Nuts & Volts - May 2004

Q&A

above the corner. Thus, the resistive divider sees mainly the fundamental, while the capacitive divider sees the higher frequency harmonics.

To generate the perfect square wave, the fundamental and all harmonics must remain in the correct amplitude and phase. One simple adjustment of the probe’s 15 pF capacitor makes this a simple calibration.

The circuit you presented as Figure 7 on page 80 could be simply modified to generate a 1 kHz square wave by replacing the 1 MHz crystal with an RC combination.

These are fun things to do ... I really enjoy your articles!

Carl Baumgaertner Engineering Dept.

Harvey Mudd College

10X Probe

9M

15pF

Compensation Adjustment

Oscilloscope Input

20pF

1M

Figure 10

Typical Oscilloscope Probe

Dear TJ,

I believe you dropped a zero in answering the February 2004 question about the necessary filter capacitor size for a 5 V power supply desired to have 100 mV ripple at 1 A (“Power Supply Design 101”). The correct answer is 83,000 µF, not 8,300, as stated. Your formula is correct, but your computation is in error.

You can do this in your head if you remember that, for 1 V of ripple at 1 A current, you need 8,300 µF. 1 V x 1 A = 8,300 µF, which you can simply scale to other values. For example, 1 V ripple at 0.5 A is 8,300/2 or 4,150 µF. The constant 8,300 µF for 1 V x 1 A formula, of course, is only for full-wave rectification at 60 Hz.

volt of potential. A capacitor of C farads with V volts across its plates will contain Q coulombs of stored charge (Q = CV).

The energy stored in a capacitor is equal to the work done to charge it (U = 1/2 CV²). Where stored energy U is in joules. When the plates are physically separated, the stored energy is increased and so is the potential (voltage), but not the charge (C), which remains constant.

Where I tripped up was the energy of one electron — which equals . 62 x 10-19 coulombs — is the reciprocal of a coulomb charge. I’m sorry for the mix-up. NV

Jack Smith Clifton,VA

Dear TJ,

Your answer to the “Basic Electronics 101: Capacitance” question in the March 2004 issue says that the charge increases when you pull the plates apart. This is not true.

The voltage increases, but the number of extra protons on one plate and extra electrons on the other plate does not change. When you pull the plates apart, you have more potential energy, just like lifting a ball higher off the ground gives it more potential energy. The charge on the plates stays the same, as does the mass of the ball.

Viktors Berstis Austin, TX

Viktors,

I fell into the same trap that many people do — I unintentionally exchanged the term “potential” for “charge.” By definition, charge is a number of electrons measured in coulombs, where one coulomb (1C) equals 6. 24 quadrillion ( 6. 24 x 10¹⁸) electrons. Voltage — also called electromotive force — is a quantitative expression of the potential difference in charges between two points in an electrical field. One coulomb equals one