How does this complementary multivibrator work?

xljin2006

How does this complementary multivibrator work? [Copy link]

 

As shown in the picture.

maychang

The NPN tube Q1 and the PNP tube Q2 must be turned on and off at the same time.

After power-on, current flows through resistor R1, but capacitor C1 is not charged and the voltage across both ends is close to zero, so Q1 cannot be turned on, Q2 cannot be turned on, and the current in speaker S is very small.

As C1 charges, the voltage across both ends rises, the current (charging current) through C1 decreases, Q1 turns on, Q2 also turns on, the voltage across the speaker S approaches the power supply voltage, and C1 discharges.

Then as C1 discharges, Q1 turns off and the entire circuit starts over.

maychang

From another perspective, Q1 is a common emitter amplifier, and Q2 is also a common emitter amplifier. The output (collector) of the common emitter amplifier is inverted with the input (base), so after two inversions from the base of Q1 to the collector of Q2, it is a common-phase amplifier, and the gain is very high. From the output of Q2 to the base of Q1, capacitor C1 is connected, so relaxation oscillation is generated, and the frequency is determined by C1 and R1.

xljin2006

1. Why does C1 discharge when the voltage across speaker S is close to the power supply voltage?

2. It is said that the resistance of R1 cannot be too small, otherwise the circuit will not vibrate. If the speaker is replaced with resistor R2, R2 cannot be too large, and R1>>R2 is required. Why is this?

maychang

xljin2006 posted on 2023-6-12 22:34 1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? 2. It is said that the resistance of R1 cannot be too small, otherwise the circuit will not vibrate. If the speaker...

[1. Why does C1 discharge when the voltage across the speaker S is close to the power supply voltage? ] The moment when C1 starts to discharge is the moment when Q1Q2 turns from the off state to the on state. Because Q2 is saturated and turned on, the voltage across the speaker S rises from close to zero to close to the power supply voltage. Before this moment, C1 has been charged by R1 and S to a voltage across it exceeding 0.7V (Q1 emitter junction voltage drop), so after the voltage across S rises to close to the power supply voltage, C1 discharges through the Q1 emitter junction.

maychang

xljin2006 posted on 2023-6-12 22:34 1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? 2. It is said that the resistance of R1 cannot be too small, otherwise the circuit will not vibrate. If the speaker...

[2. It is said that the resistance of R1 cannot be too small, otherwise the circuit will not vibrate]

If R1 is too small, it is not just a problem that the Q1 cannot oscillate. Instead, the base current of Q1 is continuously supplied by R1. The power flows through the emitter junction of Q2 and then through the collector-emitter junction of Q1 to ground. This current may be large enough to damage Q1.

maychang

xljin2006 posted on 2023-6-12 22:34 1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? 2. It is said that the resistance of R1 cannot be too small, otherwise the circuit will not vibrate. If the speaker...

[If the speaker is replaced with resistor R2, R2 cannot be too large, and R1>>R2 is required. Why is this? ]

If R1 is not much larger than R2, for example, R1 is close to R2, then after power-on, the base potential of Q1 is higher than 0.7V, Q1 is continuously turned on, Q2 is also continuously turned on, and the circuit cannot oscillate.

xljin2006

maychang published on 2023-6-13 06:43 【1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? 】 The moment when C1 starts to discharge is Q ...

I can't read the entire reply, please send it again.

maychang

xljin2006 posted on 2023-6-13 11:06 I can't read the reply in full, please post it again.

Copy it once and use a larger font.

[1. Why does C1 discharge when the voltage across the speaker S is close to the power supply voltage? ]

The moment when C1 starts to discharge is the moment when Q1Q2 turns from the off state to the on state. Because Q2 is saturated and turned on, the voltage across the speaker S rises from close to zero to close to the power supply voltage. Before this moment, C1 has been charged by R1 and S to a voltage across it exceeding 0.7V (Q1 emitter junction voltage drop), so after the voltage across S rises to close to the power supply voltage, C1 discharges through the Q1 emitter junction.

Can you see clearly?

xljin2006

maychang published on 2023-6-13 11:31 Copy it once, and use a larger font. [1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? ] C1 starts to discharge...

I thought C1 started to discharge when Q2 went from the saturation region to the amplification region, but I don't know why.

xljin2006

xljin2006 posted on 2023-6-13 17:58 I thought C1 started to discharge when Q2 went from the saturation region to the amplification region, but I don't know why.

The voltage of the collector of q2 drops, and is coupled to the base of q1 through c1. After positive feedback, the collector of q2 drops to about 0. The base of q1 is -(3-0.7)=-2.3v. Then the power supply passes through r1 and the speaker charges c1, and a new cycle begins. What I don't understand is:

xljin2006

xljin2006 published on 2023-6-13 18:12 The collector voltage of q2 drops and is coupled to the base of q1 through c1. After positive feedback, the collector voltage of q2 drops to about 0. The base of q1 is -(3-0.7)=-2.3 ...

After q2 is saturated, why does it exit saturation? Why does its collector voltage start to drop? Why doesn't it rise? Or remain the same?

xljin2006

maychang published on 2023-6-13 11:31 Copy it once, and use a larger font. [1. When the voltage across the speaker S is close to the power supply voltage, why does C1 discharge? ] C1 starts to discharge...

When the absolute value of the voltage across the capacitor increases, it is charging, and vice versa, it is discharging.

maychang

xljin2006 posted on 2023-6-13 18:19 When the absolute value of the voltage across the capacitor increases, it is charging, and vice versa, it is discharging.

When Q1Q2 is turned off, capacitor C1 has been charged by resistor R1. The voltage direction of both ends is positive on the left and negative on the right. The right end is close to zero because the voltage across S is very small. When Q1Q2 is turned on, the voltage across S changes from very small to close to the power supply voltage. The potential of the right end of C1 rises to close to the power supply voltage, and the potential of the left end is the voltage drop of Q1 emitter junction. Obviously, capacitor C1 needs to discharge through Q2 and Q1 emitter junction. During this period, the current direction in C1 is opposite to that during the period when Q1Q2 is turned off.

maychang

xljin2006 posted on 2023-6-13 18:15 After q2 is saturated, why does it exit saturation? Why does its collector begin to drop? Why doesn't it rise? Or remain the same?

[After q2 is saturated, why do we need to exit saturation? 】

Q2 goes from off to saturated conduction because C1 is charged. C1 is charged during the off period of Q1Q2. As the voltage across C1 gradually rises, the base potential of Q1 rises from close to zero to enough to turn on Q1. The conduction of Q1 makes Q2 turn on, and the conduction of Q2 makes the voltage across S rise. Therefore, the potential at the right end of C1 rises. The voltage across the capacitor cannot change suddenly, and the potential at the left end also rises. The potential at the left end of C1 rises, making Q1 turn on further, and Q2 turn on further, until Q1Q2 are both saturated. This process is very fast and takes only a very short time. This is positive feedback.

After Q2 is saturated, the voltage across S no longer increases, and C1 discharges through the emitter junction of Q2 and Q1 (described in the previous post). When C1 discharges, the voltage across the two ends gradually decreases, and the charge stored in the capacitor is not enough to saturate Q1. The collector current of Q1 will decrease, causing the collector current of Q2 to also decrease, the voltage across S decreases, and the potential of the right and left ends of C1 decreases. This process is also very fast, taking only a very short time, and the increase and decrease of the voltage at each point is exactly the opposite of the charging process of C1.

maychang

xljin2006 posted on 2023-6-13 18:15 After q2 is saturated, why does it exit saturation? Why does its collector begin to drop? Why doesn't it rise? Or remain the same?

[After q2 is saturated, why does it exit saturation? Why does its collector begin to drop? Why doesn't it rise? Or remain the same? ]

Q2 exits saturation, which means that Q2 collector current starts to decrease from the maximum value. As Q2 collector current decreases, of course collector potential decreases. Q2 collector current increases, and its collector potential will increase. Q2 collector current remains unchanged, and collector potential will be [maintained].

xljin2006

maychang published on 2023-6-13 18:34 Capacitor C1 has been charged by resistor R1 during the off period of Q1Q2, and the voltage direction of both ends is positive on the left and negative on the right. The right end is close to zero because the voltage across S is very small. Q1Q ...

I'm a newbie, how can I display the snippet to reply?

maychang

xljin2006 posted on 2023-6-13 19:15 I am a novice, how can I display the fragment to be replied?

[How do I display the fragment I want to reply to? ]

First highlight the part you want to reply to, copy the part you want to reply to, and paste it into your reply text.

xljin2006

This is simulated using Multisim. I don't know why it doesn't oscillate? It seems to be a speaker problem. The collector of q2 always outputs 3.013v, and the base of q1 always outputs 621.286mv.

maychang

xljin2006 posted on 2023-6-14 11:42 This is simulated using multisim. I don't know why it doesn't oscillate? It seems to be a problem with the speaker. The collector of q2 always outputs 3.013v, and the base of q1 always outputs 621.286m ...

[This is simulated using multisim, why is it not oscillating? It seems to be a problem with the speaker. ]

The speaker has two ends. Where is the other end of the speaker in the simulation picture?