# Amplifying voltage

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#### chris414

##### New Member
Okay this is the problem:
I am using a joystick (just a variable resistor) to control the speed of a stepper motor. The variable resistor ranges from 0 to 100kΩ - the motor does not move when the variable resistor is at 50kΩ. Currently I'm using a potential divider and op-amp to detect whether the joystick is moved left or right, which then makes the motor turn clockwise or anti-clockwise. But I need to be able to control the speed of the motor depending on the resistance of the variable resistor.

The speed of the stepper motor is determined by clock pulses created by a 555. The way i thought it could work would be by taking the voltage across the variable resistor and putting into Pin5 of the 555 to vary the time between pulses. The problem is that the potential difference across the variable resistor only ranges from 0 to 0.6V. How can I "amplify" this voltage?
Ie. if the voltage across variable resistor = 0.4V, voltage to 555 = (0.4/0.6)*supply voltage.

Also, this doesn't take into account that the resistance gets less when I move the joystick left, and greater when I move it right... perhaps there is a better way to do this?

Thanks

#### Grossel

##### Well-Known Member
You described your amplifying problem too little detailed. It's easy to make an dc amplifier circuit containing an opamp, but I need to know the maximum and minimum input voltages and the representative output voltage in both cases.

It's all math

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#### chris414

##### New Member
Okay I'll try be a bit clearer:

- Input voltage ranges between 0 and 0.6V.

- I need the output voltage to range from 0 to 5V.

The thing is that since I am controlling the speed of oscillations on a 555, I need the output voltage to be in proportion to whatever the input voltage is (input voltage determined by an analog joystick, effectively a variable resistor in a potential divider with another resister). Ie. I can't just have a "high or low" voltage coming out.

Thanks

#### Grossel

##### Well-Known Member
That means you have to amplify the signal 8 1/3 times. It's pretty easy using a opamp. Easy math tells that using a non inverting opamp amplifier you can use resistors having following values: R1=15k and R2=110k.
To remind: The amplification is 1 + R2/R1.

However you may use a variable resistor for R2, adjusting the amplification so that the output hits exactly on 5V when input voltage is maximum. Hint: Put a 20k variable resistor in serie with a 100k resistor.

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#### chris414

##### New Member
Thanks for the help

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