Electronic – How does this circuit in an electric mosquito swatter work

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I had an electric mosquito swatter lying around so I decided to open it up and check out how it works.

I understood some parts of the circuit however I did not clearly understand how the rest of it works.

I have posted the circuit of the mosquito swatter. (I did notice that there was a similar question on the forum but the circuitry differed and it was not yet solved.)

Here's the circuit:

Mosquito swatter

Things I did not get:

  1. Why did they use the capacitor and resistor in parallel as input for the full bridge rectifier?
  2. Why is the battery just directly connected to the output of the full bridge rectifier.
  3. How is the (6-pin) transformer connected internally? If I can understand this I am guessing I can understand how the transistor works.
  4. What is the configuration at the output of the transformer?

Note:

  1. Below the main circuit there is a diagram that shows the interconnection between the transformer pins when I used a multi-meter to check for continuity, note the transformer was still on the pcb.
  2. The 0 and 1 at the output is just to indicate that the 0 goes to outer 2 meshes and 1 goes to inner mesh.
  3. The LEDs are red.
  4. Some of the components did not have values written on them like the capacitor C6 so I removed them and checked them using a component tester so the values may be slightly off the standard values.
  5. I numbered C6 out of order sorry about that.
  6. The resistor parallel to C1 is 675kOhm and not 675Ohm.
  7. The resistance between Xmer pins 1-2: 2 Ohm; 3-6: 0.8 Ohm; 4-5: 245 Ohm.

Additional working note:

When the device is not charging the NO switch is closed and the push button is pressed this seems to charge the meshes for the next electric swat.

Best Answer

  1. The resistor across the 240 VAC mains series cap, C1 serves to drain residual charge, to prevent possibility of a small shock. Depending where in the AC cycle the device is unplugged, the charge on the cap could be as much as 1.414... times the RMS (nominal) AC mains voltage, ~340 VDC.
  2. Why not? If it's sealed inside a case, double-insulated, there's little chance of shock from touching battery and ground. Oh... do you mean is the 2.4 V battery getting 240 VAC? No, because C1, a 0.49 μF capacitor, has a reactance of ~6,500 Ω at 50 Hz, and ~5,400 Ω at 60 Hz, limiting the voltage and current across the battery. That would produce a charge rate of ~ 36mA at 50 Hz. BTW, I would not leave the swatter plugged in permanently, since that might be more current than good for long-term trickle charge on a small NiCd or NiMH cell.
  3. You can check winding connectivity with an ohmmeter, but basically, the emitter and the B+ are connected to pins 1 & 6, the primary winding; pins 2 & 3 are feedback windings, sending signal to base to cause oscillation; and 4 & 5 are high-voltage secondary output.
  4. The diode-capacitor combination on the output serve as a voltage multiplier. In this device, it triples the voltage from the secondary to charge C6, which stores enough power to permanently incapacitate an arthropod (give it "a short, sharp shock, from a cheap and chippy chopper", as G&S put it).

BTW, the Cockcroft–Walton generator uses voltage multiplication to generate megavolts from lower-voltage AC. No, I don't suggest building a basement particle accelerator or X-ray machine.

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