I'm trying to hook up my LCD on a breadboard and a schematic diagram involving a potentiometer is a bit confusing.

enter image description here

Full Data Sheet: http://www.lcd-module.de/fileadmin/eng/pdf/doma/dip203-4e.pdf

I guess the easiest way for me to see this is, if that potentiometer is replaced with an actual resistor (say 2K), how would this schematic look?

  • \$\begingroup\$ It wouldn't. A pot is two resistors. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 1 '16 at 23:03
  • \$\begingroup\$ you could use a few select on test resistors , assuming there was no aging effect, but a low cost trimmer is optimal for contrast ratio. It would be like designing a bias voltage to exactly match the MOSFET threshold. It just isn't that precise. \$\endgroup\$ – Tony Stewart EE75 Oct 1 '16 at 23:11
  • \$\begingroup\$ That circuit is using all three terminals of potentiometer? I assumed one of the terminal isn't used. But this is exactly why I asked this question - I've never seen a potentiometer schematic like this, so I simply have no idea how to connect that to a breadboard. \$\endgroup\$ – Xiagua Oct 1 '16 at 23:19
  • \$\begingroup\$ See my answer - connection is as in example B. R2 is your pot with 2 terminals used and R3 is internal to the LCD module \$\endgroup\$ – Russell McMahon Oct 1 '16 at 23:45

In this case the "potentiometer" is used as a 2 terminal variable resistor with the 3rd terminal not used. Technically it is then NOT a potentiometer as a potentiometer is a resistive divider which a variable "potential" is 'tapped off'. The term is commonly interchangeable with "variable resistor". In this case a single resistor of say 2K could be substituted.

What happens here is

  • Either the impedance or resistance inside the meter acts as a divider to adjust an internal voltage as the external resistance is varied OR

  • The resistor controls a current which is used for internal contrast adjustment.

Both the above can be thought of as different ways of looking at the same thing.


A & B below show two ways of achieving the same result.

In case A the voltage from a "wiper" on the 3 terminal potentiometerR1 is supplied to the target system. The internal resistance needs to be high compared with the resistance of R1 so it will not affect its voltage significantly OR the effect of the internal resistance needs to be allowed for.

In case B R2 acts a a variable resistor and in conjunction with internal resistor R3 forms a potentiometer to vary the internal voltage.

While both arrangements allow the internal voltage to be varied, the two cases are not exactly equivalent. In A the voltage can be reduced to zero and if R1 has linear resistance with position then the supplied voltage will also be linear with position.
In B the voltage is Vin/2 when R2 = R3, Vin/11 when R2 - 10 x R3 and Vin /101 when R2 = 100 x R3. ie the voltage can never fall to zero for non-infinite values of R2 and the decrease in voltage slows with increasing R2 so the voltage is non linear with position.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Beautiful exposition. :) \$\endgroup\$ – EM Fields Oct 2 '16 at 0:00

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