About electronics: Monostable Multivibrator

Thursday, September 13, 2007

Monostable Multivibrator

Introduction

In the previous experiment, you activated the R and S inputs to the latch circuit by turning a switch on, then off again. While this works, sometimes it would be nice to accomplish the same purpose with a single input action. That is, we'd like to be able to use a single input action to cause a digital pulse to be generated.

The circuit you'll demonstrate in this experiment is a variation in the bistable multivibrator circuit you have already seen in action. The difference is that we're going to modify the circuit so that once switched to the Set state, it will delay, then reset itself with no further intervention. This will give us the behavior we need.

For the circuit to behave this way, it needs to have one stable state (Reset), while the other state is not permanently stable. In practice, the Set state is quasi-stable in that it can be retained for a set period before the circuit reverts back to its stable state. In this experiment, we'll find out just how this can be accomplished.

Schematic Diagram

As you can see in the schematic diagram to the right, the monostable multivibrator is very similar in design to the bistable multivibrator you have already demonstrated. The primary difference is the use of a capacitor (C in the schematic) as one of the cross-coupling elements. The resistor is still present (R in the schematic), but now connects the base of Q2 to +5 volts instead of to the collector of Q1.

Of course, the capacitor will take a certain amount of time to charge, but once it does so it will carry no current, and Q2 will be turned on by the current through its 15K base resistor. This in turn holds the Q output at logic 0. This output is also applied as before, holding Q1 off. Assuming the T (Trigger) input is also quiescent at logic 0, Q3 is also off and the circuit will remain indefinitely in this state.

At this point, C is charged to just about +5 volts (less V_BE of Q2), with the Q1 collector connection being positive. The circuit will remain in this state until a logic 1 signal is applied to the T input.

When an input signal is received at T, Q3 turns on and pulls the left end of capacitor C down to ground. Since the capacitor voltage cannot change instantaneously, this forces the right end of C to -5 volts, immediately turning Q2 off. This in turn applies a logic 1 to Q1's input, turning Q1 on. At this point, the input to T can be discontinued; the Q output is logic 1 and Q1 will remain on.

Under these circumstances, the left end of C remains locked to ground through Q1's collector. But the right end gradually charges through R, Q2's base resistor, towards +5 volts. However, it never gets there; as soon as this voltage allows Q2's base to become forward biased, Q2 turns on and turns Q1 off again. This returns the circuit to its quiescent state.

Thus, this circuit cannot maintain a logic 1 output indefinitely; this is not really a stable state for this circuit. The circuit has only one stable state (Q = 0). It is therefore known as a monostable multivibrator.

The duration of the quasi-stable state (Q = 1) is determined by the two components R and C. Because the capacitor only charges to half the total range (from -5 volts to 0, while charging towards +5 volts), the duration of the output pulse is 0.693RC, where 0.693 is the natural logarithm of 2, R is in ohms, and C is in farads. For the component values shown here, the timing interval is 0.693 × 15,000 × 0.0001 = 1.04 seconds. So this circuit will produce a 1-second pulse each time it is triggered.

If the T input has already returned to logic 0, C will rapidly recharge through the 1K collector resistor and be ready for another input trigger signal. If T remains at logic 1, C will remain discharged until T drops again to logic 0. Then C will fully recharge in about 0.5 second and be ready for another trigger signal.