Hello. Please see the attached file for my question. What are the power ratings for R1 (60 ohms) and the 12 volt zener diode. Please show the math with your answer. Thank you for any help you can give me. ;
I do not agree. If the zener is clamping the voltage across it to 12 volts then there must be 6 volts across the 60 ohm resistor so the current through the resistor must be 100 mA. (I = 6/60 = 0.1 amps) So the resistor will dissipate 0.1 x 6 = 0.6 watts. The LED current will be about 9 mA ( volts /1000 ohms = 0.009 amps. ) (Assuming the blue LED forward voltage drop is about 3 volts.)
As the quoted current for the fan is 100 mA and the LED current is 9 mA then the total current will be about 109 mA. (It will be a bit less than this as the fan must be running from a bit less than 12 volts as the voltage drop across the 60 ohm resistor will be
60 ohms x 0.109 amps = 6.54 volts. (So the resistor will be dissipating 6.54 volts x 0.109 amps = 0.71 watts. As the voltage on the fan (And across the zenner) is 18 - 6.54 = 11.46 volts the zenner should not be conducting so it will not be dissipating any power.
You would have to look at the graphs on the data datasheet for the particular zener being used to see if it passed any current at this voltage. In practice when the battery was new the voltage would probably be a bit more than 18 volts so in that case the zener could conduct. In practice I don't think the zener serves any purpose but if it is fitted then I think it should be rated to deal with about 100 mA in case the fan fails open circuit. In that case it would dissipate 1.2 watts. 12 x 0.1 amp.) I would choose a 2 watt rating 60 ohm resistor even though a 1 watt device should be good enough.
A 9V name-brand alkaline battery has a detailed datasheet. It shows 700mAh for a Duracell or 800mAh for an Energizer when the load current is only 2mA and the voltage has dropped to 4.8V.
With your load of 110mA then voltage drops so quickly that you will see the fan slowing down and the LED dimming.
I think the batteries will get fairly hot as they are draining so quickly.
I don't know if my circuit will work or not but I would appreciate your replies. (See attached files.) I was thinking that a 7812 would do a better job? Maybe one of you guys could post a circuit that would run the fan with batteries for an hour or so. Thank you for your help.
The batteries are too weak to power the zener diode, fan and LED. A 7812 voltage regulator IC can replace the 60 ohms resistor and zener diode but it will have the same problem caused by using batteries that are too weak.
Ten AA alkaline batteries will allow the fan and LED to work maybe double or triple the time than the two 9V batteries will last.
I agree with AG. The type of 9 volt batteries that you planned to use to use are totally unsuitable for for that load. Even using 10 alkaline AA cells seems a very expensive solution. as they would probably cost £8 to £10 pounds to run it for a few hours. An LM7812 is a better solution than a resistor and zener diode but an LDO type regulator would extend the battery life as the LM7812 had a dropout voltage of about 2 volts. LDO's have a lower dropout voltage. As you have a very limited electrical knowledge I have ruled out the use of Li-ion cells as these require well designed protection circuits for them to be used safely. One way that you could use them is to change to a 5 volt fan and supply it from a power bank. One of these will have all the battery protection circuits built in.