How to Choose a Resistor for Your Design

Take a look at a circuit board and chances are you’re going to find a resistor or two. Most boards today use surface-mount device (SMD) technology, so the components are almost too small to see sometimes, but they are on there, I promise. How do engineers decide which resistors to use in the design? Sometimes it depends on how you want that portion of the circuit to perform, as in the case of an op-amp. Other times it’s to prevent too much current from passing through a given point in a circuit, which is why they are often called current-limiting resistors. Maybe you want a simple way to divide the voltage or current. The reality is that there are numerous ways to use resistors, and oftentimes, the defining the resistor value is up to you.

A pile of carbon-film resistors. Image from
A pile of carbon-film resistors. Image from Sparkfun.

Resistors are aptly named. They resist the flow of current from one place to another. As long as the current has to pass through that resistor, we can control the amount of current passing by with a very high degree of precision. This is the basic application of a resistor and the foundation of Ohm’s Law, which states that the current (I) passing through a component with impedance (resistance) is equivalent to the voltage drop across that component (V) divided by it’s impedance [(R), i.e. I = V/R]. (Here’s a neat interactive site on Ohm’s Law) This can be rearranged to determine the value of any of the three as long as two others are known. Joule’s Law is related to it, and it is used to determine the amount of energy (J) released by a conductor (like a resistor) with a current (I) passing through it for a given length of time, i.e., J = I² * R * t. Divide J by t and you get the amount of power consumed by that device in Joules/second, or Watts (W). (Thought exercise — if an incandescent light bulb consumes 100W, how much resistance does the filament have? Through some rearranging and substituting, we get W/V = I. Then V/I = R. 100W/120V = 0.8333 A. Then 120V/0.8333A = 144Ω.)


To start, let’s look at a bi-polar junction transistor, or BJT. Transistors in general are usually three-terminal components that act like electronic switches. When you apply a small current to the base pin of a BJT, the “switch” is closed and a much larger current is allowed to flow between the other two pins. Only a small amount of current is needed at the base pin and anything more is just wasted, so we can use a relatively high resistor. Most circuits operate in the 3.3-12V range, and 5-15mA of current isn’t out of the question to get a BJT to operate, so if we apply a resistor somewhere in the 1kΩ-10kΩ range, we should be good. I would design the circuit with a 10kΩ resistor first off without even thinking about it, simulate it with Multisim to see exactly what was going on, and then make any changes as needed from there. This is especially useful if your current source is a microcontroller, which are very sensitive to high current drain (>40mA) on the output pins. (For more on BJTs, click here to view my Instructable.)


Another common place to find resistors is in series with diodes, specifically LEDs. LEDs are very particular when it comes to how much maximum current can be applied. When you get an LED, you will want to know three things: (1) the forward voltage or voltage drop, (2) the forward current, and (3) the source voltage. (1) and 2) come from the manufacturer, so be sure to find that data. (Some suppliers will list the data right on the product page, but you can always find it in the datasheet.) The source voltage is up to you and is determined by the needs of your circuit, or in my case whichever battery I have laying around. Looking at the datasheet for a typical RGB LED, we see that the forward current is 20mA for each color. We also see that the forward voltage is 2.0V for red and 3.2V for both green and blue. (Note — these values are pretty typical for single color LEDs of the same color.) We know that to use the blue portion of the LED, we need at least 3.2V, so let’s use 6V for fun. The forward voltage value tells us that the LED will “consume” 3.2V of the 6.0V available, leaving 2.8V for us to do something with. If we don’t, the LED will burn up. That’s where the resistor comes in, but let’s also look at the forward current needed for the blue LED. We only need 20mA. If only there was a way to relate current and voltage to find resistance. I know, Ohm’s Law! With some rearranging, we get R = V/I, so 2.8V/20mA =  140Ω. Now, that is a “typical” value for this resistor, but in reality nearly all LEDs will work just fine within range of current values, and by extension resistor values. Also, 140Ω isn’t a common value for a resistor. If we use a 150Ω resistor, we get 18.67mA passing through the LED, which is perfect. However, my experience has been that LEDs will work just fine with as little as 5mA, in which case we would need a 560Ω resistor. The only difference is that the LED will shine dimmer. (We can use this knowledge to build a dimmer switch for our LEDs!)


Voltage and current dividers are crude but effective ways to cut a higher supply voltage or current to a required lower value. (You could easily use a voltage divider to drop a large supply voltage (>9V) for an LED so that the resistor needed doesn’t have to be a high-power resistor.) Voltage and current dividers require nothing more than a simple calculation, multiplying your supply current or voltage by a ratio of two or more resistors to determine the output. (For a good in-depth tutorial on divider circuits, see the section at


Operational amplifiers, or op-amps, use resistors to determine the gain, or increase, of the circuit. Again, it is nothing more than a simple ratio of two resistors. For filters using op-amps, the resistor acts as a current flow controller. You can build any active op-amp filter with resistors and capacitors, and there is a time constant related to the chosen values of R, C and the cutoff frequency f, specifically ∝ 1/RC. As you need a faster cutoff frequency, your needed value for R will decrease. (For more on filters, has a really good section. For more on op-amps, click here for my Instructables.)


The 555 timer is one of the oldest and most widely used ICs on the market. Set up as an astable multivibrator, or oscillator, the output is a square waveform and the duty cycle and period are fully adjustable according to the value you choose for your resistors and capacitors. For example, a large resistor will slow down the rate at which the capacitor discharges, increasing the duty cycle and/or period. (For a basic tutorial on the 555 timer, click here for my Instructable.)


As mentioned, there are too many uses for resistors to cover them all here. Hopefully this has has clarified the question, if only so far as to say that it really depends on what you want the resistor to do. Unfortunately, that doesn’t answer the question very clearly, but much of what I do comes from experience, so start with some established circuits that you know work, and then start playing around, especially with circuits that are tolerant of a wide range of resistor values, like the op-amp and 555 timer circuits. Getting a good simulator like Multisim will greatly benefit you as well since you can change things simply and easily and then monitor every possible output parameter in real time.


Now get out there and make something!


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