Choosing a Current-Limiting Resistor

Choosing a Current-Limiting Resistor


Current-limiting resistors are placed in a circuit to ensure that the amount of current that flows does not exceed what the circuit can safely handle. When current flows through a resistor, there is, in accordance with Ohm's Law, a corresponding voltage drop across the resistor (Ohm's Law states that the voltage drop is the product of the current and the resistance: $V=IR$). The presence of this resistor reduces the amount of voltage that can appear across other components that are in series with the resistor (when components are “in series,” there is only one path for current to flow, and consequently the same amount of current flows through them; this is explained further in the information available via the link in the box to the right).

Here we are interested in determining the resistance for a current-limiting resistor placed in series with an LED. The resistor and LED are, in turn, attached to a 3.3V voltage supply. This is actually a rather complicated circuit because the LED is a nonlinear device: the relationship between the current through an LED and the voltage across the LED does not follow a simple formula. Thus, we will make various simplifying assumptions and approximations.

In theory, an ideal voltage supply will supply any amount of current necessary to try to maintain its terminals at whatever voltage it is supposed to supply. (In practice, however, a voltage supply can only supply a finite amount of current.) An illuminated LED will typically have a voltage drop of about 1.8V to 2.4V. To make things concrete, we'll assume a voltage drop of 2V. To maintain this amount of voltage across the LED typically requires approximately 15 mA to 20 mA of current. Once again for the sake of concreteness, we'll assume a current of 15 mA. If we directly attached the LED to the voltage supply, the voltage supply would try to establish a voltage of 3.3V across this LED. However, LEDs typically have a maximum forward voltage of about 3V. Attempting to establish a voltage higher than this across the LED is likely to destroy the LED and draw a great deal of current. Thus, this mismatch between what the voltage supply wants to produce and what the LED can handle can damage the LED or the voltage supply or both! We thus want to determine a resistance for a current-limiting resistor that will give us the appropriate voltage of approximately 2V across the LED and ensure the current through the LED is approximately 15 mA.

To sort things out, it helps to model our circuit with a schematic diagram, as shown in Fig. 1.

Figure 1. Schematic diagram of a circuit.

In Fig. 1 you can think of the 3.3V voltage source as the chipKIT™ board. Again, we generally assume an ideal voltage sources will supply any amount of current needed for the circuit, but the chipKIT™ board can only produce a finite amount of current. (The Uno32 reference manual says the maximum amount of current an individual digital pin can produce is 18 mA, i.e., 0.0018 A.) To ensure the LED has a 2V voltage drop, we need to determine the appropriate voltage across the resistor, which we'll call $V_R$. One way to do this is to determine the voltage of each wire. The wires between components are sometimes called nodes. One thing to keep in mind is that a wire has the same voltage across its entire length. By determining the voltage of the wires, we can take the difference in voltage from one wire to the next and find the voltage drop across a component or across a group of components.

It's convenient to start by assuming the negative side of the voltage supply is at a potential of 0V. This, in turn, makes its corresponding node (i.e., the wire attached to the negative side of the voltage supply) 0V, as shown in Fig. 2. When we analyze a circuit, we are free to assign a signal ground voltage of 0V to one point in the circuit. All other voltages are then relative to that reference point. (Because voltage is a relative measure, between two points, it typically doesn't matter what point in the circuit we assign a value of 0V. Our analysis will always yield the same currents and the same voltage drops across the components. Nevertheless, it is common practice to assign the negative terminal of a voltage supply a value of 0V.) Given that the negative terminal of the voltage supply is at 0V, and given that we are considering a 3.3V supply, the positive terminal must be at a voltage of 3.3V (as is the wire/node attached to it). Given that we desire a voltage drop of 2V across the LED and given that the bottom of the LED is at 0V, the top of the LED must be at 2V (as is any wire attached to it).

Figure 2. Schematic showing node voltages.

With the node voltages labeled as shown in Fig. 2, we can now determine the voltage drop across the resistor as we will do in a moment. First, we want to point out that in practice one often writes the voltage drop associated with a component directly next to a component. So, for example, we write 3.3V next to the voltage source knowing that it is a 3.3V source. For the LED, since we are assuming a 2V voltage drop, we can simply write that next to the LED (as shown in Fig. 2). In general, given the voltage that exists on one side of an element and given the voltage drop across that element, we can always determine the voltage on the other side of the element. Conversely, if we know the voltage to either side of an element, we then know the voltage drop across that element (or we can calculate it simply by taking the difference of the voltages to either side).

Because we know the potential of the wires to either side of the resistor (Wire1 and Wire3), we can solve for the voltage drop across it, $V_R$:

\[VR = \left( {Wire1{\rm{ }}Voltage} \right) - \left( {Wire3{\rm{ }}Voltage} \right).\]

Plugging in the known values, we obtain:

\[VR = 3.3V - 2.0V = 1.3V.\]

Having calculated the voltage drop across the resistor, we can use Ohm's Law to relate the resistor's resistance to the voltage. Ohm's Law tells us $1.3\mbox{V} = IR$. In this equation, there appear to be two unknowns, the current $I$ and the resistance $R$. At first it might appear that we can make $I$ and $R$ any values provided their product is 1.3V. However, as mentioned above, a typical LED might require (or “draw”) a current of approximately 15 mA when it has a voltage across it of 2V. So, assuming $I$ is 15 mA and solving for $R$, we obtain

\[R = V/I = 1.3V/0.015A = 86.67\Omega .\]

In practice, it can be difficult to obtain a resistor with a resistance of precisely 86.67 Ω. One could, perhaps use a variable resistor and adjust its resistance to this value, but that would be a somewhat expensive solution. Instead, it often suffices to have a resistance that is about right. You should find that a resistance on the order of one- to two-hundred ohms works reasonably well (meaning we ensure that the LED is not drawing too much current and yet the current-limiting resistor isn't so large that it prevents the LED from illuminating). In these projects we will typically use a current-limiting resistor of 220 Ω.

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