h-bridgetransistors
I'm using the following circuit: Applied to the motor is 12V, and either high (5V) or low (0V) to each transistor through a 1kΩ resistor. The transistors are TIP 142 and TIP 147. The diodes are 1N5817.
When I apply 0V to R2, R1, and 5V to R3, the motor runs fine; voltage across it is 8-10V. However, whenever I connect 5V OR 0V to R4, the voltage across the motor is 1-2V, and the transistor at R3, starts getting really hot and smoking.
Why does applying anything to R4 cause this?
With the switches closed you would be trying to turn the relay on. As your circuit stands it doesn't work because the emitter voltage could never rise enough to power the relay. Even if it could it means you would be powering the relay most of the time - a waste of energy. Also, if the door/window was opened momentarily as soon as it was closed the alarm would stop.
By adding a second transistor Q2 (PNP) and a push to break switch you can improve your circuit (and get it to work).
The closed switches turn Q1 OFF (base connected to ground). As there is no current flowing through the 1K resistor Q2 is also OFF. No current can flow through the relay. The only current to flow is through R1 so the circuit will only draw 1.2mA.
If any switch is opened Q1 is turned ON and draws a current through R2 and R3. The voltage across R2 will be clamped by the emitter-base of Q2 to about 0.6V and will turn Q2 ON. With Q2 turned ON its collector current will turn the relay ON.
The diode (D1) is connected across the coil of the relay to prevent damaging Q2 from the back emf when the relay is turned OFF.
The relay switch over and closes the circuit between its common and normally open contacts.
The reset switch (push to OPEN) keeps the relay turned ON even if the door/window switches are closed after opening. It LATCHES the relay. Just omit the reset switch if you don't want this feature.
You'll notice that in Figure 2 it states "Ic/Ib = 10".
This means that to saturate the transistor you should drive it as if the Hfe is only 10, not the almost 100 in Figure 1. Note also that figure 1 is the typical figure not the guaranteed value. The guaranteed value in the data sheet is only 10 at 3A collector current.
You need to provide more base drive.
The PIC however will not be able to drive the required 180mA output. You need a darlington transistor such as a TIP120 or these days a better solution is to use a MOSFET. The TIP120 would have a saturated voltage drop of ~1V and so would require a heat sink. A mosfet would have a very low drop and would work without one.
Even there you have to be careful as many MOSFETs will require 10V drive, more than the PIC can provide. For this there are so called Logic Level FETs that switch with only 5V on the gate.
Something like FQP30N06L would work.
Best Answer
Q1 and Q3 turn ON when you connect +5V to their resistors. Q2 and Q4 turn OFF when you connect +12V to their resistors.
Connecting 5V to R2 or R4 will result in the transistor turning on.
So, if you connect 5V to R3 and less than 12V (or whatever the power supply voltage is) to R4, both transistors turn on creating a short circuit.
If you want to turn off Q2 or Q4, connect their resistors to +12V.
If you want to control the transistors with 5V (a microcontroller or whatever) then you can add two more ttransistors that control Q2 and Q4, like this:
Connecting 5V to R6 will turn Q4 on, connecting 5V to R5 will turn Q2 on.