Please help analyze this bistable circuit

静谧321

Please help analyze this bistable circuit [Copy link]

 

Please help me to explain the specific functions of this circuit; what are the uses of D1 and D2? What is the use of switch S and what happens when the switch is pressed? What components are needed to add if the function is unstable when the switch is pressed?

maychang

In a bistable circuit, in any stable state, one of VT1 and Vt2 must be saturated and turned on, while the other is turned off. Assume that VT1 is saturated and VT2 is turned off. Then the collector voltage of VT1 is very low and the collector voltage of VT2 is close to the power supply voltage.

As a result, the voltage across capacitor C1 is close to the power supply voltage (switch S is not pressed), and the voltage across capacitor C2 is close to zero.

maychang

When the switch S is pressed, the voltage between the lower end of capacitor C1 and the ground is zero. The voltage across the capacitor cannot change suddenly, and the voltage on the upper end of C1 is negative instantly. The diode D1 is turned on, pulling down the base voltage of VT1, and VT1 is turned off.

However, the voltage across capacitor C2 is close to zero before S is pressed. After switch S is pressed, the voltage at the lower end of C2 is zero, and the voltage at the upper end of C2 is also close to zero. VT2 is originally off, and the change in the voltage at the upper end of C2 has no effect on the off state.

maychang

VT1 is turned off, and the collector voltage is close to the power supply voltage, causing R3 to supply VT2 base current, turning it on and entering saturation. This process causes the bistable circuit to flip.

maychang

"When I press the switch, the function becomes unstable. What components do I need to add?"

Instability may be caused by the switch contacts shaking and closing and opening multiple times.

It is possible to consider adding a monostable circuit. The first closing of switch S causes the monostable circuit to flip (the opening of switch S during the quasi-stable state will not affect the quasi-stable state), and the bistable circuit flips, but the second closing of the switch has no effect.

The temporary stabilization time of the monostable circuit cannot be too long, otherwise it will affect the flipping frequency of the bistable circuit.

静谧321

maychang published on 2019-12-6 12:49 "When pressing the switch, the function will become unstable. What components need to be added?" Instability may be caused by the jitter of the switch contacts, multiple times...

Is there a design flaw in the switch position of this circuit?

静谧321

maychang published on 2019-12-6 12:49 "When pressing the switch, the function will become unstable. What components need to be added?" The instability may be caused by the jitter of the switch contacts, multiple times...

I am a little confused.

maychang

Quiet 321 posted on 2019-12-6 12:54 Is there a design flaw in the switch position of this circuit?

“Is there a design flaw in the switch position of this circuit?”

The switch is in the correct position.

As long as it is a mechanical switch, there will be contact jitter. It will be very expensive to eliminate the jitter of mechanical contacts.

maychang

Quiet 321 posted on 2019-12-6 12:57 I am a little confused.

“A bit confused”

The article may mean that you want to connect a capacitor in parallel across the switch.

After connecting a capacitor in parallel, the jitter can be eliminated because the capacitor is charged to the power supply voltage when the switch is opened, and the capacitor is discharged after the switch is closed. The discharge time is very short (the resistance is basically zero when the switch is closed), while it takes a certain amount of time for the capacitor to charge after the switch is opened.

LuJianchang

A capacitor is connected in parallel at both ends of the button, and debouncing can be achieved through the delay of charging and discharging of the capacitor.

静谧321

maychang posted on 2019-12-6 14:58 "A little confused" The meaning of this article may be that you want to connect a capacitor in parallel at both ends of the switch. After connecting a capacitor in parallel, the jitter...

Thank you for your answer

静谧321

LuJianchang posted on 2019-12-6 15:08 Connect a capacitor at both ends of the button, and the delay of the capacitor's charge and discharge can achieve debounce.

Thank you for your answer.

静谧321

maychang published on 2019-12-6 14:58 "A little confused" The meaning of this article may be that you want to connect a capacitor in parallel at both ends of the switch. After connecting a capacitor in parallel, the jitter...

I simulated it and found that this diagram does not realize the bistable function.

maychang

Quiet 321 posted on 2019-12-6 22:58 I simulated it and found that this diagram does not realize the bistable function.

“I simulated it.”

What does "simulation" mean? Simulating on a computer? Or actually building a circuit to test it?

maychang

Quiet 321 posted on 2019-12-6 22:58 I simulated it and found that this diagram does not realize the bistable function.

"This graph does not realize the bistable function"

So how does this circuit behave? Is it not stable but jumping around? Doesn't it flip when I press the switch?

FYzj

maychang published on 2019-12-6 12:43 When the switch S is pressed, the voltage between the lower end of capacitor C1 and the ground is zero. The voltage across the capacitor cannot change suddenly. The voltage at the upper end of C1 is negative instantly, and the diode D1 is turned on, which turns VT1 ...

Hello, I want to ask, if VT1 is turned on, shouldn't D1 be in the on state to charge C1? If S is not pressed, can VT2 be turned on? If it is turned on, does C1 discharge through R5 and R3 to provide voltage to the base of VT2?

maychang

FYzj posted on 2020-2-27 11:31 Hello, I would like to ask, if VT1 is turned on, shouldn’t D1 always be in the on state to charge C1? If S is not pressed, can VT2 be turned on? ...

Let's talk about the second question first: "If S is not pressed, can VT2 be turned on?"

When VT1 is saturated and turned on, VT2 cannot turn on itself. Because when VT1 is saturated, the voltage between the collector and the emitter is very small, usually less than the voltage required for VT2 to enter the base-to-emitter conduction, and there is no current in R4. Similarly, when VT2 is saturated and turned on, VT1 cannot turn on itself.

VT1 and VT2, only one of the two tubes can be turned on and the other can be turned off. This is the origin of the name "bistable".

maychang

FYzj posted on 2020-2-27 11:31 Hello, I would like to ask, if VT1 is turned on, shouldn’t D1 always be in the on state to charge C1? If S is not pressed, can VT2 be turned on? ...

"If VT1 is turned on, shouldn't D1 be always in the on state to charge C1?"

When capacitor C1 is fully charged, there is no current in D1. So D1 is not always in the on state. In addition, the charging of C1 (negative on top and positive on bottom) is not mainly through diode D1, but through resistor R5. D1 only works when S is pressed. When S is pressed, C1 should be said to be discharged.

maychang

FYzj posted on 2020-2-27 11:31 Hello, I would like to ask, if VT1 is turned on, shouldn’t D1 always be in the on state to charge C1? If S is not pressed, can VT2 be turned on? ...

"If it is turned on, does C1 discharge through R5 and R3 to provide voltage to the base of VT2?"

VT2 goes from cut-off to conduction because C1 discharges to cut off VT1. After VT1 is cut off, VT2 obtains base current through resistors R1 and R3 and turns on.

FYzj

maychang published on 2020-2-27 13:53 "If it is turned on, does C1 discharge through R5 and R3 to provide voltage to the base of VT2?" VT2 goes from cut-off to conduction by C1 discharge to make V ...

Thank you teacher for your answer, thank you