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Full Wave Rectifiers
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- Опубликовано: 24 май 2018
- This electronics video provides a basic introduction into full wave rectifiers which converts a sine wave AC signal into a pulsating DC signal using two diodes, a load resistor, and a center tap transformer.
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Hantek Handheld Oscilloscope: amzn.to/3B5iu9d
Center Tapped Transformer: amzn.to/2UH0voK
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you have to divide the peak voltage by 2, as it is center tap.
In the question 120 Vac is Rms value, so when we multiply it with root 2 we get peek value not the Vdc value. In order to find Vdc=Vavarage we have to divide Vpeak value to pi, which is equal do Vdc value.
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On the center tap transformer , if I am not mistaken in a full wave rectifier the waveform is 180 degrees out of phase , which means the D1 and D2 turn on , on both positive swings of the sinewave resulting in 2 positive halves on the output.
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just realized the full and half wave formulas are listed below this one is solid
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Because the waveforms are 180 degrees out of phase we can have a hot and neutral line or return which is 180 degrees out of phase. That means the relationship between a V.O.M which measures peak to peak value is measuring in digital form or analog form from HOT line to NEUTRAL or Black line to white line we can so measure 120 VAC in phase, proved on an oscilloscope we will measure peak to peak value in phase as 60 +VP to 60 -VP right in phase. That relationship of the sinewave is on one line and to the return in a circuit. That shows the phase loop.
well explained
To my understanding also in electricity there has to be a load to draw current to it.
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I have a center tap transformer it was for microtek ups 1400. I want to use it for for battery charging and I don't want to disturb its design. The challenge is that the center is positive I wonder how the diagram of electifire should be away from the this one you have just explained.
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Love the video, it helps a lot. But 1 quick clarification: You said the peak voltage is 12*sqrt(2) = 16.97, but when calculating average voltage you use 16.27 (16.7 - 0.7) as the peak voltage. What is the new value (16.27)?
up
That's the output voltage. Vo,pk = Vs,pk - Vdiode = 16.97 - 0.7 = 16.27
peak voltage the resistor will deal with.
ya, quite confusing when calculating Vaverage you use (16.27).. y dont you use (16.97) instead? anyone explain please
@@nurhafizahaziz6785 We use the 16.97 V if we are solving wuth the "Ideal diode model" which ignores the parrier potential of the diode (0.7 V in silicon diodes).
But the correct way, or most common way is to replace the diode with a 0.7 V battery then adding or subtracting it from the original Vp.
How is Vp 120VAC? that the RMS value correct? should 120 be multiplied by 1.414 for Vpin?
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I'd like to see this proved in a lab. I am getting close to 15.5 vdc on the output. What value resistor are you using? Thanks.
What type of setup were you using? Were you working with a physical circuit, or with a simulation tool like Multisim?
@@PunmasterSTP Yes, a physical circuit.
@@ThomasACarlos Hmmm…if you have access to a simulation tool, it might be worthwhile to double-check using that. What components were you using? (i.e. load resistor, diodes, transformer, AC source, etc.)
@@PunmasterSTP Unfortunately, 1 year later, I do not have my notes. Sorry.
@@ThomasACarlos Oh it's quite alright, I was just curious. I like to reply to old comments just to see how people have been doing, and start some interesting conversations. How has school been going/how did it go?
when do we use full wave rectifier and half wave rectifier?
I think if you need to make DC current (such as for electronic circuits) but you start off with an AC current source (like the electricity from wall outlets).
What is the point of having a center tap transformer? I understand that the bridge rectifier allows for the full secondary voltage at the output vs this type of full-wave rectifier which reduces it to half, but I still cant figure out the purpose of dropping half the voltage before it reaches the output. Thanks for the videos.
I think that in a way, you don't necessarily "drop" more of the voltage with a (non-bridge) full-wave rectifier compared to a half-wave rectifier. In a half-wave rectifier, you can think of it as going from V = 0 to V = Vpeak, while in a directly center-tapped full-wave rectifier, you'd go from V = -Vpeak/2 to V = +Vpeak/2. The range is still the same (Vpeak, or the max voltage across the entire secondary coil), but instead of larger half-waves, you get smaller half-waves twice as frequently.
@@PunmasterSTP makes much more sense thanks!!
@@chrisreale4326 I’m glad my explanation could help! A lot of times I try to reply to things to lock in my understanding. I also find it interesting to reply to very old comments, to see how people have been doing and to start up some conversations.
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What about when both diodes are flipped to reverse Bias, I have a practice problem like that and Im so confused how it isn't an Open Circuit
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So why doesn't a center tap short circuit the inductor?
what must be the value of the resistor?
You need to know the is current this video for exp the full wave rectifier
Vr = 16.27 v
The transformation prob
Its given is or ip current
Your diodes are backwards. Your diagram shows the center tap as being always negative.
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that's cool but help us to see step by step how to measure it
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electronics are not much difficult........
I stopped watching 7 mins in, the explanation of the full wave rectifier with a center tapped transformer was way wrong
You also need 4 diodes to make any kind of dc rectifier
and that's not even getting to the missing capacitors and resistors to make this work
Thank you very much
Thank you sir