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I bought a digital mounting thermometer 12 volt,for a riding lawnmower with an air-cooled engine have to put in a zener diode,to control the starting voltage of about 55 voltage for a split-second.The thermometer draws 3mA and maximum voltage 28 volts.How do i install this diode.A diagram would be appreciated. LEO

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You also need to know the minimum operating voltage from the lawnmower. Then you can calculate the value of the series current-limiting resistor so that the thermometer gets its 3mA and the zener diode gets a couple of mA.

Assume the minimum operating voltage is 14V. Then the resistor will have 14-12=2V across it and have 3 + 2= 5mA through it and have a value of 2V/5mA= 400 ohms.Use 390 ohms.

When the lawnmower gives 55V then the resistor will have 55-12= 43V across it and it and have 43/400= 108mA through it. Therefore the power in the resistor is 43 x 108mA= 4.6W. Use 10W since a 5W resistor will be very hot if you need to crank the engine for a long time..
The zener diode will have 12V across it and 108 - 3= 105mA through it and therefore will dissipate 12V x 105mA= 1.26W. Use 2W.

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It is my understanding that the spike that is generated during starting is as a result of turning the starter motor off... after all, you are "unloading" an inductive circuit (the motor plus the inductance of the wire to and from the battery to the motor). To size the resistor properly, you need to figure out what would be a reasonable number of times to try starting the motor, and then calculate the RMS value of the spike... yes, the 10W resistor and 2W diode would/will work, but they are also overkill.

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