Production of 74HC00 multivibrator

1. Circuit and working principle

The circuit is shown in the figure below. 74HC00 is a four-one-two input NAND gate.

If one input of the two-input NAND gate is connected to a high level, or the two inputs are short-circuited, its output will be inverted with the remaining input or the two short-circuited inputs, which is equivalent to an inverter. In the circuit shown in the figure below, if the ① pin of IC1A and the ⑤ pin of IC1B are high (K1 is pressed, K2 is disconnected), then IC1A can be regarded as an inverter with the ② pin input and the ③ pin output, and IC1B can be regarded as an inverter with the ④ pin input and the ⑥ pin output. Its transfer characteristics are shown in the figure on the right. Due to the negative feedback of R1, if the voltage of the ② pin is low and the ③ pin outputs a high voltage, the ② pin level is pulled up through R1; if the voltage of the ② pin is high and the ③ pin output is low, the ② pin level is pulled down through R1, and the compromise stops at the center point C. The output is 100% fed back to the input, which is equivalent to folding the lower left triangle part to the upper right corner according to the dotted line. The intersection point C of the dotted line and the transfer characteristic is the working point of the inverter, which is approximately equal to 1/2VCC. Point C is located at the center of the steep slope of the transfer characteristic. In this example, a 1mV change in the 74HC00 input can cause the output to change by as much as 1V.

Since IC1's ③ and ④ are pressed continuously, the signal output from its ⑥ pin is in phase with that from the ② pin but with amplified amplitude. In the figure, C1 plays a positive feedback role. As long as the voltage of the ② pin has a slight fluctuation, such as increasing by 0.1mV, the voltage of the ③ pin will decrease by 100mV, and then be inverted by IC1B, and the output voltage of the ⑥ pin will increase by more than 1V. This voltage change is sent back to the ② pin through C1, so that the voltage of the ② pin continues to increase until VCC+0.7V. At this time, the protection diode inside IC1 is turned on, so that the input voltage cannot be high, and the inverter working point stops at point D in the right figure. Point D is located on the horizontal line of the transfer characteristic, and the input change has almost no effect on the output. At this time, the ② pin of IC1 is high level, the ③ pin is low level, and the ⑥ pin is high level. Resistor R1 is connected between the ② and ③ pins. The ③ pin is the output end, and the internal resistance is very low. The ② pin is the input end, and the internal resistance is extremely high. The potential difference between ② high and ③ low makes the direction of the current I on R1 as shown in the left figure. The starting voltage of the discharge is VCC+0.7V, and the final voltage of the discharge is 0V.

The actual discharge stops near point C (1/2VCC). It takes about 1.1R1C1=1.1×(2.2×l0(6))×(0.1×10(-6)≈0.25s to discharge from VCC+0.7V to 1/2VCC.

At this time, pin ② becomes low, and is amplified by IC1A → pin ③ becomes high → IC1B amplifies → pin ⑥ quickly becomes low → C1 → pin ②. The positive feedback continues until the voltage of pin ② drops to -0.7V. At this time, the protection diode inside IC1 is turned on, so that the input voltage cannot be lowered, and the inverter operating point stops at point E. Point E is on the horizontal line of the transfer characteristic, and input changes have almost no effect on the output. The state at this time is ② low, ③ high, and ⑥ low. R1 charges C1. The charging starting voltage is -0.7V, and the charging final voltage is VCC.

It takes about 1.1R1C1=0.25s to charge from 0.7V to 1/2VCC, then the charging stops, enters positive feedback, and turns to the working point D. In fact, the circuit works in the D and E states for a long time, and the time in C is very short, so the output is a square wave, and one cycle is about 0.5s. Square waves have more harmonics than sine waves, and sound more pleasant. The output signals of many music films are composed of equal-amplitude square waves of different frequencies. If the amplitude can change with the beat, it will sound better. Similarly, IC1C's (13) pin = high, IC1D ⑨, ⑩ are connected in parallel, and can also be regarded as two inverters, generating a square wave oscillation with a period of 0.5ms. That is 2kHz. Because the electro-acoustic conversion efficiency of the buzzer is the highest when the resonant frequency is around 2kHz, it sounds the loudest.

When the capacitor is 1000pF, the resistance is about 250kΩ. In the figure below, adjust the 500kΩ potentiometer to the center to find the maximum volume point. You can also try using a fixed resistor of 240kΩ to 270kΩ for R2.

Because the resistance of the buzzer is about 40Ω, and the output impedance of IC1 is about 1kΩ, IC1 cannot drive the buzzer directly, so it needs to pass through Q1 for current amplification. IC1C⑧ pin outputs a high level of 3V, and the voltage of Q1 base is 0.7V when it is turned on. R3=1Ω, Q1 base current is (3-0.7)/1k=2.3mA, Q11 amplification factor is 50, and collector current is 115mA. However, a 40Ω buzzer only needs 70mA to drive, and the voltage across the two ends reaches 2.8V. So where does the current of 115-70=45mA go? Q1 amplification factor is 50, which means that Q1 is in the linear amplification area Ic/IB, and in the saturation area, IG/IBF drops, and the tube voltage drop of Q1 is very low.

Control function of the AND-NON logic: Pin 1 of IC1A is usually grounded through R4, pin 3 outputs constant high, pin 4 = pin 3, and pin 6 outputs constant low. (13) pin = pin 6 = low, pin 8 is low, and the buzzer does not sound. The entire circuit consumes very little power.

After K1 is pressed, the (13) pin is high level, IC1D and IC1C generate a 2Hz square wave, and control the IC1D (13) pin. When the (13) pin is high level, IC1D and IC1C generate a 2kHz square wave through R3 and Q1 to drive the buzzer; when the (13) pin is low level, IC1D and IC1C stop oscillating, the ⑧ pin outputs a low level, and Q1 is turned off. As a result, the buzzer emits an intermittent beep twice per second. IC1B ⑤ pin is usually connected high through R5 and works normally. After K2 is pressed, ⑤ pin is low, and IC1A and IC1B stop oscillating. (13) pin = ⑥ pin is always high, and the buzzer emits a continuous beep.

2. Assembly of general circuit boards

No matter how many components there are during assembly, careful planning should be done before inserting the components on the circuit board, and cross-wiring should be avoided as much as possible.

As shown in the figure below, the pin arrangement of IC1 is counted counterclockwise from the notch. The 7 pins at the bottom are ① to ⑦, R1 is close to ① to ③, C1 is close to ⑥, ⑦ is ground, and ⑦ is close to R4, K2, and the E pole of Q1. The 7 pins at the top are (8) to (14), R2 is close to (13) to (11), C2 and R3 are close to ⑧, the B pole of Q1 is close to R3, BP is close to the C pole of Q1, etc. After this layout, with IC1 as the center, the upper and lower bare copper wires or resistor pins are connected radially on the plane of the general circuit board, and the connection lines can basically not be staggered. When checking, also use IC1 as the center and check one pin at a time.

3. Debugging components

After all installations are completed, do not rush to power on the test. First, connect an ammeter in series between the power supply + and the VCC of the test board. Under normal circumstances, the current is extremely small. If the current is greater than 500μA and unstable, it means that the input terminal is suspended or the solder joint is poor. You can check it like this: touch the metal part of the screwdriver with your body, and use the screwdriver head to touch each pin of 1C1 in turn. If the current changes significantly, it means that there is a problem with the wiring of the pin. If the current is normal, set the ammeter to 100mA (200mA). If the current changes but there is no sound when K1 is pressed, check R3, Q1, and BP; if the current is very small and there is no sound when K1 is pressed, check IC1.

1. Check the BP, use tweezers to short-circuit the C and E poles of Q1 for a moment, the current will change, and the buzzer should make a "ka-ba" sound. If not, it means that the BP is broken or the line is not connected.

2. Check the transistor. Use tweezers to connect VCC or ground momentarily at the end of R3 near IC1. Connect back and forth. The buzzer should make a "click" sound. If not, it means the transistor is broken, connected in reverse, or the wire is not connected (a momentary short circuit generally does not damage the IC).

3. Check IC1, press K2, and use a digital multimeter to measure the voltage. The (8)~(12) pins of IC1 should be neither high nor low. For example, in the case of a 3V battery, the measured value is about 1.5V. If ⑧ is always low, (9), (10), and (11) pins are always high. If (13) is high, it means that K2 works when pressed. If (12) is low, it means that the negative feedback of R2 does not work. Use the red test lead to short-circuit the (11) and (12) pins, and the black pen to ground. The voltmeter shows that it is neither high nor low. R2 may be poorly soldered. If (13) is low, it means that K2 does not work when pressed. Check pins ⑥ and ⑤. If pin ⑤ is high, it means that switch K2 is broken or not connected properly. If pins ⑥ and ⑤ are both low, it means that the IC is broken or the solder joint is short-circuited to the ground. When K1 is pressed, the voltage reading of pin ⑥ is unstable, sometimes high and sometimes low, indicating that the 2Hz oscillation is normal; if it is always high or low, you can also short-circuit pins ② and ③ with the red test lead to check the fault. The method for detecting when K1 is pressed is similar.