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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.

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  • \$\begingroup\$ Not really i checked it does work the LED seems to have a low turn ON current. \$\endgroup\$
    – Seb
    Jun 5 '20 at 19:45
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  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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  • \$\begingroup\$ I am puzzled as to why they triple the voltage with a voltage multiplier rather than just use 3 times as many turns on the transformer winding which I would have thought would be cheaper. \$\endgroup\$
    – abligh
    Jun 6 '20 at 5:33
  • \$\begingroup\$ Won't the resistor across C1 along with removing residual charge also reduce the overall impedance to the input of the FBR. In this case 674 ohm || 0.47uF should give an overall impedance of 602ohm. And if that is the case then the battery would be getting a really large voltage. I tried to simulate this on this site: mosquito swatter cap dropper simulation \$\endgroup\$
    – Seb
    Jun 6 '20 at 6:29
  • \$\begingroup\$ @Seb Double-check the resistor accross the c1. It will be 100kOhm or more. You probably misread the code and it is probably 670kOhm \$\endgroup\$
    – fraxinus
    Jun 6 '20 at 7:23
  • \$\begingroup\$ @abligh it is easier to make a transformer with less windings and ticker wire. It is also less prone to insulation breakdown and easier to find off the shelf one (made for a mobile phone charger, for example). You can also use diodes w/ less reverse voltage rate. \$\endgroup\$
    – fraxinus
    Jun 6 '20 at 7:27
  • \$\begingroup\$ @fraxinus you were right it is in kOhm, that clears my doubt, sorry about that I will edit it in my question. However i made the changes in the simulator yet it does not indicate any large voltage drop that would allow the battery to safely charge. \$\endgroup\$
    – Seb
    Jun 6 '20 at 7:42
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Why did they use the capacitor and resistor in parallel as input for the full bridge rectifier?

That is a "capacitive dropper circuit"

It is basically a cheap way to make a low voltage, low current supply directly from the high mains voltage.

Why is the battery just directly connected to the output of the full bridge rectifier.

It is part of the capacitove dropper circuit and used as a device to limit the voltage for the rest of the circuit to around 2.4 V. In most capacitive dropper circuits a zener diode is used.

How is the (6-pin) transformer connected internally? If I can understand this I am guessing I can understand how the transistor works. What is the configuration at the output of the transformer?

I suggest that you search for "transformer oscillator circuit" and look at how those circuits are drawn. Then realize that the transformer has 3 windings, possible between pins 1 and 2, pins 3 and 6 and pins 4 and 5. You can measure this with an ohm meter while the circuit is off.

Then re-draw your circuit so that it looks more like the examples you found.

This circuit is at mains voltage when connected to the mains so don't touch anything when it is connected to the mains voltage!

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  • \$\begingroup\$ Generally in a capacitive dropper the capacitor should be in series with the resistor so that the capacitor can offer high resistance to the input voltage at low frequencies and the resistor can reduce the inrush current.However in this case both are in parallel so won't it reduce the overall impedance. Aslo the wiki link to the [capacitive dropper](link:en.wikipedia.org/wiki/Capacitive_power_supply) has a similar arrangement at the input side of the FBR however it does not explain the significance of keeping it in parallel. \$\endgroup\$
    – Seb
    Jun 6 '20 at 7:20
  • \$\begingroup\$ Capacitive dropper needs both parallel and series resistor in order to work properly and safely. \$\endgroup\$
    – fraxinus
    Jun 6 '20 at 7:31
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    \$\begingroup\$ @Seb How about different resistors having different functions. Also the value of a resistor matters. You cannot compare a series resistor there to a parallel resistor in another similar circuit. A resistor in series with the capacitor limits the inrush current, such a resistor will have typically a value of less than 1 kohm. The resistor in parallel with the capacitor is there to discharge the capacitor when unplugging the device from mains voltage so that a charged capacitor cannot zap the user. Such a resistor is typically 1 Mohm.... \$\endgroup\$ Jun 6 '20 at 11:18
  • \$\begingroup\$ ... also in good designs there should be two such resistors in series as the voltage across the capacitor can reach up to 400 V DC which is above the rated voltage for most standard resistors. Two resistors in series will divide the voltage to safe values. \$\endgroup\$ Jun 6 '20 at 11:19
  • \$\begingroup\$ You said that the battery is used as a device to limit the voltage for the rest of the circuit to 2.4v however unlike the zener diode which is passive the battery will provide 2.4v and offset any input ac voltage which would mean that the ripple voltage at the output of the FBR should get offset by 2.4v. What i don't get is how can it act like a zener? \$\endgroup\$
    – Seb
    Jun 6 '20 at 14:43
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Diodes D1,D2,D3 and D4 have been copied wrongly.they all have to be reversed or flipped over(anode instead of cathode)

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  • \$\begingroup\$ Checked it twice, that's is exactly how they are connected on the board, while removing the circuit from the swatter bat however i also ended up removing the battery so i am not exactly sure if the battery is connected in the correct way in my circuit. \$\endgroup\$
    – Seb
    Jun 7 '20 at 15:33
  • \$\begingroup\$ Whenever the device is plugged to the mains 220v or 230v ac supply,the bridge rectifier will charge the battery. \$\endgroup\$ Jun 7 '20 at 15:36
  • \$\begingroup\$ Yes but the battery seems to have 0.8 v only when i check it using a multi meter and check the voltage across it in dc mode when the circuit is live. but when the circuit is OFF and i remove the battery it indicates 2v across the battery. \$\endgroup\$
    – Seb
    Jun 7 '20 at 15:39
  • \$\begingroup\$ Yes if you reverse the battery it works. \$\endgroup\$ Jun 7 '20 at 15:42
  • \$\begingroup\$ It seems you were feeding the battery with reverse polarity.that is why it was reading 0.8v. \$\endgroup\$ Jun 7 '20 at 15:45

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