In Canada I was taught... AC is like sticking your paddle in the water and stirring it back and forth... There's motion, but you won't go anywhere. Half wave rectification is like paddling a canoe, 50% in the water, 50% out. Full wave rectification is a pedal boat.
That's very interesting. I had to run it on an Excel spreadsheet to prove it to myself. A halfwave of 119V is indeed 84V RMS. A raw average is only 53. The math conflicts with the intuitive understanding. I have to think about this some more.
Rectified vs. rms AC are quite different. For the AC thru the variac the peak voltage will be 1.41 * 65 volts (we’re stepping it down) while the rectified AC will be 1.41 * 110 volts (simply clipping part of the mains AC). Filaments are quite nonlinear so the larger voltage differential wins out.
Man, thanks so much John, these videos are so helpfull, been working in electrics and solar for a long time and was not aware of the ac plus dc feature, cheers, all the best. And a happy new year!!!!
John, I just wanted to tell you that I've been following your videos for many years. Thanks for all the good content! I really like your style. And I think you should, if you want to, make some really "basics" videos.
Interesting, I tried this myself and as expected the bulb is dimmer with the diode, are you using a Variac to make the half-wave Vpeak equal the line voltage Vpp? Also if there's a DC component I would expect a capacitor after the diode
Hi John, I use those ready-made LM386 prints quite a lot, but I notice that this is not the best! there is quite a bit of sensitivity missing and that is what I never hear from you? I put a BC in front of it and it is immediately a lot louder, I can imagine that this will also be the case with that TDA. maybe a challenge for you? ( I always do it simple BC109C 2 resistors 4K7, 470K, 2 capacitors 4u7...0,1uF.... ) give your view on/about that? ( I often use this for small / simple DIY FM receivers )
It's much easier to understand if you picture a square wave of +/- 120V. Square waves are effectively DC that swaps polarity. Peak, RMS, and average (in absolute value) is all the same (120V). Rectifying it creates a 120V pulse train of 50% duty cycle. It's half the _power_. But half the power isn't half the voltage (in RMS), but 1/SQRT(2) of the original voltage. (The square of the voltage is half the square of the original voltage) In this case, rectified 120V becomes 84.9V. If you want to do the RMS calculation, it's SQRT( (120^2 + 0^2)/2). The square root of the average of the voltage squared, or Root Mean, Squared. The real eye opener is how a "true RMS" meter behaves. It's not measuring the "power" in the wave form (which is the whole point of RMS measurements), but it's passing the signal though a DC blocking capacitor first. A 120V pulse train of 50% duty cycle becomes a 60V square wave, or 60V RMS. (Still 120V peak-to-peak) Only "AC+DC" mode on the meter gives you the correct answer about the power dissipation you'd get if you put the signal into a resistor.
Looking at the dc reading... VDC = Rectified VAC peak * 0.636, where 0.636 is the AC average after rectification without a smooting cap. Therefore VDC = 120 * 0.707 * 0.636 = 54VDC (without a smooting capacitor) Also, the ac reading is incorrect on the dmm because it's trying to read the true rms, but the voltage after the diode is actually dc.
Hi, how about in our country we are using line to line unlike yours using line to ground. It is applicable? Or we need to series a diode in both lines? Hope this will be answered. Thanks
So, you are feeding one bulb with a stepped down 65VAC and the other with half-wave of 120VAC, sure the last one is brighter. At first I thought the only difference is diode in series. It just division by 2 to calc the half-wave RMS voltage, but you have to divide Vpeak-peak that is 120VAC(rms) x √2. So, 120√2 ÷ 2 - 0.7 ≈ 84.15
Unfortunately John most of us we have "simple" meters, or older "classic" FLUKE (like my trusty 77) NOT ABLE to measure DC+AC like that 121GW EEVBlog or higher level and cost BRIMEN meters
In Canada... AC is like sticking your paddle in the water and stirring it back and forth... There's motion, but you won't go anywhere. Half wave rectification is paddling. 50% in the water, 50% out. Full wave rectification is a pedal (paddle) boat.
I wonder if this would work on a old school soldering iron. Maybe you could try that with a temp probe to show the same iron get hotter when you put the diode in line vs. when you bypass it with a switch? I guess im just not smart enough to wrap my head around to taking half the waveform of voltage away and having a brighter bulb . Maybe you could go more into depth of this phenomenon and help us feeble-minded people be able to understand this😅. Maybe put a current meter in line and show d.c. and or a.c. current increase using the same bulb ,with and without the diode in series.
@@JohnAudioTech so when you put it in low mode it really makes the iron hotter just like the lightbulb gets brighter with the diode inline ...Right???? 😆 .Maybe i better watch again and listen closer about the variac situation and light bulb wattage rating cause its 🐠
@@jimhoward1655 No. It bypasses diode in high mode, letting the full AC wave through. Some people misunderstand the video. The video was comparing half the line voltage with half wave rectified line voltage.
@@JohnAudioTech thank you! i didnt catch that fact, i feel much better. Makes sense now,peak would be smaller so .707 of lower peak would be less and dimmer bulb 💡
The cap or the rectifier in the bulb on the left has failed. I'd be curious to see a thermal comparison of the bases as well as how much current each draws (and perhaps even how they behave when supplied with DC)
The Bulb-only circuit is being fed with 65VRMS. The bulb & diode circuit is being fed with 120VRMS. After the diode, the bulb is still being fed with 85VRMS, because the peak of 120VRMS would be 170V and half-wave RMS = Vpeak/2 = 170/2 = 85. Applying a higher voltage across the same load would result in higher current and therefore higher power.
@@JohnAudioTech no I'm not, just thought you are connecting them to the same voltage and saying the one with the diode is brighter. Brightness is power, so I meant same voltage half-wave rectified can't deliver more power (brightness) than full wave. In fact your video is confusing as you digress too much and don't specify at the beginning the conditions of measurement, voltages,...etc.
Yeah, no. Incandescent light bulb's brightness is dependent on the AVERAGE power the filament dissipates. This is neither RMS or peak voltage and so a DVM will mislead you.
In Canada I was taught...
AC is like sticking your paddle in the water and stirring it back and forth...
There's motion, but you won't go anywhere.
Half wave rectification is like paddling a canoe, 50% in the water, 50% out.
Full wave rectification is a pedal boat.
Bro you are killing it on videos lately
That's very interesting. I had to run it on an Excel spreadsheet to prove it to myself. A halfwave of 119V is indeed 84V RMS. A raw average is only 53. The math conflicts with the intuitive understanding. I have to think about this some more.
"I have to think about this some more."
As you absorb this kind of knowledge it "becomes" intuitive.
@@CraftAero I can't just file it away as a "fact". I have to understand it. That's what's eluding me.
Rectified vs. rms AC are quite different. For the AC thru the variac the peak voltage will be 1.41 * 65 volts (we’re stepping it down) while the rectified AC will be 1.41 * 110 volts (simply clipping part of the mains AC). Filaments are quite nonlinear so the larger voltage differential wins out.
Man, thanks so much John, these videos are so helpfull, been working in electrics and solar for a long time and was not aware of the ac plus dc feature, cheers, all the best. And a happy new year!!!!
John, I just wanted to tell you that I've been following your videos for many years. Thanks for all the good content! I really like your style. And I think you should, if you want to, make some really "basics" videos.
Very interesting John!
Fascinating, nice new year teaser !!
Measure the current 😜
Interesting, I tried this myself and as expected the bulb is dimmer with the diode, are you using a Variac to make the half-wave Vpeak equal the line voltage Vpp?
Also if there's a DC component I would expect a capacitor after the diode
Good question, I will try it with 230 volts. Are these real light bulbs or are they energy-saving bulbs with built-in electronics?
Incandescent bulbs. I should have clarified that.
I often wonder why some are such dim bulbs.
Hi John, I use those ready-made LM386 prints quite a lot, but I notice that this is not the best! there is quite a bit of sensitivity missing and that is what I never hear from you?
I put a BC in front of it and it is immediately a lot louder, I can imagine that this will also be the case with that TDA. maybe a challenge for you?
( I always do it simple BC109C 2 resistors 4K7, 470K, 2 capacitors 4u7...0,1uF.... ) give your view on/about that? ( I often use this for small / simple DIY FM receivers )
this is the moment when the scope takes over
It's much easier to understand if you picture a square wave of +/- 120V. Square waves are effectively DC that swaps polarity. Peak, RMS, and average (in absolute value) is all the same (120V). Rectifying it creates a 120V pulse train of 50% duty cycle. It's half the _power_. But half the power isn't half the voltage (in RMS), but 1/SQRT(2) of the original voltage. (The square of the voltage is half the square of the original voltage) In this case, rectified 120V becomes 84.9V. If you want to do the RMS calculation, it's SQRT( (120^2 + 0^2)/2). The square root of the average of the voltage squared, or Root Mean, Squared.
The real eye opener is how a "true RMS" meter behaves. It's not measuring the "power" in the wave form (which is the whole point of RMS measurements), but it's passing the signal though a DC blocking capacitor first. A 120V pulse train of 50% duty cycle becomes a 60V square wave, or 60V RMS. (Still 120V peak-to-peak) Only "AC+DC" mode on the meter gives you the correct answer about the power dissipation you'd get if you put the signal into a resistor.
Why don't you answer comments anymore?
It does take a lot of time and he is putting out more videos😊
Looking at the dc reading...
VDC = Rectified VAC peak * 0.636, where 0.636 is the AC average after rectification without a smooting cap.
Therefore VDC = 120 * 0.707 * 0.636 = 54VDC (without a smooting capacitor)
Also, the ac reading is incorrect on the dmm because it's trying to read the true rms, but the voltage after the diode is actually dc.
Hi, how about in our country we are using line to line unlike yours using line to ground. It is applicable? Or we need to series a diode in both lines?
Hope this will be answered. Thanks
Voltage across the load is what matters.
@JohnAudioTech Thanks
So, you are feeding one bulb with a stepped down 65VAC and the other with half-wave of 120VAC, sure the last one is brighter. At first I thought the only difference is diode in series.
It just division by 2 to calc the half-wave RMS voltage, but you have to divide Vpeak-peak that is 120VAC(rms) x √2.
So, 120√2 ÷ 2 - 0.7 ≈ 84.15
Unfortunately John most of us we have "simple" meters, or older "classic" FLUKE (like my trusty 77) NOT ABLE to measure DC+AC like that 121GW EEVBlog or higher level and cost BRIMEN meters
V(rms) = sqrt((V(ac)(rms))^2 + (V(dc))^2)
V(full wave) = sqrt(65^2 + 0) = 65v
V(half wave) = sqrt(65^2 + 50^2)
= 82v
In Canada...
AC is like sticking your paddle in the water and stirring it back and forth...
There's motion, but you won't go anywhere.
Half wave rectification is paddling. 50% in the water, 50% out.
Full wave rectification is a pedal (paddle) boat.
I wonder if this would work on a old school soldering iron. Maybe you could try that with a temp probe to show the same iron get hotter when you put the diode in line vs. when you bypass it with a switch?
I guess im just not smart enough to wrap my head around to taking half the waveform of voltage away and having a brighter bulb . Maybe you could go more into depth of this phenomenon and help us feeble-minded people be able to understand this😅. Maybe put a current meter in line and show d.c. and or a.c. current increase using the same bulb ,with and without the diode in series.
I have a soldering iron that has a hi/low switch that put the diode in series in low mode for half the power.
@@JohnAudioTech so when you put it in low mode it really makes the iron hotter just like the lightbulb gets brighter with the diode inline ...Right???? 😆 .Maybe i better watch again and listen closer about the variac situation and light bulb wattage rating cause its 🐠
@@jimhoward1655 No. It bypasses diode in high mode, letting the full AC wave through. Some people misunderstand the video. The video was comparing half the line voltage with half wave rectified line voltage.
@@JohnAudioTech thank you! i didnt catch that fact, i feel much better. Makes sense now,peak would be smaller so .707 of lower peak would be less and dimmer bulb 💡
The cap or the rectifier in the bulb on the left has failed. I'd be curious to see a thermal comparison of the bases as well as how much current each draws (and perhaps even how they behave when supplied with DC)
These are incandescent bulbs.
@@JohnAudioTech Difference in the filament's behaviour then? A thermal image and a resistance measurement across the base?
Just wrong usage of instruments, some or some functions are just not capable to deliver the correct measurement. Physic’s laws are always there ……
sorry, half wave can't provide more power than full wave. r.m.s of full wave is Vm/sqr(2), for half wave is Vm/2.
You confuse voltage with power. Thank about that.
@@_-Skeptic-_
Being a "skeptic" is good, being a student is even better.
The Bulb-only circuit is being fed with 65VRMS. The bulb & diode circuit is being fed with 120VRMS. After the diode, the bulb is still being fed with 85VRMS, because the peak of 120VRMS would be 170V and half-wave RMS = Vpeak/2 = 170/2 = 85. Applying a higher voltage across the same load would result in higher current and therefore higher power.
@@xraytonyb Yep.
@@JohnAudioTech no I'm not, just thought you are connecting them to the same voltage and saying the one with the diode is brighter. Brightness is power, so I meant same voltage half-wave rectified can't deliver more power (brightness) than full wave. In fact your video is confusing as you digress too much and don't specify at the beginning the conditions of measurement, voltages,...etc.
Incandescent bulb has inductance and resintance in AC😅
AC "resistance" is better referred to as impedance.
Yeah, no.
Incandescent light bulb's brightness is dependent on the AVERAGE power the filament dissipates. This is neither RMS or peak voltage and so a DVM will mislead you.
One bulb is LED and the other is incandescent that is your problem.
Both are incandescent.
Lol, the "phenomenon" is not a magic trick.