IT = CV is a very basic and useful relationship that all electronics students should master. It is akin to Ohm’s Law, but something I was never taught in tech school –I had to learn it myself much later –now I use it at least weekly. Every electronics instructor should get a hold on this one and make sure his students get this simple concept –perhaps some already teach this –if so, good for them.
The problem with tech schools
Excellent professors are few and far between and my tech school, even though highly rated, typified this shortcoming. The problem, I believe, is that many engineers who do not fare well in industry find a home in academia where they perpetuate mediocrity –and these are definitely not ‘circuit-heads.’ My hat goes off to those bright instructors who truly teach, inspire and bring out the best in their students, but I recall only a few that even came close –in my case, these excelled at conveying absolute truth and dispelling abundant electronic misconceptions.
The problem with coulombs
We have all touched upon this unit of electrical charge, but at best it is simply an obscure unit of measure –while we may understand the concept, we simply do no not think in coulombs. We think in volts, amps, ohms and seconds, but not coulombs –it is simply too intangible.
An important equality
Q = C * V Where Q = charge in coulombs, C = capacitance in farads, and V = volts.
…perhaps poorly defined as volt-farads…
Q = I * T Where Q = charge in coulombs, I = current in amps, and T = time in seconds.
…perhaps better defined as amp-seconds…
Therefore: IT = CV (Note that the unit of charge has disappeared.)
Solving for the four specific variables
I (in amps) = CV /T (here, it represents a fixed current flow –a current source)
T (in seconds) = CV /I (here, it represents an elapsed time of change in time)
C (in farads) = IT /V (here, it represents a fixed value)
V (in volts) = IT /C (here, it represents a change in voltage)
What is a voltage source?
Most understand this –it is a voltage that is essentially unchanged over a range of load currents.
Common examples include batteries or fixed voltage power supplies.
What is a current source?
A current source is a little more mysterious and less understood –it is a current flow that is essentially unchanged over a range of load voltages.
Examples include a current-limited power supply, constant current diode, a photo-voltaic solar panel, diode leakage, or the input current to an op amp integrator. (I, myself, use a current limited power supply to simulate solar panels.)
Most simply, it may be defined as the current flowing through a fixed resistor when a fixed voltage is placed across it.
Examples of how this is applied
1. If 1amp flows into a 2000uf capacitor for 2seconds, what is the voltage developed?
V = IT /C = 1A * 2sec /0.002Farad = 1000V
2. How much capacitance is required to limit its voltage change to 1V when the capacitor discharge current is 100mA and the elapsed time is 8.33mS?
C = IT /V = 0.1A * 0.00833sec / 1V = 0.000833farads = 833uf
3. If a leakage current charges a 0.47uf capacitor to 3V in 60sec, what is the leakage current?
I = CV /T = 0.47E-6farads * 3V /60sec = 2.35E-8A = 0.0235uA
4. If a charging current of 100uA is applied to a 0.1uf capacitor, how long does it take to charge to 8V?
T = CV /I = 0.1E-6 * 8V /100E-6A = 8E-3 = 8mS
The real world
Unfortunately, in the real world, currents are rarely constant. However, approximations are often quite accurate if the current does not vary more than roughly 2:1. In the following case, it is simply averaged.
Problem: Ignoring normal design rules, calculate the approximate frequency of this TLC555 astable oscillator, then compare with experimental measurements.
R1 = 100K ±1%, C1 = 0.1uf ±1%, Vcc = 12V
Vthesh low = 4V, Vthresh high = 8V
First, determine average capacitor charge /discharge current
Iavg = (12V — Vavg) /100K = 6V /100K = 60uA
Second, determine charge period between the two thresholds (4 to 8V)
T = CV /I = 0.1E-6uf * 4V /60E-6 = 6.66E-3 = 6.66mS
Since this is half of the full cycle period, the full cycle = 2 * 6.66mS = 13.33mS
F = 1 /period = 1 /13.33mS = 75HZ
Measured frequency = 71.84HZ
Accuracy = (71.84 – 75) / 75 = -0.042 = -4.2% (not bad for an approximation…)
Note that the math could be further manipulated to eliminate the voltage as its effect in this case cancels out.
Undocumented words and phrases (for our ESL friends)
circuit-head –idiom that I coined –someone who eats, drinks and sleeps circuits –akin to the more commonly used ‘motor-head’ idiom that aptly describes those who eat, drink and sleep engines and fast cars…