EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: eev_carl on September 07, 2020, 02:05:48 pm
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Hi,
I have this circuit wired up and the scope is giving me the correct half-sine wave shape. However, the voltages seem off to me. I'm pretty sure it's because of my misunderstanding of the ground on the oscilloscope. I'm using an F-44X (12.6VCT). The secondary center tap isn't connected to anything. I'm using the B lead as ground. Both channels' ground clips are plugged in there.
The yellow is the secondary voltage and the blue is the rectified voltage.
Thanks,
Carl
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What do you expect the voltages to be and why?
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Looks fine for a 12.6V secondary transformer with 15% regulation.
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What do you expect the voltages to be and why?
12.6 V because VCT is measured between the two taps. My scope shows 40Vpp.
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Sure looks wrong to me. Scope says it’s about double what it should be.
12.6 * 1.15 * 1.414= 20.48 peak to peak
Why the scope says 41.2 for the full sine wave and 20.4 for half wave leaves me scratching my head. Do you have a DMM to confirm the secondary voltage?
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I think those 12.6V is an AC RMS value. The peak
voltage under full load is 35.5Vpp (12.6*1.41*2).
Under no load it is higher.
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Why the scope says 41.2 for the full sine wave and 20.4 for half wave leaves me scratching my head. Do you have a DMM to confirm the secondary voltage?
Yes. My DMM is 14.68V. I figure the variance is due to the test circuit.
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No, your DMM shows the AC RMS value, not Vpp.
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Thanks. so that's 14.68 * 2^1/2 * 2 = 41V
From my original post why is the RMS value for the blue channel not 14.68 ?
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It is half wave rectified, i.e. current flows only half the time.
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You have built the circuit and connected the scope properly to demonstrate a half wave recitifier--no problems there.
As I think you've figured out so far, the peak to peak voltage of a full sine wave is about 2.818x the RMS value. Your transformer is rated at 12.6Vrms, the center tap is irrelevant because you've left it open. Since you have almost no load, the open circuit voltage will be higher, in this case 14.6Vrms. You haven't displayed the scope measurement of Vrms for the sine input, but it would likely be about 14Vrms or so. b/t/w, none of the theoretical values are going to match your scope perfectly here because there is significant distortion of the sine waves and your scope isn't all that accurate--scopes generally aren't. But the values will be in the ballpark.
The half wave will have a maximum voltage of about 0.6V less, due to the diode drop, and a Vp-p of half of the input minus the 0.6 volt for the diode drop, which is what your scope is showing. To talk about RMS values for anything other than a sine wave or a square wave, you'll need to do some math, or at least understand that math is needed. You are showing an RMS value for your rectified output of a little less than half the Vp-p value, in theory a perfect half wave would have a Vrms = Vp-p/2.
https://www.youtube.com/watch?v=A2SMI31EgMA (https://www.youtube.com/watch?v=A2SMI31EgMA)
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Thanks all! On to week 3. BJT posts OTW