EE 322

Lab 6: THE 555 TIMER

The purpose of this lab is to investigate the use of the 555 timer integrated circuit as an oscillator and as a monostable multivibrator for generating pulses and delays.

The 555 timer uses two comparators and a digital flip-flop to control an output driver and a `discharge' transistor (see sketch below). The comparators switch state at 1/3 and 2/3 of the supply voltage (Vcc), called the `trigger' and `threshold' levels respectively. When the trigger input (Tr) falls below (1/3) Vcc the output voltage goes high and the discharge transistor behaves like an open switch. When the threshold input (Th) rises above (2/3) Vcc the output goes low and the discharge transistor shorts the discharge terminal (pin 7) to ground.

Astable or relaxation oscillator

When connected as shown below, the 555 functions as an `astable' (unstable) multivibrator or `relaxation' oscillator. C charges up through RA and RB until its voltage VC reaches the threshold level (2/3) Vcc. The discharge transistor then turns on connecting the junction between RA and RB to ground and discharging (relaxing) C until VC = (1/3) Vcc, whereupon the discharge transistor turns off and the process is repeated. The period of oscillation is given by

T = 0.69 ( RA + 2 RB) C.

  1. Design your circuit to have an oscillation period T = 2 ms, with RA = RB and C = 0.1 uF. Construct and test the circuit, using the dual-trace capabilities of your scope to look at both the output and the capacitor voltage (be sure to use d.c. coupling and keep track of 0 VDC). Make a hard copy of or sketch the waveforms and explain them. How does the observed period/frequency compare with the theoretically expected values? What is the purpose of the 0.01 uF capacitor on the control voltage terminal?

  1. Why is the output pulse at a high level longer than it is at a low level? Does the ratio of the charge and discharge times agree quantitatively with your explanation?
  2. Short out resistor RB. How does this affect the waveforms, and why? The slope of the discharge waveform can be used to be used to compute the current drawn by the internal discharge transistor of the 555. Derive an expression for calculating the discharge current and evaluate (Hint dV/dt=I/C for a capacitor). Assuming that the discharge transistor has a beta of 100, what base current is supplied by the internal RS flip-flop of the 555?
  3. Restore RB to its original value and replace RA by a 470 ohm resistor. What do the waveforms look like now, and why? Why do we not short out RA?
  4. Vary Vcc between +5 +15 volts to see how the output frequency depends upon the supply voltage. Does it? Use a frequency counter to measure the frequency at 5, 10, and 15 V.

Voltage-Controlled Oscillator

Often a circuit requires a voltage-controlled oscillator, that is, an oscillator who's frequency depends on an input voltage. The frequency of the astable multivibrator above depends only on the values of RA, RB and C. By using a voltage-controlled current source to charge C, one can adjust the charging rate and hence the frequency. If a current source charges the capacitor with a constant current I, as in the circuit below, the time for the capacitor to charge from VTh to VTr is

tch = C (VTh-VTr)/I = C Vcc /(3 I).

The circuit below will discharge virtually instantaneously, so the frequency of oscillation will be

fosc = 1 /(tch+tdis) =1/ tch = 3 I/ (C Vcc).

  1. Return Vcc to 5 V for the 555 timer.
  2. The voltage-controlled current source below is from Figure 4.11A of Horowitz and Hill. Build this current source. Use +-15 V power supplies for the op-amp. Select R such that the frequency of oscillation will vary from 5 kHz to 10 kHz as Vin is varied from 10 V to 5 V.

  1. Now use this current source to charge the capacitor in the 555 timer circuit. Record and plot fosc vs. Vin.

Monostable Operation

The circuit shown below produces an output pulse whenever the trigger input (Tr) falls below (1/3) Vcc. The duration of the output pulse is T = 1.1 RA C. After this time the output remains `stable' at 0 volts until the next trigger pulse occurs.

  1. Design the circuit to have a 1 ms pulse length and trigger it with short-duration, negative-going pulses as shown.
  2. The input trigger signal can be obtained by appropriately setting your pulse or signal generator. Look at the pulse from the signal generator with the scope to adjust it. A good pulse signal would be 100 Hz going from 5 V to 0 V for 0.1 ms each cycle.
  3. Look at the output and capacitor waveforms. Sketch or copy and explain.
  4. Measure the output pulse width and compare with its theoretical value. Vary the width of the trigger pulse to determine its effect upon the output pulse width, and describe the results.
  5. The monostable can be used to generate long time delays by increasing the values of RA and C. Try generating a 10 second delay (pulse) using a l00 uF electrolytic capacitor. Trigger the circuit manually as shown below. The switch can be just a wire you touch to ground to start the 10 second pulse.
  6. How does the pulse length compare with theory? (Note: You can use a long time base and the `scroll' mode on your digital scope to watch the capacitor charge and to measure the time delay.)

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