# How to analyze the output voltage of this potesiometer schematic?

Status
Not open for further replies.

#### electronion

##### Newbie level 6
I am confuse how to analyze the output voltage of this schematic. do you know how much the out voltage if the potensiometer turn from 0% to 100%?. I've tried to simulate it in Electronic Workbench. and the result for 0% potensiometer turn is 5 Volt, for 50% is 3.32 Volt. How could it be 5 Volt for 0% and 3.32Volt for 50%?
Can you help me?

#### Attachments

• Untitled.jpg
7.9 KB · Views: 73

#### udhay_cit

##### Full Member level 6
The value of potentiometer can vary from 0ohms to 10KOhms (assume it R1). The bottom is fixed 1Kohms (Assume it as R2). Assume the VCC=10V.
The R1 variable resistor will give you 10Kohms resistance for 0% position & 0ohms resistance for 100%.

The formula for output voltage( Vout) is Vout=VCC * [R2/(R1+R2)]

Condition 1 : (Assume R1=0%) 10Kohms. Here Vout= 10 * [1K/(1K+10K)] = 0.909V (Approximately 1V)

Condition 1 : (Assume R1=100%) 0ohms. Here Vout= 10 * [1K/(1K+0)] = 10V

So the 0-100% variation of pot will give you 0.9v-10v variation...
The voltage divider has 10:1 configuration....

Regards
Udhay

#### hemnath

Don't get complicate. Apply ohm's law, Which states I = V/R. if ur Vcc = +5V. current through the string, I = 5/11KΩ => 0.4545mA .Take upper resistor as R1 and lower resistor as R2
1. If you moving the adjustable end of POT towards +Vcc(i.e., fully turned to one side), then it becomes,
voltage drop across R1, Vr1 = I * R1 => 0.4545mA * 10K => 4.54V. and drop across R2, Vr2 = 0.45V.
+5V------^^^^^^^^--------^^^^^--------gnd
---------10K---------------------1K-----------
------- R1------------------------ R2 -----------

2. If POT is exactly at center, then it becomes, take a rough sheet and draw, split the POT into two, as

+5V------^^^^----^^^^--------^^^^^--------gnd
------------5K--------5K-------------1K-----------
------- --- R1----------------------- R2 -----------
Vr1 = 0.4545mA * 5K => 2.2727V
As in your figure, it shows you are not measuring w.r.t ground. But i think while you simulating, all the voltages are w.r.t ground.

3. When POT is moved fully to another side, drop is zero .
+5V------^^^^^^^^--------^^^^^--------gnd
-------------------10K---------------1K-----------
--------------- R1 ----------------- R2 -----------

Best wishes

Status
Not open for further replies.