Why does the capacitor connection reduce into 110 V instead of 99 V?

Question

I saw the following voltage divider on YouTube:

Simple convert 220V to 110V

And I saw it is a so-called voltage divider using capacitors: Capacitive Voltage Divider as an AC Voltage Divider

And I thought to implement the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Based upon the site, I find with a 10 µF capacitor and 220 V 50 Hz AC the voltage across C1 is 99 V and not 110 V.

Based on the site's formulas, the C1 + C2 "resistance" is about 637 Ω.

The reason why I need to do this is to make an el-cheapo heatgun on a leftover 120 V soldering iron that I cannot use because in Greece where I live the power is in 200 V AC and if I plug it in it may cause harm to myself and be fried as well.

So I want to ask:

1. Is a good idea to use a voltage divider as power supply to soldering iron as heating element?
2. Do I need smaller capacitors?
3. Will electrolytic capacitors fit the job for my application?

The soldering iron is a 60 W one. The idea is to implement something like this:

How to Make Hot Air Gun from Soldering Iron

In order words, to place a cheap 12 V pump as air flow and pass it from a soldering iron tip.

Answer 1

Why?

1. Because the two capacitors are not identical due to tolerance.
2. Because the top capacitor is loaded with something that you don't show in the schematic diagram, and the bottom one is not

Is a good idea

Absolutely not!. That circuit may be good enough to light an LED, certainly not for a load of any significant power, such as a heat gun.

The only time we use capacitor dividers is at radio frequencies (megaHerz, gigaHerz), and rarely at high power.

Do I need smaller capacitors?

No, you need to start from scratch and use a transformer instead.

Will electrolytic capacitors will fit the job for my application?

No. You're working with AC, and electrolytic capacitors are polarized (work at DC). *

(*) There are some non-polarized electrolytic capacitors but they are only good for low power audio applications.

Answer 2

1. No, on the contrary, that is an extremely bad idea.

2. No, you don't need capacitors at all.

3. No, electrolytic capacitors would be an extremely bad idea too.

You really don't adapt a 60W 110V device to 220V mains with a capacitive voltage divider made of electrolytics. Use a transformer.

Electrolytic capacitors are typically polarized capacitors that cannot handle voltage in reverse, so that's why they are wrong type.

Also a 10uF capacitor has 318 ohms of impedance at 50 Hz. A capacitive divider with 10uF capacitors would thus have 159 ohms output impedance to load. So it would not work to power a 110V 60W load which requires about 0.55A of current, and can be assumed to be just a load of 200 ohms, so the capacitive divider with 159 ohms can't drive such a high load.

Answer 3

Well, when we have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

It is not hard to see that:

$$\underline{\text{V}}_{\space\text{out}}=\frac{\text{R}\text{||}\frac{1}{\text{j}\omega\text{C}_1}}{\frac{1}{\text{j}\omega\text{C}_2}+\left(\text{R}\text{||}\frac{1}{\text{j}\omega\text{C}_1}\right)}\cdot\underline{\text{V}}_{\space\text{in}}=\frac{\text{C}_2\text{R}\omega}{\text{R}\left(\text{C}_1+\text{C}_2\right)\omega-\text{j}}\cdot\underline{\text{V}}_{\space\text{in}}\tag1$$

So, the amplitude is given by:

$$\left|\underline{\text{V}}_{\space\text{out}}\right|=\frac{\text{C}_2\text{R}\omega}{\sqrt{1+\left(\text{R}\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}}\cdot\left|\underline{\text{V}}_{\space\text{in}}\right|\tag2$$

In your example we know that \$\left|\underline{\text{V}}_{\space\text{in}}\right|=220\sqrt{2}\space\text{V}\$, \$\text{C}_1=\text{C}_2=10\cdot10^{-6}\space\text{F}\$, \$\text{R}=\frac{605}{3}\space\Omega\$ and \$\omega=100\pi\space\text{rad}\$. So the amplitude for the output voltage will be:

$$\left|\underline{\text{V}}_{\space\text{out}}\right|=\frac{10\cdot10^{-6}\cdot\frac{605}{3}\cdot100\pi}{\sqrt{1+\left(\frac{605}{3}\cdot\left(10\cdot10^{-6}+10\cdot10^{-6}\right)\cdot100\pi\right)^2}}\cdot220\sqrt{2}=$$ $$13310 \pi \sqrt{\frac{2}{90000+14641 \pi ^2}}\approx122.115\space\text{V}\tag3$$

Answer 4

For a small and thrashbin-sourced project, the capacitive divider is pretty much a viable option.

The caveat is to know what you are doing.

If your heat gun is only a heater element and does not contain a control logic or something else that can depend on voltage and fears of voltage spikes on power-on, you may as well go with a single capacitor.

The heating element is 240 Ohm (60W at 120V). The total resistance should be twice that (480 ohm). The capacitor has to be sqrt(3) * 240 Ohm at 50Hz, so it gets more or less 7.7uF.

The voltage over the cap will be ~300V peak and the reactive power will be like 105W (both are important when sizing the capacitor).

Be sure to parallel the capacitor with 100k .. 1M resistor in order not to get refreshing shocks from the power plug.