# Electronic formulas simplified – Calculations cheat sheet

Formulas and calculations in electronics may seem intimidating at first. You can actually accomplish a lot in electronics without doing much math by learning some shortcuts.

• Common LED and resistor limitations calculated
• Resistor voltage divider

## Common LED and resistor limitations calculated

• Most LEDs (specifically 3-5mm through hole) should have the current limited to no more than 20mA.
• Most through hole resistors are 1/4W (0.25) maximum but should be kept at 1/8W or lower. You can easily find higher wattage resistors, which will cost more and are larger.

Current through an LED needs to be limited to 20mA of current or less. The LED needs help doing this with a resistor. It’s plenty fine to used a lower current, the LED just won’t be as bright.

### 1KΩ resistor passes 1mA per volt across it

First we will see how much a 1000 ohm (1KΩ) resistor limits current, and how much heat (power) it generates. The 2 formulas we need for this is from ohms law for current = I = V/R and power = P = VI

• 3.3V/1,000Ω=0.0033A (3.3mA)             3.3V x 0.0033A = 0.011W (rounded)
• 5V/1,000Ω = 0.005A (5mA)                    5V x 0.005A = 0.025W
• 10V/1KΩ = 0.01A (10mA)                      10V x 0.01A = 0.1A (getting close to 1/8W, recommended limit for 1/4W resistor)
• 20V/1KΩ = 0.02A (20mA)                       20V x 0.02A = 0.4W (WAY TOO HOT for 1/4W!)

This gives us a few quick rules to remember as a starting point for picking a 1/4W resistor in any circuit.

• 1KΩ resistor gives us 1mA of current for each volt across it.
• 1KΩ resistor should not have much more than 10V across it.
• Doubling a voltage across a particular resistor, and thus the current, quadruples the wattage.

#### Forward biased LEDs block voltage

Now we also need to factor in when an LED is being protected by a resistor. LEDs (and other diodes) are almost certainly going to block a noticeable amount of voltage (Vdrop) from reaching the resistor. That means that the resistor actually sees less than the power supply voltage (Vcc) when it is connected in series with a diode/LED. Thus the current through a series resistor and LED is based on (Vcc – diode Vdrop)/R = I

• Forward biased red LEDs typically block about 2V (a bit less at low current)
• Forward biased green/blue LEDs typically block about 3V (a bit less at low current).
• Silicon based rectifier diodes typically block about 0.6-0.7V while forward biased.

• 5V-2V = 3V    …   3V/1000Ω = 0.003A (3mA) through red LED and 1KΩ resistor in series @ 5V supply.
• 10V-2V = 8V    …   8V/1000Ω = 0.008A (8mA) through red LED and 1KΩ resistor in series @ 10V supply.
• 5V-3V = 2V   …    2V/1,000Ω = 0.002A (2mA) through green LED and 1KΩ resistor in series @ 5V supply.
• 10V-3V = 7V   …    7V/1,000Ω = 0.007A (7mA) through green LED and 1KΩ resistor in series @10V supply.

Measuring voltage across individuals components is a vital skill for quick videos 5 by electronzap

##### Half the resistance = Twice the current

Lighting a red LED with a 5V supply and 1KΩ resistor means you will have about 3mA of current flowing through them. That’s not very bright for a red LED. We would like at least twice that current, or better yet, about 4 times that current to get fairly bright.

• 5V supply-2Vdrop = 3V … 3V/500Ω = 0.006A (6mA) … 0.006A x 3V = 0.018W
• 5V-2V = 3V … 3V/250Ω = 0.012A (12mA) … 3V x 0.012A = 0.036W
• 10V-2V = 8V … 8V/500Ω = 0.016A (16mA) … 8V x 0.016A = 0.128W (about the most we want to warm a resistor)
• 20V-2V = 18V … 18V/2,000Ω =  0.009A (9mA) … 18V x 0.009A =  0.162W a bit warmer than we like, better to go a bit higher in resistance.

So now we know, that with a 5V supply, it is probably best to have about 250Ω (220 is common and will work well) resistor protecting it. While at 10V, a 500Ω (470 and 510 are probably more common, and will work well) will get a little bit warmer than we’d like, but will provide a nice amount of current.

##### Resistor voltage divider

We noticed above (explained in text, and shown in videos) that about 2V of the total power supply voltage does not make it to a resistor when in series with a red LED. This is a property of electronics circuits, explained in Kirchoffs voltage law, where the total voltage always gets split up among series components. More detailed diagrams at 004 Voltage dividers-trimpot-LDR light dependent resistor-fixed resistor
Electronics course 4 Voltage dividers using fixed light dependent LDR and trimpot resistors

Vout = Vcc x R2/(R1+R2) for 2 series resistors voltage divider.

• More positive side resistor is usually labeled R1 (for 2 series resistors)
• More negative side resistor is usually labeled R2 (for 2 series resistors)
• Vout is the voltage where the 2 resistors are connected. It’s in relationship to ground (negative side of power supply).

Resistors do not block a certain amount of voltage like diodes do. Instead they divide up the total voltage (after subtracting semiconductor voltage drops, if there are any) based on their percentage of resistance to the other resistive components.

Power supply voltage minus any semiconductor drop(s) then divided between resistors proportional to resistance.

• 1/2 supply voltage is easy- 2 equal value resistors in series.
• 1/3 and/or 2/3 supply voltage is easy- 3 equal value resistors in series.
• If the wiper of a trimpot potentiometer voltage divider is 1/4th of the way towards the positive rail, then the trimpot will output 1/4 of the power supply voltage. Half way = half the power supply voltage. 3/4th of the way, then 3/4 of the power supply voltage. All the way to the positive rail, then the full power supply voltage. All the way to the negative rail, then it will output 0V.
• Remember that current taken from the output to power a load will drop the output voltage even more. If the voltage must be maintained, then use a power amplifier (transistor, op amp) wired as a voltage follower/buffer.

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