Quantcast Negative counters diodes D1 and D2 in the positive-counter circuit ( view (A) of figure 4-43) will cause the circuit to respond to negative pulses and become a negative-counter circuit. Diode D2 conducts during the time the negative pulse is applied and current flows in the opposite direction through R1, as was indicated by the arrow. At the end of the negative pulse, D1 conducts and discharges C1. The current through R1 increases with an increase in pulse frequency as before. ">

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Negative Counters

Reversing the connections of diodes D1 and D2 in the positive-counter circuit ( view (A) of figure 4-43) will cause the circuit to respond to negative pulses and become a negative-counter circuit. Diode D2 conducts during the time the negative pulse is applied and current flows in the opposite direction through R1, as was indicated by the arrow. At the end of the negative pulse, D1 conducts and discharges C1. The current through R1 increases with an increase in pulse frequency as before. However, if the voltage developed across R1 is applied to the same control circuit, as shown in view (A) of figure 4-44, the increase in current will be in a negative direction and the amplifier will conduct less. Thus, the effect is opposite to that of the positive counter.

Step-by-Step (Step) Counters

The STEP-BY-STEP (STEP) COUNTER is used as a voltage multiplier when a stepped voltage must be provided to any device which requires such an input. The step counter provides an output which increases in one-step increments for each cycle of the input. At some predetermined level, the output voltage reaches a point which causes a circuit, such as a blocking oscillator, to be triggered.

A schematic diagram of a positive step counter is shown in view (A) of figure 4-45. For step counting, the load resistor of the positive-counting circuit is replaced by capacitor C2. This capacitor is relatively large in comparison to C1. Each time D2 conducts, the charge on C2 increases as shown in view (B). The steps are not the same height each time. They decrease exponentially with time as the voltage across C2 approaches the input voltage.

Figure 4-45A. - Basic step counter and waveforms.

Figure 4-45B. - Basic step counter and waveforms.

As long as C2 has no discharge path, the voltage across its terminals increases with each successive step until it is equal in amplitude to the applied pulse. The voltage across C2 could be applied to a blocking-oscillator circuit to cause the oscillator to pulse after a certain amount of voltage is applied to it.

The circuit in figure 4-46, (view A) and (view B), may be used as a frequency divider. When used in this manner, Q1 is used as a single-swing blocking oscillator that is triggered when the voltage across C2 becomes great enough to forward bias Q1. At other times, the transistor is cut off by the bias voltage developed in the section of R2 that is between the ground and the slide.

Figure 4-46A. - Step counter as a frequency divider and waveforms.

Figure 4-46B. - Step counter as a frequency divider and waveforms.

The action of the counter can best be understood by referring back to figure 4-45. Assume C2 is 10 times larger than C1 and the peak voltage is 10 volts. C1 will assume 9/10 of the positive input voltage at T0, while C2 will assume only 1/10, or 1 volt in this example. At T1 the input will drop in a negative direction and D2 will be cut off. The cathode of D1 will become more negative than its anode and conduct, discharging C1. The charge on C2 will remain at 1 volt because it has no discharge path. At T2 the second pulse will be applied. The 1-volt charge on C2 will oppose the 10 volts of the second pulse, and the applied voltage for the capacitors to charge will be 9 volts. C2 will again charge 10 percent, or 0.9 volt. This is in addition to the initial charge of volt. At the end of the second pulse, the voltage on C2 will be 1.9 volts. At T3 the third pulse will be 10 volts, but 1.9 volts will oppose it. Therefore, the applied voltage will be 10 - 1.9 volts, or 8.1 volts. C2 will charge to 10 percent of 8.1 volts, or .81 volt. The voltage on C2 will become 1 + .9 + .81, or 2.71 volts. Successive input pulses will raise C2 by 10 percent of the remaining voltage toward 10 volts until the blocking oscillator works. If the oscillator bias is set so that Q1 begins conduction at 3.8 volts, this will continue until 3.8 volts is exceeded. Since the fourth step is 3.5 volts and the fifth is 4.1 volts, the 3.8-volt level is crossed at the fifth step. If the oscillator goes through 1 cycle of operation every fifth step and C2 is discharged at this point, this circuit would be a 5-to-1 divider. The circuit can be made to divide by 3, 4, or some other value by setting the bias at a different level. For example, if the bias is set at 2.9 volts, conduction will occur at the fourth step, making it a 4-to-1 divider.

The counting stability of the step counter is dependent upon the exponential charging rate of capacitor C2. As C2 increases to higher steps, the voltage increments are less and less. If the ratio becomes too great, the higher steps become almost indiscernible. For this reason, accuracy decreases as the ratio increases. When you desire to count by a large number, 24 for example, a 6-to-1 counter and a 4-to-1 counter are connected in cascade (series). A more stable method of counting 24 would be to use a 2:1, 3:1, 4:1 counter connected in cascade. Most step counters operate on a ratio of 5 to 1 or less.

Q.27 What is the difference between a positive counter and a step counter? answer.gif (214 bytes)




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