A very useful electronic circuit is one which is set into one of two stable states by one input switch and will not change its state until the other input switch is pressed. An example of a use of such a circuit is a burglar alarm fitted to a car, the alarm comes on when the burglar opens the door and does not go off until the owner presses a reset switch within the car.
Such a circuit may be made with two NAND gates
connected as shown in Figure 1.
The
two inputs are two simple push switches, they will be LOW when the switch is pressed
but spring open to become HIGH when the switch is released.
We will call the
inputs S (set) and R (reset) and the two outputs P and Q. The other inputs to the NAND
gates are labelled B and C.
Initially suppose that P is LOW (0) and Q is HIGH
(1). Now if S is pressed the output P becomes HIGH (1) and the output Q becomes
LOW (0) since C becomes 1. The input B therefore becomes LOW (0).
Releasing S and then pressing it again will not alter the output states. This can
only be done by pressing R.
| S | R | B | C | P | Q |
| 1 | 1 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 |
A
circuit using a NAND gate bistable that will act as a fire alarm is shown IN Figure
2.
To start with the buzzer is off and so the output from gate A must be 0. This
means that the output from gate B must be 1. The thermistor is cold, its resistance is
high and so the two inputs to gate A are both 1. When the thermistor is warmed its
resistance falls and the lower input to gate A becomes 0. Gate A now switches and its
output becomes high (1) and the buzzer sounds.
See if you can show that
cooling the thermistor will not switch the alarm off.
To do this the reset switch R
must be pressed.
In the section on transistors we considered the transistor astable multivibrator. Figure 3 shows the NAND gate version of this useful circuit.
We will call the output voltage from the
circuit V. Imagine that initially the input to A is high, this means that the output from A is
low and the output from B is high. If the output from B is high the input to B must be
low.
The capacitor C therefore charges up through R and the potential at the
input to B rises. When this potential is high enough B switches making the output to B
low. This means that the input to A is low and therefore the output from A is high. C now
discharges through R until B switches to a high output; the process then repeats itself.
The switching rate and therefore the output frequency depends on the values of R and
C.