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.


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!

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

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

  2. 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 )

  3. 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.

  4. 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.

  5. 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!

  6. 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.


    • 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!

  7. 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).

  8. 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.

  9. 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

  10. 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



    • 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”.

    • 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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