A Look at Ohm's Law

Two properties affect the voltage at VAD2: current and resistance, and Ohm’s Law explains how it works.  Ohm’s Law states that the voltage (V) across a resistor is equal to the current (I) passing through it multiplied by its resistance (R).  So, if you know two of these values, you can use the Ohm’s Law equation to calculate the third:

                       V = I x R

Voltage (V) is measured in units of volts, which are abbreviated with an upper-case V.  Current (I) is measured in amperes, or amps, which are abbreviated A. Resistance (R) is measured in ohms which is abbreviated with the Greek letter omega (Ω).  The current levels you are likely to see through this circuit are in milliamps (mA).  The lower-case m indicates that it’s a measurement of thousandths of amps.  Similarly, the lower-case k in kΩ indicates that the measurement is in thousands of ohms.

In some textbooks, you will see E = I × R instead. E stands for electric potential, which is another way to say “volts."
Let’s use Ohm’s Law to calculate VAD2 with the phototransistor, letting two different amounts of current flow through the circuit:

  • 0.92 mA, which might happen as a result of fairly bright light
  • 0.15 mA, which would happen with less bright light

The examples below show the conditions and their solutions.  When you try these calculations, remember that milli (m) is thousandths and kilo (k) is thousands when you substitute the numbers into Ohm’s Law.

Example 1: I = 0.92 mA and R = 2 kΩ

Equation 1: 0.92mA and 2 kOhm

Example 2: I = 0.25 mA and R = 2 kΩ

Equation 2: 0.25mA and 2 kOhm

Your Turn – Ohm’s Law and Resistor Adjustments

Let’s say that the ambient light in your room is twice as bright as the light that resulted in VA3 = 3.1 V for bright light and 0.5 V for shade.  Another situation that could cause higher current is if the ambient light is a stronger source of infrared.  In either case, the phototransistor could allow twice as much current to flow through the circuit, which could lead to measurement difficulties.

Question: What could you do to bring the circuit’s voltage response back down to 3.1 V for bright light and 0.5 V for dim?
Answer: Cut the resistor value in half; make it 1 kΩ instead of 2 kΩ.

  • Try repeating the Ohm’s Law calculations with R = 1 kΩ, and bright current I = 3.1 mA and dim current I = 0.5 mA. 

Does it bring VA3 back to 3.1 V for bright light and 0.5 V for dim light with twice the current?  (It should; if it didn’t for you, check your calculations.)