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Re: Urgent question regarding zapper handles
 

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Hulda Clark Cleanses


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analogkid Views: 1,264
Published: 15 years ago
 
This is a reply to # 799,728

Re: Urgent question regarding zapper handles


Brief contact between the handles should not damage this circuit. Here's why, based on some data and some conservative assumptions.

Based on the circuit on the zapperplans website, the output device is an unbuffered CMOS inverter, with a 1K resistor in series. Based on the device data sheet on Texas Instrument's website, the typical output current with a 10V supply is 2.6mA at an output voltage of 9.5V. This gives an equivalent channel resistance of about 192 ohms for the P-channel FET. When you touch the handles together, the total circuit is the 9V battery, the 192 ohm channel resistance, and the 1K output resistor. The current through this string is about 7.55mA. While this is more than the 2.6mA typical spec (no max spec is given), it is not that far above the typical output current with a 15V supply (6.8mA), and well within the manufacturing tolerance of these kinds of chips.

Another approach is to consider the power dissipated in the device. The absolute max power allowed is 100 milliwatts (mW) for any output transistor. 7.55mA through 192 ohms calculates out to a power of 11mW, only 11% of the max allowed. Even better, since the output is actually a square wave the average power dissipated is only 1/2 as much.

An extra level of output protection comes from the fact that the transistors are FETs. The FET's operate on the edge of saturation, so they are somewhere between constant current and constant resistance devices, as opposed to the constant voltage characteristic of bipolar transistors. This acts as a kind of self-limiting protection.

BTW, typical 555 based zapper circuits are immune to external shorts because the 555 (either bipolar or CMOS) can source way more current than an external short can draw.

ak
 

 
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