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There may be instances when it is necessary to design a simple pulse generator circuit using a bipolar junction transistor, BJT, as the basic active component.
Understanding how to design a pulse generator circuit using a BJT adds another tool to the electronic circuit designers toolbox.
These pulse generator circuits are easy to design and can be used in a variety of scenarios where they can provide an easy solution to a problem.
Being transistor based, the circuits are straightforward and the electronic components are easily available - there is no need for a specific integrated circuit.
The simple pulse generator circuit uses a CR network as the basis of its operation.
Using a series capacitor and a resistor to ground as the network, this acts as a differentiator. The rising edges pass a 'spike' through the capacitor as the capacitor charge changes to accommodate the different voltage across it.
However as the charge changes to accommodate the new voltage, so the voltage at the output falls away.
This CR network forms the basis of the transistor pulse generator circuit design, using the capabilities of the transistor, the pulse becomes more of a square wave pulse in the way one is normally conceived, and the negative going pulse is removed from the output.
Simple transistor pulse generator circuit
It is very easy to create a simple pulse generator circuit design using just two transistors and a handful of other electronic components.
Although the design is not as sophisticated as some may be, it suits very well for a number of uses, and it uses one transistor and only a few other electronic components.
This simple circuit design gives a negative going pulse from a positive going input - this may be suitable for some applications, and it has the advantages of great simplicity.
This basic pulse generator circuit uses a differentiator to provide the pulse and then it uses a transistor to provide the pulses switching between virtually the rail voltage and 0V, while inverting the pulse.
For this circuit, typically the value of R2 » R1. Also the pulse width will be of the order of τ = R2 C.
It should be remembered that a circuit like this will be a little unpredictable in terms of the actual time for the pulse. A transistor pulse generator like this will offer satisfactory levels of performance for some applications, but not where a high level of performance is needed.
Against, this, the transistor based pulse generator circuit is very easy to realise.
Non-inverting pulse generator
It is often the case that a positive going pulse is required to be produced from a positive going edge. This requires a slightly more complex circuit which uses two transistors and a handful of other electronic components.
iI is this scenario that is more likely to be required, and therefore it will be investigated in a little more detail.
Essentially the circuit consists of two transistor switch circuits that are capacitively coupled.
For the sake of the explanation of the circuit, let's assume the rail voltage is 5 volts as this ties in nicely with the traditional logic standard and many signals conform to this standard.
In this circuit a rising edge is applied to the first transistor. The first resistor R1 acts to define the base current required. Typically it might be in the region of 4k7 to 10k for a 5 volt TTL style input signal. A 4k7 resistor would limit the current to around 1mA and this would enable a collector current of around 50mA for a transistor with a β or hFE of around 50.
The transistor TR2 is normally held in saturation by the current passing through resistor R3. Typically this resistor might also be between 4k7 and 10k, although a higher value will enable the pulse from the previous stage be defined more accurately by R2 and C1. However it should be sufficient to ensure that TR2 is ON and remains in saturation.
In terms of its operation, when the input to the overall circuit is low, TR1 is OFF and the collector is HIGH, i.e. at the rail voltage.
In this state, the capacitor C1 is charged up with the positive potential on the terminal connected to TR1. As the terminal connected to the base of TR1 is about 0.7 volts above ground because the base has current flowing into it via R3.
The potential across the capacitor will be the rail voltage less the base emitter voltage of TR2, i.e. 4.3V if a 5 volt rail is used.
If a positive step input is applied to the input of TR1 via the resistor R1, this turns TR1 ON, bringing it into saturation. The voltage at the collector terminal falls to the transistor saturation voltage which is about 0.2 volts above ground.
This means that the base of TR2 is temporarily brought down to a voltage of the capacitor voltage less the saturation voltage of TR1, i.e. 4.3 - 0.2 = 4.1 V.
This turns of TR2, because the base voltage is below its turn on voltage, and as a result the collector of TR2 rises in voltage to the rail voltage.
However the capacitor C1 cannot hold the base of TR2 below 0V for long as the base current flows through R3 and this side of the capacitor starts to charge up towards the rail voltage through R3.
As the base voltage rises above 0.7 V, the transistor TR2 starts to conduct again and the voltage on its collector falls to VCEsat above ground.
This has a time constant of C1 . R3 (capacitance in Farads and resistance on Ohms). The time constant is given the letter τ, although the exact triggering point will have an impact on the time the collector of TR2 remains high.
It is necessary to remember that the rail voltage used for this form of pulse generator must not be above about 7 volts, otherwise the negative voltage to which the base of TR2 is exposed as the capacitor drives the base emitter junction negative could cause a catastrophic breakdown of the junction.
This simple pulse generator circuit takes a rising edge of a waveform and converts it into a short pulse that can be used with a variety of logic functions. In order to use it with logic devices it may require the edges to be speeded up, and this can be achieved using a buffer or two inverter gates. Although these introduce a delay, this is ot likely to be an issue as the transistor circuit itself will be relatively slow.
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