5V to 3.3V Voltage Divider

Introduction

In many circuit applications there is a requirement to drop 5 Volt down to 3.3 Volt.

For example, a microcontroller may not be able to tolerate input voltages greater than 3.3V. However, the output device can only provide 5V.

In this case it becomes necessary to divide +5V down to +3.3V.

One way to do this is with a voltage divider network that consists of two resistors in series as shown in the picture below.

Calculator

This tool calculates the resistor values for a maximum +3.3V drop across resistor R2.

Enter either R1 or R2 in milliOhm/Ohm/kiloOhm/megaOhm. The tool will calculate the other resistor value. 


	

			

		

						

				

				

                

			

			
						

		

		






Formula





Ohm’s law is used to calculate the voltage drop across each resistor. The current through each series resistor is the same as the total current through the circuit. 





Therefore the total current Itotal is given by





Itotal = Vin/(R1+R2) = V1/R1 





V1=Vin*R1/(R1+R2)





Itotal = Vin/(R1+R2) = V2/R2





V2=Vin*R2/(R1+R2)





In this case, V2 = +3.3V and Vin = +5V. 





How to Pick Resistor Values





In general, there are many combinations of resistor values that will give the same result. For instance, 





    

  • R1 = 1 kΩ and R2  = 1.94 kΩ 
    • 




  • R1 =10 MΩ and R2 = 19.4 MΩ
    • 






Both combinations provide +3.3V at the output.





However, if you’re trying to interface to an Analog-to-Digital Converter (ADC) like the ATmega328P for instance, there is a requirement that the input impedance be lower than 10 kΩ. 





Higher values of input impedance will increase the sampling time and result in errors as the input signal won’t settle to a stable value fast enough.





On account of this requirement, the better set of values for this application is R1 = 1 kΩ and R2  = 1.94 kΩ. The effective impedance is 660 Ω (using this calculator)





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