For the uA78M05C it says 7-25V (for the original). I have an old power source with an output of 7.5V. Since I do not want to burn unnecessary energy I will try to power the Nano directly with the 7.5V and the power relay over a 5v voltage regulator.

Tomorrow I have time to test this configuration.

Since you seem determined to ignore all advice and do it the hard way then I'll just wish you good luck...you're going to need it.

Steve

I will just suggest you do some reading up on voltage regulators, the type used on Arduino boards. Voltage regulators of this type have what is called a dropout voltage which is typically about 2 volts. They also have a current limit. The Arduino Uno and Nano both use a low drop out voltage 5.0 volt regulator. The suggested input voltage is ~ 7.0 to 12.0 volts. Yes, I know what the data sheet says for the regulator. The reason the voltage is capped on an Arduino board is because once you understand how a regulator of this type works you will understand things like power and heat. If I apply 12 volts to a 5.0 volt regulator on an Arduino board do you wonder where the remaining 7.0 volts goes? How about if the total load on the regulator is 500 mA (0.5 Amp)? How much power (heat) needs dissipated?

Good design practices are about learning how stuff works and understanding data sheets. No, a voltage divider won't work for all the well covered reasons.

Your best solution is using a buck converter and those modules are inexpensive and plentiful as was suggested. You may want to read a little and rethink your design plans.

Ron

Of course I do not ignore sound advice. I started completely wrong and tried to adjust a broken circuit with the parts laying around.

My current design is completely changed: SMPS (230V -> 5/6/9V DC adjustable) which powers the Nano and a LM7805 voltage regulator (6.15V -> 5.01V for the opto, mosfet and relay due to the higher switching current. The Nano alone can not produce the necessary current to switch my relay (~700mA Songle SRD-05-C), therefore I think the voltage regulator is necessary to ensure stable 5V supply.

So far my Nano switches runs normally (blinking status LED with 1.75s) and also switches the Opto after a button push. Including the mosfet and relay I can successfully switch the relay. I will monitor temperature, functionality and current at certain nodes.

I will post an updated schematic either to complete this question or for further help if something burns up.

Hi,
Can you post a diagram as to how you will connect your project please so we can see what you are thinking of doing.
A picture is worth a thousand words.

Please note, the Vin of 7V to 12V, is immediately regulated down to 5V on the NANO board, there is no reason to specifically provide a supply to Vin.
You can supply 5V directly to the 5V pin.


Thanks.. Tom... :slight_smile:

You don't seem to be listening, let's see if I can re-iterate it...

The Arduino and your peripherals all run on 5V. Therefore, you drive them all in paralel from a decent 5V supply. The only reason that the barrel jack exists is because 9V PSUs were more common than 5V ones at the time, so an under specified SMD regulator was provided to allow their use. Whenever you have any choice, FORGET THE BARREL JACK exists.

As was mentioned earlier, if you are feeding 12V in, where to you think 7V goes? If you are feeding 9V in, where do you think 4V goes? The answer is HEAT. Lets say you are pulling a total of 200mA from the power supply at 5V. With 12V in, you need to dump (W=EI) 70.2 = 1.4 WATTS of heat from the onboard device which will fold back pretty pronto because it doesn't have a proper heatsink and cannot dissipate anything like that.

To sum up, you need ONE 5V supply, that's it. Feed everything directly at 5V and make your life simple.

I now had time to test the functionality and drew a schematic.

The problem with my power supply is that it is not stable enough. There are three main lines needed (5V,12V,24V) in the whole setup (~5 commercial devices and 4 custom boards). There are some peripheral boards which do not constantly draw power. I ended up with voltages between 4.5-5V and the Nano shut down. For this reason I increased the mainsline to 6.0V and build voltage regulators on the custom boards for the parts that need accurate 5V. I also do not want to drive the voltage regulators with unnecessary high voltages.

After some testing I concluded:

  • Heat of the voltage regulator is non noticable (~1V difference)
  • Mosfet and Opto switches (far) under the limits
  • The power draw from the power relay in the switching moment does not effect the 5V from the regulator

I do not see a major problem with this design other than the oversized parts. Maybe you could look over the schematic check for major design flaws.

6V is not enough for the input of the 7805 nor the onboard Nano regulator. You should be running the Nano from 5V regulated. Sort the fault, a sticking plaster over the symtoms is not adequate.

You don't seem to actually want any responses unless the answers are what you want to hear, so I'm bowing out now.

v3xX:
I ended up with voltages between 4.5-5V and the Nano shut down.

No.

Hi,
Read post #17.

Look at the data sheet for the LM7805, you need more than that capacitor as bypass for it to operate properly.

Tom... :slight_smile:

Hi,
This would be better;

Tom.... :slight_smile: