Effect of High Voltage AC/DC on Human Body




How does human body react to AC versus DC at high voltages? Watch the video below:




Don’t try to touch any sort of high voltage yourself of course, you may die if you know what I mean!

The experiment clearly shows that AC is much more hurtful and dangerous than DC voltage. Like I mentioned the capacitive property of body makes it way more conductive towards AC current. A capacitor’s impedance is proportional to reverse of frequency and drops as the frequency is increased. But at zero the capacitive impedance is equal to infinity, which leaves the body resistance that can be around 100 kOhm over dry skin, or much smaller across tissues under the skin. And that’s why AC is much more dangerous than DC.

In fact based on Wikipedia where I get most of my knowledge from, an AC voltage of below 50VAC is considered extra low voltage, while a DC level below 120VDC is considered extra low voltage. That confirms the more dangerous nature of AC.

Now let me explain to you how the circuit I made works. Below is the schematic again:

Booster CircuitThis is some sort of booster circuit. The first stage consists of C1 and D1 that work together to create a DC shift on the AC signal input. If you look at the plot below you will see the AC input as the black line. When the input rises, D1 is off as it is reverse polarity biased and the capacitor acts like a short for AC and so V1 will follow the input. But as the input tries to go negative and pull V1 down with it, it will forward bias D1 and will turn it on.



When D1 turns on, it will clamp V1 close to zero, or more accurately -0.7V for a regular diode. But the input voltage keeps going down. This means that a current will run through C1 and D1 that will charge C1 close to the peak of the AC signal, which in my case was close to 170VDC.

Remember that a 120VAC has a peak of 170V. 120VAC is the effective voltage, or root mean square value.

Now  when the AC signal reaches its minimum at the negative peak, it will try to rise at which point the current through D1 flips and turns D1 off. So now again the charged capacitor acts as an AC short and rises all the way up to the positive peak, which added to the charge on the capacitor, will add up to twice the amplitude of the AC signal or in my case, 2 x 170V = ~340VDC.

plot-v1

Now the second part of the circuit which is C2 and D2 is basically a peak detector, or a peak hold circuit. C2 is initially not charged. Every time V1 rises above C2, D2 is forward biased and turns on, current flows into C2 and charges it up. And every time V1 drops, the current flips that turns D2 off and so C2 will just hold its voltage.

plot-v2So V2 will end up being charged to a DC level almost double the AC signal amplitude. Of course this is considering no load on the voltages, and also there is some voltage drop over the diodes. With loads, the voltages will drop.

This circuit can be repeated and cascaded to create even higher voltage. Every similar circuit will add the equivalent of the AC signal peak to peak voltage to the output. I’ll leave it to you to figure out how to cascade it.

Just remember to rate your capacitor properly and don’t connect them backwards. And don’t mess with any high voltages!



71 thoughts on “Effect of High Voltage AC/DC on Human Body

  1. Hello
    I can’t understand at all how this circuit works 🙁 can someone help explain in details ?

    Best regards.

    • Ok. This is how I was able to understand it . I will be making references to the circuit above.

      First, the diodes. they basically are to activate a certain part of the circuit during a certain time interval.

      In the positive cycle (imagine you are pushing the positive charges manually in the top wire), D2 is activated and current flows through the components C1, D2,C2.

      Next, in the negative cycle, (this is where the charging happens) the diode D2 is deactivated and D1 is activated. hence , you are now pulling charges away from the top wire and the current flows through D1,C1. here, the capacitor C1 is charged to the peak voltage and store a “potential ” to surge 170 V once unleashed. notice in the negative cycle, the capacitor C1 has the right plate as positive and left plate as negative. This means , the C1 would like to discharge in the right direction.

      In the second positive cycle , you are pushing charges back into the top wire , and this time , you are assisted with the additional voltage of the charged capacitor C1 ( which also want’s to discharge in the same direction as you are trying to push the positive charges through). Now you have double the voltage you are supplying through the components D1, C2 . This is where C2’s
      job begins. It holds the 340 V peak voltage across it’s terminals and doesn’t let it go. In every full cycle, notice that only the half wave reaches C2. It can be noted that the voltage across C2 is DC. The diodes D1 and D2 kinda act like a rectifier circuit.

      Overall , this means that the input voltage is 170 V AC, and we get double the output 340 V DC . hence the name “voltage multiplier”
      If anyone finds a flaw in my explanation , pls do point it out.

      Thank you for your patient reading.

  2. Hi Mehdi
    Please help me destroy an electrical instrument connected to home wiring by injecting high voltage into one of the plug in home. The high voltage device should be protable.
    With regards
    Psycho Santosh

  3. Hi Mehdi,
    Have been following your channel for many many years now. Thanks for your videos. I hope you will make a short video of your personal life.
    I was wondering if you were ever interested in tube amplification and sound engineering. Would you make a video about filter capacitors, power tubes and a simple amp circuits.
    Another request, do you know about dielectric breakdown and could you make videos about it? For example : PTFE material etc. I think this will be complimentary to your existing videos.
    Thanks,
    Naren (Indian living in France)

  4. Hello,
    I have a 48v solar panel setup connected to 4x12v dc batteries in series connected to and inverter which converts to 220v ac and connects to a breaker from which a room is powered.
    I was thinking if it is possible to wire the system to 4x12v batteries in parallel, get rid of the inverter and connect the system directly for 12v lights around the house.
    Is such thing feasible, safe and worth it? Would be great to have your views on it.

  5. What if the input voltage is 220v AC ? What would be the output voltage ?

  6. I understand the first part of the circuit, but not how the second part gets the 340VDC. If the peak of the AC is 170V where does the extra voltage come from to the second part?

  7. I have doubt, what happens if a capacitor is charged with positive potential and then a negative potential(polarity changed) is applied to the capacitor won’t it cancel each other ?

      • on the second thought, i think it might be cancelled out. As mehdi connected an electrolytic capacitor as C1 , it should have blown up due to reverse polarity.

  8. Hi Mehdi,
    I wants to know that how much load current we should put on this circuit

  9. Hi Mehdi,

    How is it that your bulbs continuously glow brighter across the capacitors? I understand higher voltage will push more current through the bulbs and they will glow brighter but there is no way the capacitors can continuously provide as much power as the direct AC line. So what gives? If this is true then I could just cascade the circuit to get brighter and brighter bulbs.

    • the capacitor can provide the bulb continuously when it is charging continuously.
      simple answer isn’t it ?

  10. Hi mehdi
    Big fan of your work.

    Can i use this circuit to charge a single camera flash capacitor?

  11. Pingback: Making a Jacob’s Ladder, to Celebrate a Million Subs! | ElectroBoom

  12. Could this be used as a power supply for an electromagnetic disk launcher? Asking because I’m trying to make a launcher, and it requires around the highest voltage yours makes.

    • just look for it as a diode multiplier source. It serves for low power only, but can kick you as f*ck. (only multiplyable for 10 times )

  13. Dear Mehdi,
    i want to power up a treadmill DC motor (180 VDC) can this method work ??

    • Voltage multiplier good for small currents but can not drive a dc motor properly. You can use a transformer.

  14. Hello Mehdi,
    May I use a 1n4001 diode for my circuit instead of 1n4004? They are only a few numbers apart from each other, so I am assuming that it would work. Sorry if I may sound stupid, but in my defence, I am only 16.

    (Big fan by the way!)

    ~ Sincerely,
    Min Ho

    • Your capacitor is good, your diode is bad, it can take very little voltage. You need >400V. So use 1N4004. Also depends on how much current you draw from it too. Or diode will blow.

      • So, if I touch a 170V DC capacitor it will not hurt me ? I am building a capacitor bank for a project.

        • It feels differently depending on where you touch it on your body. But just touching it with your fingers you will be fine.

      • Sir can i use In4007 diode instead of in4004 in the that you made for 170v dc and 340v dc. I have 240v Ac here and i am using 400v 220uf capacitors and In4007 diode. Is it ok???
        Hoping for a rly
        Thanks

  15. Hello Mehdi

    As usual, your videos are super great but I want to ask one thing.
    On your first voltage diagram, there is a point when you have the most negative input voltage value. That is the point when capacitor C1 charged to 170 VDC. After that input voltage is raising. Till it gets to zero, the diode D1 should be open, but in our situation it is closed. So my question is: The Diode is closed because the voltage from capacitor now affects the diode D1 ?
    Thank you

    • When the voltage reaches peak negative and starts rising again, right then D1 turns off. Because the capacitor C1 is fully charged to 170V and if D1 remains on, C1 would discharge, by a current reverse of what it was charged in reverse direction through the D1 diode, which is not possible as D1 blocks reverse current, so it turns off. And C1 keeps 170V in it.

      C1 may discharge if an output load is attached drawing current from it. So every time the input goes to its negative peak, D1 turns back on and recharges C1.

  16. Hey mehdi can I use this circuit to make a 12v power supply? And if I can,can you help me 😀

  17. I’m struggling to understand how C1 can’t charge in reverse through D2 and C2.

    • D2 can’t send charge back into C1, it will be off against reverse current. D2 only turns on of V1 is higher than V2

    • For a no load case like what I showed, you could use 1N4004 and >200V, 100uF capacitor

        • so 12 volts from a battery with plenty of amps first connected to an oscillator circuit to produce ac the a hundred capacitor diode connect in this way with a large spark gap at the end would work as a tazer without the need for a transformer?

  18. Hi Mehdi,
    I wanted to know how exactly the offset in the v1 created.
    And what do you mean by ‘not connecting the capacitors backward’. Here the first capacitor’s polarity is reversed so we need to connect its negative side to the supply , ri8
    Thnx 🙂

    • The offset in V1 is created because D1 only allows current to flow one way through it, and so the current though C1 can only flow back into the supply when its voltage is negative and D1 is on.

      And this is the reason C1 polarity is as shown, because it only charges when supply is negative and teh D1 side is shorted to ground through D1, so negative towards the supply!

  19. Mehdi,

    A comment followed by a question.

    I just saw your homemade guitar video and I laughed my ass off!

    Now for the question, you’re old enough to remember CRT televisions (I hope). In the above comments someone indicated DC is less dangerous that AC. If that’s the case then why in the old CRT type tvs was it always explained to not mess with the “tube” inside the screen? Did it have to do with the fact that it was a type of capacitator? Also, like PCs the inside of the old CRT tvs were they direct current?Yes, as you can tell I know absolutely nothing about electricity except to not stick your fingers in sockets nor be near anything electric and plugged in near water.

    cheers!

    • in old tube TVs the DC voltage was boosted up to a few thousand volts. So although DC is less dangerous, its super high level is definitely lethal. That’s why you don’t mess with them!

      And thanks for the comment!

  20. I’d be curious to see what the current readings are for AC vs DC. The claim that ‘AC hurts due to the capacitive properties of Medhi’ seems to indicate more current should flow in AC then in the DC case.

    It’s also a bit counter intuitive (at least for me). I’d expect initial touch of the DC line to replicate ~half of an AC cycle, (ie some initial shock+pain as the capacitor charges), but that doesn’t seem to happen. Maybe for weird biology reasons (the initial charge may be short/small enough pulse not to trigger sense, where as 60Hz AC is on average 20V/ms change /semi-constant capacitive flow).

    • Touching a DC, that initial pulse is much faster than 60Hz and I assume It doesn’t feel as bad as 60Hz.

      • I wonder then if a 60hz 120V square wave is as bad as the equivalent sine wave.

          • but 120v ac at 100khz you would not feel right due to nerves not sensing them-i wonder if you would only feel the pain when your body starts cooking after a minute or so?

  21. I am having a little trouble understanding the circuit. Why does C1 charge up the way it does? I would think when D1 is on it would just act like a short to the AC like it does when D1 is off.

    • You’re right, when D1 is on it will short the AC, which means C1 charges up. When the AC direction changes, D1 turns of and so C1 doesn’t lose it’s charge anymore and keeps the AC peak.

      • I guess my question is really why does C1 charge the way it does? If it keeps acting as a short to the AC then how does it get a DC charge? Is it because C1 is polarized?

        • “Acting like a short” is relative. It is not a dead short and has some impedance for 60 Hz. Which means the current through it will be limited and the voltage across it can rise when the diode is conducting.

  22. Hello M. Mehdi
    The output voltage V2 is a DCsignal i couldn’t use it to add the equivalent of the AC signal peak to peak voltage to the output! Can you give me a hint please??

    • Doesn’t really matter with no load on the output. But the more load you have on the output the larger capacitors you would need. If you want to turn a lamp on through it like I did, you need over 1mF

  23. Hi
    I love your website… I haven’t laughed so much in ages. best thing on internet.
    Im thinking of your Capacitor DC voltage experiments… I try to make home made lasers and have had the F*** moment when touching 350V DC on Capacitor. I also have burns on my hand from another silly 10,000v shock when I was showing its Amps not Volts that Kill….. Remember not to be earthed to Ground

    Regards

    Sean

    • Hah, I guess that shows you that it is the voltage that generates the amps that kills! 🙂 “It’s the amps not the volts” is a wrong statement. That’s why they say “danger, high voltage”.

  24. Dear Mehdi,
    What effect does the capacitance of the capacitors have on the circuit? Will it output more current? Thanks.

    • Yes, for more power transfer to the output, and keeping the voltage high you need less impedance, which means increase the frequency and/or the capacitance.

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