why a 4 diode circuit contains higher frequency than in 2 diode circuit?

Question

hi, i was just wondering if you could help me in answering the question why a circuit with 4 diode has a higher frequency than a 2 diode circuit..

thanks for that info. it gave me an idea, but it still did not answer my question.

i'm talking about a full wave rectifier. the one with 4 diodes is usually called bridged rectifier. the other one is a full wave rectifier with 2 diodes.

in terms of frequency, the question is, why a 4 diode rectifier has higher frequency than that with a 2 diode rectifier.

or is it really true that a 4 diode rectifier has higher frequency than that with a 2 diode rectifier?

thanks for that info. it gave me an idea, but it still did not answer my question.

i'm talking about a full wave rectifier. the one with 4 diodes is usually called bridged rectifier. the other one is a full wave rectifier with 2 diodes.

in terms of frequency, the question is, why a 4 diode rectifier has higher frequency than that with a 2 diode rectifier.

or is it really true that a 4 diode rectifier has higher frequency than that with a 2 diode rectifier?


i do appreciate science, except when i find a hard time understanding some concepts. ^-^

Answer 1

Think of an AC waveform at 60Hz...
If you only use one diode (or two), you will get only the "positive" half of the wave. Everything below will be lost. This is still 60Hz (Hump and flat, hump and flat). Now if you use a 4 diode system, you will get the positive and the recitified neg half. This now looks like it is at a freq of 120Hz because you get twice as much (hump and hump, Hump and hump). This is typically called the 120Hz ripple, because you would generally filter it with a large capacitor to get a DC form.

Answer 2

A 2-diode, full wave rectifier with a center tapped transformer has the same output voltage waveform as a 4-diode, full-wave bridge rectifier. Both have the same ripple frequency, twice the frequency of the AC input. The ripple frequency is twice the input frequency because one half cycle is "turned over," so that there are two output peaks per cycle of input.

Answer 3

You can tell immediately that the diode will conduct, just by looking at the schematic, because the 4.7kohm leg that the diode is on will have more current due to less resistance, so the diode will be 'biased on'. You are definitely on the right track. Current through a diode varies with the applied voltage, which will change the diode's PN junction resistance, and current is absolute throughout the entire circuit. So, in order to find the total current, you have to know the total resistance and that diode has some resistance, which will affect your total current. Unfortunately, I can't remember how to find the diode's resistance. Let me do some research, and I will get back with you. :-(

Answer 4

Full-wave, 4-diode bridge rectifier will have a ripple freq.of twice the input. So will a full-wave 2-diode rectifier, as used on a center-tapped secondary. Only a half-wave, 1-diode rectifier has ripple the sme frequency as the input.

Bank on it.

Answer 5

Draw a sine wave, cover everything below the 0 line, and observe the result. You see a peak, space, peak, space, peak, space.... This is half wave rectification.

Now take the peaks from below the 0 line and draw them as a mirror image above the line so that were one peak ends at the zero line the next begins. There are no spaces, only a continuous stream of peaks. This is full wave rectification.

Frequency in Hertz is the number of times the signal path repeats itself in one second. A 60 Hz signal comes out of a half wave rectifier at 60 Hz (peak and space are one wavelength). A 60 Hz signal comes out of a full wave rectifier at 120 Hz (just peak no space).

Answer 6

Look at and compare the waveforms from a half wave, and full wave, rectifier. A picture is worth a thousand words:

Answer 7

are you asking about a full wave rect or a bridge? the freq also will be affected by the freq characteristics of the diode..example: 1N34 and 1N21 are both detector diodes. but vast differences in freq charateristics. the 1N34 is fine for AM radio detectors (100KHz-30 MHz) then as the RF carrier freq goes up from HF to SHF, etc the parameter requirements change (radar,ECM,etc) your homework is to research the "Esaki diode". report is due Monday morning.