Tension divisor

Hi,

I've a basic electronic question. Basically, I try to connect an Arduino proMicro(ApM) 32u with a RFID-RC522, classic situation.

The Pro Micro provide 5VDC and the RFID need 3.3V. Ok, it's work with the 5VDC directly from the proMicro, BUT, I want to understand what is happen.

So to realize my circuit, I start by simplify the circuit with 2 resistances, R1= 1K Ohm and R2= 2K Ohm, given to me Vref = (5V*2000)/(1000+2000) = 3.333333V. So, I'm proud of me, connect everything and it doesn't work :frowning: .

Then I returned to my drawing table, take my multimeters and start to take mesurements. The VRef DROP to 2.3v when I connect the RC522 board. When I unplug it, my Vref is back to 3.3 :o

New Idea, what is the resistance of the board? In the specsheet of the board, the current used it's between 13 and 31 mA, my conclusion is I can calculate R of the board (Rb).

Too simple, I've said in my mind, V = RI so R = 3.3V/0.01415A = 233,2 Ohm or 233 never mind.

The new Resistance Req = should be 209Ohm (2000233)/(2000+233) , than the new Vref = 5209/(1000+209)=0.86VDC :confused: far away from my 2.3V

So I change my R1 and R2 by a factor of 100, so R1 = 100, R2 = 200, Rb = 233, (Req = 108) and Vref = 1.61.

Again, with R1=120 and R2 = 10KOhm, Vref = 3.2V.

I wondering if this is the good approach of the problem, I dont want to burn the chip... :frowning: , impossible it work with Vcc (5V), but did I've the good approach to resolv this issue?

Any comments or proposition?

Cheers,

Martin

In general, you can't use a voltage divider to power equipment as the voltage is unregulated. Use a step down converter instead.

Since you cannot directly connect the outputs of a 5V Arduino to a 3.3V device, it is much safer to use a 3.3V Arduino instead. Then you can power both with 3.3V and connect them directly to each other.

For dropping 5 V to 3.3 V, a switchmode regulator is barely more efficient than a linear regulator.
This module includes a 3.3 V regulator and four "bidirectional" level translators which do work satisfactorily with the MRFC522 RFID module.

A voltage divider is for measurements or signals, never for power. Voltage dividers require their load
to be a constant or very high impedance. A voltage divider cannot provide a low impedance output without
wasting vast amounts of power (upto 100's of times the load).

A voltage regulator has a very low output impedance without wasting any more power than necessary
(linear regulators), or for switch-mode regulators without wasting any power (except for a few percent
switching losses). Regulators routinely have sub 1 ohm output impedance.

To provide a 1 ohm output impedance 3.3V supply from 5V using a resistive divider would waste over 5
watts (assuming the 5V supply could provide that much). It would need a 1.5 and 3 ohm resistor pair.

A linear 3.3V regulator suppling a 50mA device wastes 0.085 watts (66% efficient), a switch-mode perhaps
0.015 watts (90%), the above resistive divider would be 3% efficient by contrast - its completely unworkable.

Thank you jremington for your answer. Basic and at least, stupid question.

jremington:
In general, you can't use a voltage divider to power equipment as the voltage is unregulated.

Light up my brain please. My understand is the DC power IS regulated by definition. To avoid having DC fluctuation, I can put a condensator between the R2 poles and that's it. But I've LM317 in my hands, I will give it a chance. Otherwise, my ATX bench power generator is not the best idea :frowning:

Fell free to suggest me any forum or documentation to read about this.

jremington:
Since you cannot directly connect the outputs of a 5V Arduino to a 3.3V device, it is much safer to use a 3.3V Arduino instead. Then you can power both with 3.3V and connect them directly to each other.

The problem is the Arduino U32 pro Micro I use didn't have the 3.3V regulator.

Thank you MarkT,

MarkT:
To provide a 1 ohm output impedance 3.3V supply from 5V using a resistive divider would waste over 5
watts (assuming the 5V supply could provide that much). It would need a 1.5 and 3 ohm resistor pair.

A linear 3.3V regulator suppling a 50mA device wastes 0.085 watts (66% efficient), a switch-mode perhaps
0.015 watts (90%), the above resistive divider would be 3% efficient by contrast - its completely unworkable.

Where can I find the mathematics to calculate that lost? It's not that I want to challange you, I want to increase my understand of the electronic. Like I've posted previously, I've in my hands LM317 which can give me 3.3V power output. I will give it a chance.

Again, thank you

V = IR
P = VI

Add Kirchoff’s laws and that’s all the physics you need.

You can derive everything from these, such as series and parallel resistance combinations, voltage ratios, power dissipation in terms of V and R…

Thank you Paul__B for this recommendation,

Paul__B:
For dropping 5 V to 3.3 V, a switchmode regulator is barely more efficient than a linear regulator.

This module includes a 3.3 V regulator and four "bidirectional" level translators which do work satisfactorily with the MRFC522 RFID module.

What is the purpose of the

Paul__B:
"bidirectional" level translators

Again, thank you.

Bidirectional level translators are specifically required to interface between 3.3 V and 5 V logic level devices using “I2C” protocol. This allows data to pass in either direction. This is not necessary in interfacing to the MRFC522 RFID module and these level translators are limited in speed however they appear to be sufficiently fast to operate either sending data to or receiving it from the MRFC522.

Other situations such as driving WS2812 “individually addressable” LED chains (which require 5 V logic) with a 3.3 V device utilise a faster data transfer rate and as a result these bidirectional level translators do not work in that situation.

A particularly useful native 3.3 V microcontroller module is the inexpensive ESP8266 which you may choose to use for WiFi operation or not implement it and simply use it as a general-purpose microcontroller. Modules such as the WeMOS D1 Mini include the 5 V to 3.3 V regulator.

The only reason for deliberately using the Pro Micro is to make it a HID device emulating a USB keyboard or mouse - is that the reason for your choice in this case?

Mathematics of voltage dividers with load.

The problem is the Arduino U32 pro Micro I use didn't have the 3.3V regulator.

I linked a great source of step down regulators.

Paul__B:
The only reason for deliberately using the Pro Micro is to make it a HID device emulating a USB keyboard or mouse - is that the reason for your choice in this case?

Got it :). Keyboard. To be honnest, this circuit is to test a use case with Windows PIN with a token. So the user will just have to pass his token to send the PIN to the Windows text area.

Why? 1st, I’ve all the hardware to build 5 devices
2nd, learning curve
3rd, Having fun to hack Windows :stuck_out_tongue:

jremington:
Mathematics of voltage dividers with load.
I linked a great source of step down regulators.

jremington, I see the site proposed and this is the one I’ved used to calculate my final resistance (reference my question).

Thank you, I’ve save it in my favorite.