Rectifier Circuits

Thread Starter

Uisge

Joined Nov 3, 2009
1
I noticed a wording error in this e-book:

http://www.allaboutcircuits.com/vol_3/chpt_3/4.html

The line is:
Furthermore, the AC power source only supplies power to the load once every half-cycle, meaning that much of its capacity is unused.

In my opinion, the line should read:
Furthermore, the AC power source only supplies power to the load one half of every full cycle, meaning that half of its capacity is unused.

I know that the author meant what I have noted, but it wasn't quite stated correctly.

Cheers,
Uisge
 

Dcrunkilton

Joined Jul 31, 2004
422
I have updated the ibiblio copy for numerous corrections (not just for this thread) over the last few days. The list of affected files and images follows:
DC V1:dcmeter.sml contrib.sml 00502.png
AC V2:basicac.sml contrib.sml
Semi V3: bjt.sml diode.sml contrib.sml theory.sml 03288.png 03304.png 03392.png
Digital V4: boolean.sml contrib.sml
 

Unregistered

Joined Dec 31, 1969
0
Half-wave rectifier application: Two level lamp dimmer.

In the “Dim” switch position, the incandescent lamp receives approximately one-half the power it would normally receive operating on full-wave AC. Because the half-wave rectified power pulses far more rapidly than the filament has time to heat up and cool down, the lamp does not blink. Instead, its filament merely operates at a lesser temperature than normal, providing less light output. This principle of “pulsing” power rapidly to a slow-responding load device to control the electrical power sent to it is common in the world of industrial electronics. Since the controlling device (the diode, in this case) is either fully conducting or fully nonconducting at any given time,

--------it dissipates little heat energy while controlling load power, making this method of power control very energy-efficient. --------



This circuit is perhaps the crudest possible method of pulsing power to a load, but it suffices as a proof-of-concept application.

Note: depending on the load it may produce energy in the form of heat. Which would have to be dissipated with a heat sink especially if the wattage of the diode rating is exceeded.

The point being that with no load the diode dissipates very little significant heat. The voltage drop across the diode .6v if silicon would still apply using ohms law for watts.
 
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