I've been around electronics for a lot of time but I really didn't understand how capacitors work in an AC or DC circuits.

I know capacitor has a big role in smoothing the output of bridge rectifier. As the ripples go through the capacitor and the capacitor works as a tank that is filled upto the maximum voltage of the ripple signal and when the ripple falls to lower voltage, the capacitor is holding that max voltage and delivers it to the output, waiting for the next max ripple signal. And that's how capacitor smooth ripple voltage.

But in this circuit that have a TRIAC to be triggered with a DIAC and complete the circuit and the heater works. There's C1 I think it's a ceramic cap or something other than an electrolytic capacitor. And it works as an open circuit, so there's no current flowing through it. But my question is what its job? I learned something that in an AC circuit that has an inductive load, capacitor is very useful in bringing up the lagging voltage to a good level with the current.

Is it the job for this capacitor in this circuit?

Also what's the job of C2 & C3 ?

enter image description here

  • 1
    \$\begingroup\$ your core question ("what is the role of C1 here?) is pretty different to the question in your title. But I think the main problem is what you describe in your first paragraph: You have no idea what a capacitor actually is / what it does. So, that renders your question far too broad. I'd recommend starting with learning the basics of linear electrical circuits – otherwise explaining this will just be us reproducing a complete introduction to electronics. \$\endgroup\$ Sep 28 '19 at 10:07
  • \$\begingroup\$ OK, I can edit the title. Thanks :) \$\endgroup\$
    – R1S8K
    Sep 28 '19 at 13:46
  • \$\begingroup\$ as said, the title is a small problem – the real problem is that you don't seem to understand basics of capacitors. \$\endgroup\$ Sep 28 '19 at 16:23
  • \$\begingroup\$ I know something about them, but I don't know if I'm studying; for example, a SMPS schematic diagram. That why each capacitor is used, then I wouldn't know all of them. I may know some caps that are for filtering and smoothing the rectified output whether from the full bridge rectifier, or the outputs after the high frequency transformer. \$\endgroup\$
    – R1S8K
    Oct 1 '19 at 15:14


Capacitors will always conduct current if there is a change in voltage across it.

Ic=CdV/dt However the current is not determined by the physical size as the density of nF,uF is determined by the dielectric constant and electrolytics and batteries alike are much greater than ceramic and plastic caps since they are highly corrosive (acidic) and thus can stored more charges.

Here you will likely find they are 400 or 600 V rated plastic caps that may be able to handle > 1kV transients.


Caps have many applications in time and frequency sensitive circuits.

  • resonate with inductors to perform LPF, HPF or BPF depending on values and circuit
  • integrate current with a current-sense resistor to convert "int(I.in) to V.out"
  • differentiate voltage to create a current then sense this to convert "dV/dt to V.out"


  • C1 is a LPF to attenuate noise > 100kHz from switching current < 1us an also absorb some external noise
  • C2 C3 are LPF time/phase delay circuits cascaded with variable RC=T values
  • C3 acts like a tiny battery each cycle that activates the DIAC switch which in turn discharges C3 as it's voltage then triggers the Triac.
  • Then the Triac latches ON and no longer needs the voltage from C3.
  • This repeats every half cycle as the Triac turns OFF (unlatched) when current drops near the zero crossing.
  • \$\begingroup\$ Yep, when you mentioned LPF, HPF or BPF. That's a huge topic but actually explain the role of using caps in many circuits. It could be also used for smoothing out the voltage. Also blocking AC signals or Bypassing AC signals, a lot of things but it depends on the circuit design and what's the application of the circuit. \$\endgroup\$
    – R1S8K
    Sep 28 '19 at 15:45
  • \$\begingroup\$ Yes and the huge topic can be described in several domains; time , frequency and phase — even for a simple RC filter with up to 90% of a half cycle delay at peak 5T — leaving a tradeoff of attenuation and hysteresis \$\endgroup\$ Sep 28 '19 at 15:57

To analyze the circuit we can break it down to two parts:

1.RFI suppression LC Filter

A triac can be used to give variable AC power control by using a 'phase-delayed switching' technique, in which the triac is triggered part-way through each half-cycle. Each time the triac is gated on, its load current switches sharply (in a few microseconds) from zero to a value set by its load resistance and instantaneous supply voltage values. In resistively loaded circuits, this switching action inevitably generates a pulse of RFI, which is least when the triac is triggered close to the 0° and 180° 'zero crossing' points of the supply line waveform (at which the switch-on currents are minimal), and is greatest when the device is triggered 90° after the start of each half-cycle (where the switch-on currents are at their greatest).

The RFI pulses occur at twice the supply line frequency and can be very annoying.RFI can usually be eliminated by fitting the interface with a simple L-C filter. The filter is fitted close to the triac, and greatly reduces the rate-of-rise of the AC power line currents.

2.Variable phase-delay trigger network

As the input voltage is applied to the circuit, c1 and c2 starts charging at a rate determined by the resistance R2. Whenever the voltage across the capacitor c3 exceeds the breakover voltage of the diac, diac triggered and starts conducting. Then, the capacitor C3 starts discharging through the conducting diac into the gate of the triac.

Therefore, the triac is turned ON and passes the current to the Load. By varying the resistance R2, rate of charge in the capacitor is varied and hence the voltage at which the triac is triggered in both positive and negative half cycles of the input is controlled.


Optimizing the Triacs

Switching Regulator Noise Reduction with an LC Filter

A Beginner’s Guide to DIAC

  • \$\begingroup\$ This is a great answer on how a dimmer works but lacks detail on how the caps work in AC ccts. Can you describe what the C2 Voltage would look like? \$\endgroup\$ Sep 28 '19 at 11:20
  • \$\begingroup\$ @SunnyskyguyEE75 , Yes sir, Found this link that has a great explanation with related diagrams, I prefer to put the link here instead of copy-pasting it. link \$\endgroup\$
    – Sepehr
    Sep 28 '19 at 12:06
  • \$\begingroup\$ the link shows a far superior design with much less hysteresis using zero crossing pulse, phase shifter then pulse transformer but doesn’t answer my question, but. Still a great resource \$\endgroup\$ Sep 28 '19 at 12:11
  • \$\begingroup\$ I think C2 voltage would be something like a sawtooth signal ! It's my guess .. would anyone agree with me? \$\endgroup\$
    – R1S8K
    Sep 28 '19 at 15:41
  • \$\begingroup\$ @Sepehr Hi, in the provided link, the last circuit. I think it's more complicated than it should be ! A complicated circuit for only a light dimmer or turning on a 12V lamp. I think there are more easier circuits to turn on a 12V lamp. Do agree with me? \$\endgroup\$
    – R1S8K
    Sep 28 '19 at 15:53

Capacitors SEEM to conduct current, if the voltage across the capacitor is changing.

Unlike a resistor with its continuous flow of electrons along the resistive material (such as carbon particles, or very thin metal films), the capacitor's insulation prevents the charge moving INTO one plate of the capacitor and then moving OUT the other plate. However the extreme ability of electrons to push on other electrons, with even a minor imbalance bringing in outside electrons to achieve a balance, with that pushing occurring across the plate-to-plate insulation, causes the appearance of "current flowing thru a capacitor when the voltage changes".

Capacitors store charge. When the voltage across 1 microfarad capacitor changes by 1 microvolt, we'll have Q = C*V, or dQ = C * dV = 1uF * 1uV = 1picoCoulomb of charge movement.

WIth 6.24 * 10^+18 electrons in a coulomb, we scale that by 10^-12, to find 6.28 Million electrons will move because of that 1 microvolt change in voltage.

However, if you could mark a single electron and then track it, you will not see it move onto one plate of the capacitor and across the insulation and then exit the other plate.

Thus that 1 microvolt change will move 6.28 Million electrons onto one plate, and cause 6.28 Million different electrons to LEAVE the other plate.

JONK --- am I saying this correctly?


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