# What are some good circuits for learning about capacitors?

I'm looking to gain an in depth understanding of capacitors (timing constants, filtering, etc..) What are some good circuits to learn about capacitors? I've got all the equipment I think I need (oscilloscope, waveform generator, DMM, breadboards, etc)

• Hm, nothing is as useful to learn about basic linear components as the very basic theory behind linear networks, which you can very well learn from literally thousands of textbooks. Could you try to be more specific? – Marcus Müller Mar 22 '20 at 17:50
• – G36 Mar 22 '20 at 18:51
• @MarcusMüller I understand the theory, I work as a full time EE right now, but I don't have much experience with physically seeing the theory put to use – Sal M Mar 22 '20 at 23:41

I think the simplest are also the most informative.

Take a simple voltage divider built entirely of resistors: simulate this circuit – Schematic created using CircuitLab

The voltage ( $$\V_{Probe}\$$)at the point marked "Probe" is given by:

$$V_{Probe} = V_1 \times \frac{R_2}{R_1+R_2}$$

Compare with the two variants possible using a resistor and a capacitor: simulate this circuit

Voltage dividers work for impedances as well as for resistances.

The impedance of a capacitor is given by:

$$Z_C = \frac {1}{2 \pi f C}$$

Where:

• $$\C\$$ is the capacitance in farads
• $$\f\$$ is the frequency in hertz

You can put the impedance of a capacitor into the voltage divider formula, and calculate the voltage at "Probe" for various frequencies. Compare calculated values with values you measure using your signal generator and your oscilloscope.

Can you tell which arrangement is typically referred to as high pass filter, and which arrangement is known as a low pass filter?

• For a demonstration of time constant, try the second and third with a square wave input. Vary the frequency. – Mattman944 Mar 22 '20 at 20:18
• You left off a j from your impedance calculation. – Hearth Mar 22 '20 at 22:18
• I agree with @Hearth. Without the $\sqrt{-1}$ your formula for the impedance of the capacitor is actually the formula for the reactance instead. – Elliot Alderson Mar 22 '20 at 23:44
• @Hearth: Yes, strictly speaking the formula is for the reactance of the capacitor. Do you want to try to explain that "well, yeah, the formula says use $\sqrt{-1}$ but then ignore it and just use the part of the result without $j$?" I don't want to try to explain that to somebody just starting out. – JRE Mar 22 '20 at 23:54
• @JRE Sorry--I'd just spent several hours grading lab reports for a three-phase power systems class, so I was still in "make sure all the details are right" mode. – Hearth Mar 23 '20 at 0:44

Build an astable bipolar oscillator, using a +3 or +6 or +9 or +12 volt supply.

Examine the various waveforms. They will change, as VDD exceeds the reverse breakdown of emitter junction.

And, for the circuits suggested by "JRE", add one diode across the capacitor; explain the waveforms. DC_restoration, using one cap and one diode, is a useful trick to understand.

And for study, buy Millman & Taub "pulse digital and switching waveforms". It was a huge eye-opener for me, lo those many decades ago.

price as low as \$4.81