Hi all,
after lots of reading, Googling and ending up getting the same results over and over again, I decided to take a shot and ask the question here.
I'm trying to wrap my head around how a rectified waveform is actually the sum of an AC and DC component. I'm taking an example of a half wave rectifier and I'm talking about the unsmoothed ripple here, so no filter capacitor included.
Let's say I have a maximum secondary voltage Vp2 of 50V, applied to a load resistance of 800 ohms. The resulting maximum current Im is 50V/800, or 62.5 mA.
Idc then equals 62.5/pi (half-wave rectifier) or 19.89 mA. Irms equals 62.5/2 = 31.25 mA.
Now, a rectified current (or voltage) is actually the sum of an an AC current + DC current. Let's say I visualise the above example (see picture). If I understand it correctly, the peak-peak value of the AC component is equal to the peak-value of the rectified wave (62.5 mA in this example). The minimum value of the AC component however is (in this example) -19.89 mA, so if you add the 19.89 mA DC-component to the AC-component, you get a nice rectified result (blue waveform).
The way I understand it and cross-checked with examples and books (Principles Of Electronics among one of them), this drawing makes sense.
However, there's also the maths. As stated in various sources and books:
Iac, being the rms value of the ripple current.
Now if I make that calculation in my example (sqrt of 31.25² - 19.89²), I get an Iac(rms) of 24.10 mA (as extra proof I calculate the ripple factor Iac(rms)/Idc = 24.10/19.89 = 1.21, which is ripple factor for a half wave rectifier, so this adds up).
However, when I try to calculate the Iac(peak) value out of the rms, I get 24.10 mA x 2 (half wave rectifier), which give me 48.20 mA.
Which is different from the peak value I got when I made the drawing (should be the same
Now I know I'm doing something wrong and since the math works out, it probably is the drawing, but if that is not correct, can someone explain me how exactly the rectified output is a combination of AC and DC? Because when I draw it this way (which somehow seems to make sense , the math doesn't work out?
Thanks!
after lots of reading, Googling and ending up getting the same results over and over again, I decided to take a shot and ask the question here.
I'm trying to wrap my head around how a rectified waveform is actually the sum of an AC and DC component. I'm taking an example of a half wave rectifier and I'm talking about the unsmoothed ripple here, so no filter capacitor included.
Let's say I have a maximum secondary voltage Vp2 of 50V, applied to a load resistance of 800 ohms. The resulting maximum current Im is 50V/800, or 62.5 mA.
Idc then equals 62.5/pi (half-wave rectifier) or 19.89 mA. Irms equals 62.5/2 = 31.25 mA.
Now, a rectified current (or voltage) is actually the sum of an an AC current + DC current. Let's say I visualise the above example (see picture). If I understand it correctly, the peak-peak value of the AC component is equal to the peak-value of the rectified wave (62.5 mA in this example). The minimum value of the AC component however is (in this example) -19.89 mA, so if you add the 19.89 mA DC-component to the AC-component, you get a nice rectified result (blue waveform).
The way I understand it and cross-checked with examples and books (Principles Of Electronics among one of them), this drawing makes sense.
However, there's also the maths. As stated in various sources and books:
Iac, being the rms value of the ripple current.
Now if I make that calculation in my example (sqrt of 31.25² - 19.89²), I get an Iac(rms) of 24.10 mA (as extra proof I calculate the ripple factor Iac(rms)/Idc = 24.10/19.89 = 1.21, which is ripple factor for a half wave rectifier, so this adds up).
However, when I try to calculate the Iac(peak) value out of the rms, I get 24.10 mA x 2 (half wave rectifier), which give me 48.20 mA.
Which is different from the peak value I got when I made the drawing (should be the same
Now I know I'm doing something wrong and since the math works out, it probably is the drawing, but if that is not correct, can someone explain me how exactly the rectified output is a combination of AC and DC? Because when I draw it this way (which somehow seems to make sense , the math doesn't work out?
Thanks!