# Q: Digital potentiometer MCP4162

I am having a hard time understanding the MCP4162 Datasheet.

The application for the digi pot is have an Ardunio to control the wiper resistance so I can have a constant desired output voltage regardless of input signal. So depending on the input signal the digipot will find the ideal resistance to get close enough to the target output voltage of 1.98V.

Example: Set Output Voltage = 1.98V

Input = 1.414Vp

$Digi pot = ((\frac{V_o}{V_in})-1)*R1$

$Digi pot = ((\frac{1.98}{1.414})-1)*1000$

$Digi pot = 400\Omega$

Input = 0.5Vp

$Digi pot = ((\frac{V_o}{V_in})-1)*R1$

$Digi pot = ((\frac{1.98}{0.5})-1)*1000$

$Digi pot = 2960\Omega$

simulate this circuit – Schematic created using CircuitLab

I have read the datasheet multiple times and I still don't which register is the one I need to be changing.

The one I would assume is the 0x04 address where it's described below. The datasheet only tells you about 2 registers and what their bits do when you write into them.

and I dont understand the equation stated in the datahseet that tells you the output resistance.

• which register is the one I need to be changing - that would depend on what you want to achieve... – Eugene Sh. Jul 6 '18 at 18:21
• Sorry about that, The application I am trying to use it for is an non inverting amplifier and putting digi pot in the negative feedback loop so I can have a constant gain. – Pllsz Jul 6 '18 at 18:24
• @EugeneSh. I have inserted a schematic of what I want to do with the chip. – Pllsz Jul 6 '18 at 18:27
• Something is wrong. If your pot were a simple resistor, you would have a constant gain. I think you are trying to build an adjustable gain amplifier, but your question, as is, does not make sense. – Scott Seidman Jul 6 '18 at 18:32
• Oh man no no you're right. It doesnt make sense. I meant a constant output Voltage, regardless of input signal. – Pllsz Jul 6 '18 at 18:34