So i have a 9 volt battery, a peltier cooler 12706 that requires 12 volts, and a fan that requires 5 volts. Would it be ok to power the peltier cooler with the 9 volt battery without completely destroying it? Is it the same with the fan? and to connect the circut would it go from the battery to a switch, then split into a paralell circut to have the fan on top of the cooler connect back together and return to the other end of the battery?
I do not think you will have good luck with that. Running the peltier at lower voltage than spec means you will not get the spec'd cooling from it......this may be ok for your application....but not good if you need the specified amount of cooling. Shortened component life would be expected if inadequate cooling is supplied. Running the 5 v fan with the 9 volt battery will likely work for a short while (perhaps very short while) until something burns. Get a 12 v battery and a 12 volt fan or run the fan with a resistor in-line with it to limit current .
My 2 cents. You can always experiment ...if you can afford some fried components. Others may be much more knowledgeable on this forum.
So the peltier cooler will be fine at 9 volts but the fan will fry very quickly? In that case would i need to include a ic 7805 right before it reaches the fan and then connect it back to the main circut. But when the 5 volt wire gets re-connects with the 9 volt wire will nothing happen? I also dont want to burn the battery fast.
It is fine if the peltier works at the lower standards in terms or temperature. It is not cooling something that will become too warm.
+ 9v goes to the IC (input) and the cooler +
- 9v terminal gos to the 'ground' of the IC and the cooler -
'output' of the IC goes to the fan......then the other end of the fan gets connected to the ground (-9v wire) also.....
I do not have access to a scanner right now.....hope you can draw it....
Hope your battery has enough 'oomph' to run the cooler and fan and IC. The IC has a max output of 5v 1 amp......hope your fan requires less than 1 amp.
If you know the current draw of the motor , it might be simpler to in-line a resistor with the motor and skip the IC... (like a 20 ohm resistor if your motor draws 200 mAmp)
FYI from the 7805 spec page ( you might want to read it over)
The greater the difference between the input and output voltage, more the heat generated. If the regulator does not have a heat sink to dissipate this heat, it can get destroyed and malfunction. Hence, it is advisable to limit the voltage to a maximum of 2-3 volts above the output voltage. So, we now have 2 options. Either design your circuit so that the input voltage going into the regulator is limited to 2-3 volts above the output regulated voltage or place an appropriate heatsink, that can efficiently dissipate heat.