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Mathematics for
Basic Electronics

Copyright©2007 by Daniel B. Sedory. All rights reserved.

 

 

In order to carry out many tasks as electronics technicians, we frequently use a core group of mathematical formulas involving power, voltage, current and the resistance, capacitance or inductance of various components, but you don't need to fully understand the function of each part within a circuit to make use of the math.

Two distinct types of current may be found in electronic circuits: Alternating (AC) and Direct (DC); and both may exist within the same circuit at the same time!
I
n this chapter, we'll deal with formulas for Direct Current only.

You should, of course, already be familiar with each type of current in its most common forms: Electricity from power companies in the US is supplied to most outlets as 120 volts AC, whereas the battery used to start a car should measure about +12 volts DC (when its engine is off). The computer you're using to read this page most likely has a power supply which converts 120 VAC into at least +5 and +12 VDC; unless you're running a laptop from batteries! And apart from plasma screens, you may be viewing it on a monitor which has a high voltage power supply producing anywhere from about 7,000 to 30,000 volts DC, if not more! The operating voltage for a particular CRT (Cathode Ray Tube) depends upon its design, size of the tube, etc.; generally, newer tube designs (and laws) require lower voltages. At one time we worked on projector displays with a supply that could make 75,000 volts or more; and we had to wear radiation badges! High voltages produce X-rays!

Because DC voltages are often supplied to circuits by first converting AC into DC, remnants of the original alternating current (sometimes called AC noise) are often large enough to measure in the DC circuits of cost effective supplies. This means it's quite possible for a defective PC power supply to produce so much noise, its motherboard will no longer boot up!


Ohm's Law
A Relationship between Current and Voltage

Functioning DC circuits always have current (measured in amperes1; simply called amps by most technicians), resistance (measured in ohms2; often symbolized by the Greek letter Omega: W) and voltage (measured in volts3) associated with them.

To be correct, Ohm's Law must be stated similar to this:

For a given temperature, when the current passing through a particular conductor (which obeys Ohm's Law; not all conductors do) remains directly proportional  to varying ranges of applied voltages, then the constant of proportionality is that specific conductor's resistance.

However, electronics technicians often think of Ohm's Law as the equation:

I R = E

(where I stands for current, R stands for resistance and E stands for Electromotive force or voltage; some texts use a V instead).

Since all electronic circuits produce heat, an easy way to remember this formula might be to think of the heat produced by a FIRE then drop the 'F' from the word, and place an equal sign between the R and E to complete the RElationship. And if you can't remember that, you're going to raise your teacher's ire!  When any two of the three variables in this equation have known values, we can solve for the unknown value using one of these three formulas:

1)   E = I R
2)   I = E / R
3)   R = E / I


Physics Corner:
Unlike most electronics technicians, physicists must account for every physical unit involved in their equations based on the SI units of meter (length), kilogram (mass), second (time) and ampere; yes, current measured in amps is a base unit of physics. It's fairly easy to think of an ampere* of current as a charge of one coulomb (or 6.25 x 1018 electrons) flowing through any cross section of a wire (or component) each second, but in reality, a Coulomb is defined as 1 ampere-second. The units for potential difference (or electromotive force) and resistance are then derived from the base units as follows:

Velocity (or speed) is in: meters/second (or: m/s).
Acceleration is in: meters per second per second (or: m/s2).
Force (mechanical) = mass times acceleration, and is in:
    kilogram-meters per second per second, or newtons (N) (or: kg·m/s2).
Energy, work or a quantity of heat is in:
    joules (J) or newton-meters (N·m) (or: kg·m2/s2).
Power is in: watts (W) or joules per second (J/s or N·m/s) (or: kg·m2/s3).
Electric potential is in: volts or watts per ampere (W/A) (or: kg·m2/A·s3).
Electric resistance is in: ohms or volts per ampere (V/A) (or: kg·m2/A2·s3).

Thus, if we were to solve for the potential difference (volts) across a 47 ohm resistor having a current of 100mA (100 milliamperes), the physical units would cancel-out as follows, using equation 1) from above: E = IR = 0.1A times 47ohms = 0.1A x 47(V/A) = 4.7V.
Technicians rarely take time to balance these physical units, since they've already been proven to work out for all their commonly used formulas (as we just verified for E = IR).

____________
*Operationally, one ampere is defined as the force of attraction between two parallel wires separated by 1 meter and carrying exactly the same amount of current (which will be 1 ampere), when that force is measured to be exactly 2 x 10-7 newton; that amount of force also ensures we have a Permeability constant (m0) of 4p x 10-7 henry per meter (or tesla-meter per ampere).

Example Problems:

1. What's the voltage drop across a 1kW (1000 ohms) resistor that has 23mA (milliamps; milli- means one millionth or 1 x 10-3) of current passing through it?
Answer:   E = IR = 1000 x 0.023 = 23 Vdc.

2. How much current is flowing through a circuit which has a total resistance of only 250W and is connected to +12 volt DC power supply (and ground of course)?
A:   I = E/R = (12 / 250) = 0.048 or 48mA (48 milliamps).

All electronic circuits have some resistance; including the circuitry inside a power source, where even the wires to external terminal connectors offer some (though negligible) resistance to the current passing through them. If the internal resistance becomes substantial (perhaps a shorted component), it may be the reason why a power source (apart from connecting too great a load) can't maintain its rated current and/or voltage. The common 12V vehicle battery or 1.5V cells normally fail to do what's required of them because they build up too great an internal resistance due to chemical oxidation; in simple terms: their potential energy (as a power source) has been depleted or used up!

3. What "load" (of resistance) is required to cause a +12 volt power supply to produce 800mA of current (assuming it has the capability to do so) in it?
A:   R = E/I = (12 / 0.8) = 15W (15 ohms).

This example question brings up another very important one:
"How do we know if that 15W resistor will be able to handle 0.8A of current without burning up?" The answer lies in learning how to determine the power ratings of your components; similar to the specs for a power supply which lists its maximum current output, resistors are packaged in various maximum wattages, which we'll discuss in our next section.

Power in DC Circuits

Anything producing heat, has some amount of Power associated with it (or used by it) to make that heat; such as the power wattages marked on light bulbs for 120VAC home outlets. As a matter of fact, Work is defined as any rate of Power for a period Time; such as Kilowatt-hours (1000 watt-hours). Thus, power companies initially charge us for the actual work (a sustained power level we used over a period of time), before adding all the fees and taxes, etc. We calculate the power in DC circuits (measured in watts), using the formula:

P = I E

(where P stands for the Power in watts, and I and E still stand for Current in amps and Electromotive force in volts). And if you can't see this formula is as easy as apple PIE to remember, perhaps you need a PI (private investigator) to find the equal sign for you!

When we combine this equation (P = IE) with our first two, substituting either E or I by their appropriate formulas (IR or E/R), the result gives us two additional power equations, making some tasks a bit easier:

4)   P = I E
5)   P = R
6)   P = / R

 

Power in a Horseless Carriage
If you live in the USA this may seem odd, but in Europe, engine power is often given in watts rather than horsepower! So a little 1 horsepower motor there, would be 745.7 watts (or 0.75 Kilowatt) and could produce the equivalent power of seven and a half 100W light bulbs burning in your home. And a 300-hp automobile engine would be listed as 223.7 Kilowatts; running at full power means it would be equivalent to 2,237 100W light bulbs glowing at rated voltage!

 

Example Problems:

1. The last question of our Examples set above, had a +5 volt power supply with a "load resistor" across it, which caused the supply to produce 1 amp of current. What was its power at that time?
A:   P = IE = 1 x 5 = 5 watts.

2. What's the actual power in a 250W resistor with 48mA of current flowing through it?
A:   P = I2 R = (0.048)2 x 250 = 0.576 watt (which means we better be using more than just a half-watt resistor, or it will be overheating and may burn up! A good rule of thumb for an ordinary carbon-comp resistor is to use a component with twice the power rating you're expecting it to produce).

3. What amount of power would be used by a 27W (ohm) resistor dropping 50 volts across it (assuming it can handle these conditions)?
A:   P = E2/R = (50)2 / 27 = (2500 / 27) = 92.6 watts.

4. How much current would be passing through that 27 ohm resistor above?
A:   Using equation 2), I = E / R = (50 / 27) = 1.85 amps.

5. What total values of resistance and power would be required to place a "test load" with a current output of 500mA across a +12 volt DC power supply?
A:   Using equation 3), R = E / I = (12 / 0.5) = 24W and from equation 4), P = I E = 0.5 x 12 = 6 watts.

NOTE: Since it's rather difficult to find resistors of calculated values, you'll often need to assemble "test loads" from a number of different resistors you happen to have on-hand (and of course, making sure to use power ratings that won't overheat any of those resistors). For instance, in the last example above, you may have to compromise your desired test current a bit by using two 10W and one 4.7W resistors in series, for a total of (using equation 2):
I = E/R = (12 / 24.7) = 486mA; instead of 500mA. And if its rated current output was 0.5A, expecting it to produce even 486mA just might be too much for a poorly designed supply, at least for some period of time and/or without more than desirable noise in its output. It's never a good idea to push things to their limits; unless you have no choice and can afford to replace them, or you're purposely causing "failures" (as UL engineers do before allowing an electronic device to be "UL Approved")!

6. Assuming we have a 1A rated supply in #5 above, what are the voltage drops across each resistor and what would be some acceptable wattages for each one? Note: The current through series resistors is always the same.
A: For our three series resistors, the current remains at: 486mA, but each resistor drops the total voltage level by some amount: The two 10W resistors will each drop about, E = IR = (0.486) (10) = 4.86 volts, or 9.72 volts total, leaving a drop of (12.0 - 9.72) = 2.28 volts across the 4.7W resistor, and we see that (0.486) times (4.7) also equals 2.28 volts. Using equation 4), its power would be roughly: (0.486 x 2.28) or 1.11W, so we could use at least a 2-watt resistor there, and each of the 10-ohm drops would produce: (0.486 x 4.86) or 2.36W. So 5-watt resistors might be a good choice for them, or perhaps 4-watts if we kept them cool, but if you decide (from experience) the type of resistors you have dissipate heat better than the old carbon or ceramic type, feel free to try only a 3-watt there. As we said above, doubling the required wattage of your resistors is only a general rule of thumb, not a precise formula! But the more heat a circuit's components retain (especially within a closed container), the hotter they may get and will influence the temperature of all the components around them; which brings us to a couple factors left unmentioned above about resistors: Temperature Coefficients and Tolerances.

Resistor Tolerances

Apart from the heat produced in each resistor and the ambient temperature within a whole device, both of which can affect its resistance (the graphite in carbon composition resistors has a slightly negative temperature coefficient, which means the resistance decreases when temperature increases), designers also need to factor in tolerances when deciding what type of resistors to use. Manufacturers make resistors in bulk, meaning they can't test all of them, so they assign a 5% or 10% (or even 20%) tolerance to their resistors; meaning its actual resistance may be +/- that percentage of its listed value. If the voltage and current of particular resistors are critical to a certain device (such as a volt/ohmmeter) its designer may specify components having only 1% tolerance must be used.

So, when it comes to figuring out the maximum voltage or current a particular resistor can handle, imagine if its 20% tolerance ends up on the very low side, and you don't take that into account; it could easily burn up! For example, if we have a 330 ohm resistor that needs to drop 24 volts across it, that will produce (equation 6), P=E2/R = (24x24)/330 = 1.75 watts of power; which you might think you can get away with by using a 2 watt resistor. But what if the actual resistance was only: 330 - (330 x .2) = 330 - 66 = 264 ohms? Now you'd have: P = 576/264 = 2.182 watts of power, and maybe smoke too! And the hotter it gets, its resistance may decrease even further due to its negative temperature coefficient! Both good reasons to design circuits with double the anticipated power ratings.

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1 Named after: André-Marie Ampère; French physicist (1775–1836).

2 Named after: Georg Simon Ohm; German physicist (1789-1854).

3 Named after: Alessandro Volta; Italian physicist (1745-1827).

 

These pages are still under construction !!!

 


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Last Update: 23 March 2007. (23.03.2007)