#Pnp switch series

While Q1 is off, C1 can charge, relatively quickly through R1.Q2 being on allows C2 to discharge through Q2's collector.At this point we've swapped states: C1 is discharged, C2 is charged, Q1 is off, and Q2 is on.Once C1 fully discharges, its right plate will be pulled up to about 0.6V, which will turn on Q2.While C1 is discharging, C2 quickly charges through the lower value resistor - R4.This will allow C1 to discharge through Q1's collector. If Q1 is on, then C1's left plate (on the schematic) is connected to about 0V.To begin, assume C1 is fully charged (storing a voltage of about V CC), C2 is discharged, Q1 is on, and Q2 is off. The caps alternatively charge and discharge, which causes the two transistors to alternatively turn on and off.Īnalyzing this circuit's operation is an excellent study in the operation of both caps and transistors. By using feedback we can use a pair of transistors to create two complementing, oscillating signals.Īside from the two transistors, the capacitors are the real key to this circuit. Here's an example oscillating circuit, which we call an astable multivibrator. There are lots of ways to create an oscillator circuit including quartz crystals, op amps, and, of course, transistors. Oscillators are used in all sorts of circuits: from simply blinking an LED to the producing a clock signal to drive a microcontroller. The H-bridge has a truth table that looks a little like this: Input AĪn oscillator is a circuit that produces a periodic signal that swings between a high and low voltage. But if the two inputs are opposite, the motor will spin in one direction or the other. If both inputs are the same voltage, the outputs to the motor will be the same voltage, and the motor won't be able to spin. (Note: there's usually quite a bit more to a well-designed H-bridge including flyback diodes, base resistors and Schmidt triggers.) The value of the resistor, and voltage across it, will set the current.Ĭan you guess why it's called an H bridge? The base-emitter node can get its happy voltage drop of 0.6V, and the resistor can drop the remaining voltage.

#Pnp switch series

The series resistor between our control source and the base limits current into the base. If you supply a current over the maximum rating, the transistor might blow up. Some transistors may only be rated for a maximum of 10-100mA of current to flow through them. The diode only needs 0.6V to turn on, more voltage than that means more current. We're forward-biasing the base-emitter diode to turn the load on. Recall that, in a way, a transistor is just a pair of interconnected diodes. Don't forget to add this resistor! A transistor without a resistor on the base is like an LED with no current-limiting resistor.

You'll notice that each of those circuits uses a series resistor between the control input and the base of the transistor. In that case, it'd be impossible to turn the switch off because V B (connecting to the control pin) would always be less than V E. For example, this circuit wouldn't work if you were trying to use a 5V-operating Arduino to switch off a 12V motor.

This can cause complications, especially if the load's high voltage (V CC being 12V connecting to the emitter V E in this picture) is higher than our control input's high voltage. This circuit works just as well as the NPN-based switch, but there's one huge difference: to turn the load 'on', the base must be low. This time however, the emitter is tied high, and the load is connected to the transistor on the ground side. Similar to the NPN circuit, the base is our input, and the emitter is tied to a constant voltage.