Will the push-pull circuit be straight through?

shaorc

The step-down circuit shown in the figure

In normal operation, if there is a sudden power failure, that is, all three power supplies are gone, and then the power is turned on again

When the op amp U1 is powered on, the voltage at each location will not reach a steady state, causing the op amp to output a high impedance state.

At this time, the voltage of Vcc2 forms the base current of V1 through R1, which can make V1 work in the amplification area and thus be turned on.

The GS inter-electrode capacitor voltage of Q1 will be discharged through R2, the be-electrode of V2 and R3, thus providing base current to V2, making V2 also in the amplified conduction state.

Will there be a period of time when V1 and V2 are directly connected, short-circuiting VCC2?

maychang

"Will there be a period of time when V1 and V2 will be directly connected, short-circuiting VCC2?"

This situation is almost certain to happen in the circuit in the first post. In the figure, U1 outputs a rectangular wave, and this "shoot-through" phenomenon will occur regardless of the rising or falling edge of the rectangular wave. The reason is that the delay time from the off state to the on state of the transistor is short, while the delay time from the on state to the off state is long. The original poster can find out by checking the working parameters of various transistor switches.

maychang

The delay time from turn-on to turn-off of the transistor is related to the degree of saturation when it is turned on. The deeper the saturation of the transistor, the longer the delay time of the turn-off.

This "common conduction" phenomenon will cause a considerable current to flow through the two transistors instantaneously. For high-power transistors, this instantaneous large current may be fatal.

maychang

The second and third floors are talking about the steady state, that is, the power supply has been powered on for a long enough time and U1 outputs a normal rectangular wave.

As for powering on again after a power outage, the situation is more complicated.

maychang

Judging from the connection of Q1, D1, and L1, this is a Buck circuit.

The source potential of Q1 is constantly changing relative to the "ground" ("ground" refers to the positive terminal of D1), so the signal driving Q1 must also change with the source potential. However, this circuit cannot do this and cannot be used.

sfcsdc

I have built this kind of circuit before using 8050 and 9012, but it burned up as soon as the power was turned on.

It should be that the two tubes are turned on at the same time to generate a large current.

PowerAnts

In fact, I have always been skeptical about the saturation of the totem pole. I discussed it with Mr. Xu from IR a few years ago, and I was the one who finally convinced him.

My opinion is: for the push-pull totem pole outputted by the emitter, both transistors work in the common collector state, and saturation will not occur. Even if the high level of the IC output is VCC and the ground level is VSS, the highest potential of the two transistors is only VCC and the lowest is VSS. The collector will not be forward biased, and there will be no ts and tf time periods. The two transistors can only work in the linear region and the cut-off region, turning on slowly and turning off quickly. I suggest high-frequency tubes instead of switch tubes.

Finally, I recommend that the collector of the top tube of the totem pole be 1V higher than the IC, and the collector of the bottom tube be 1V lower than the VSS of the IC.

maychang

PowerAnts posted on 2019-7-15 13:09 In fact, I have always been skeptical about this totem pole saturation. I discussed it with IR's engineer Xu a few years ago, and I finally convinced him. My opinion is: launch...

That makes sense.

shaorc

sfcsdc posted on 2019-7-15 08:35 I have built this kind of circuit with 8050 and 9012 before, and it burned up as soon as it was powered on. It should be that the two tubes were turned on at the same time to generate a large current.

The tubes burned out when powered on for the first time?

伟林电源

sfcsdc posted on 2019-7-15 08:35 I have built this kind of circuit with 8050 and 9012 before, and it burned up as soon as it was powered on. It should be that the two tubes were turned on at the same time to generate a large current.

Why not use the paired tubes 8050 and 8550?

伟林电源

PowerAnts posted on 2019-7-15 13:09 In fact, I have always been skeptical about this totem pole saturation. I discussed it with IR's engineer Xu a few years ago, and I finally convinced him. My opinion is: launch...

You are right. In fact, the most likely cause of two-pipe direct connection should be:

VCC2>=VCC1+upper tube Vc

The Vc of the lower tube should be greater than VSS <= 1V when the MOS is turned off.

PowerAnts

Wei Lin Power published on 2019-7-22 20:53 You are right. In fact, the most likely cause of the two tubes to be directly connected should be: VCC2>=VCC1+upper tube Vc. The lower tube Vc should be higher than...

Is that the opposite? It is possible to adjust the relationship between VCC2 and VCC1. In fact, due to the high voltage, the op amp cannot be rail-to-rail, and the output voltage of U1 is 2 PN junction voltage drops lower than VCC1.

The condition for judging whether BJT will be saturated is very simple: whether the collector junction is forward biased or not! If there is forward bias, it is saturated, and if there is no forward bias, it is in the linear region.

伟林电源

You mean VCC1=VCC2, right?

PowerAnts

Wei Lin Power posted on 2019-7-23 20:24 You are talking about the premise that VCC1=VCC2, right?

VCC1=VCC2 will not be saturated, VCC2>=VCC1+upper tube VC will not be saturated

伟林电源

This direct pass condition was mentioned by Maychang.

PowerAnts

Weilin Power Supply published on 2019-7-25 23:28 This direct-through condition was mentioned by Maychang.

It's just my personal opinion, what does it have to do with others?

伟林电源

I am confused by you. My understanding comes from maychang's point of view, so there is the 11th floor.

PowerAnts

Wei Lin Power Supply published on 2019-7-26 10:59 I am confused by you. My understanding is derived from maychang's point of view, so there is the 11th floor.

Random accusations! Haha

伟林电源

This is the fact, and I have always insisted that the common condition does not hold.

maychang

PowerAnts posted on 2019-7-27 16:22 Random accusations! Haha

The original poster posted the exact same post on 21ic, where I said that under certain conditions, V1 or V2 may enter saturation, so they may be "conducted together". Weilin Power Supply based its statement on what I said there. However, what I said there was that if Vcc1 is higher than Vcc2, V1 may enter saturation.