How to coordinate the ground of the switching tube of the switching power supply and the ground of the control branch?

乱世煮酒论天下

How to coordinate the ground of the switching tube of the switching power supply and the ground of the control branch? [Copy link]

 

Figure 1 shows a BUCK circuit made with TITA-TI. OUT1 is the driving branch voltage waveform, and a power supply is used.

Figure 2 shows a BUCK circuit made with LTspice. V(out1)-V(out2) is the driving branch voltage waveform, and a power supply is also used.

Looking at this circuit topology diagram, the ground of the main circuit is the ground of the power supply, which is the negative pole DC-, and the ground of the drive circuit is the S pole of the switch tube. However, I can use this model for simulation. In the actual circuit, the chip drives the op amp and then drives the power circuit to finally reach the gate of the switch tube. The ground of the drive circuit is actually the S pole gnd of the switch tube.

So how does the ground gnd of the drive circuit establish a relationship with the negative pole DC- of the main circuit power supply?

beyond_笑谈

There is only one GND relative to the input voltage V1. V2 (VGS) is the gate drive voltage of the MOS, and MOSFET_VS is floating relative to GND (this statement is not very accurate, but it's roughly the meaning).

maychang

[How to establish the relationship between the ground gnd of the drive circuit and the negative DC- of the main circuit power supply? 】

The two are not related. As mentioned in the second floor, V2 is floating.

maychang

In the Buck circuit, the freewheeling diode D1 is always turned on for part of the time and turned off for part of the time in a switching cycle. When D1 is turned on, its upper end (that is, the source of your MOS tube) is a PN junction forward voltage drop to the "ground" in the figure (that is, the negative end of the DC input). When D1 is turned off, its upper end is approximately the DC input voltage to the "ground" in the figure. Therefore, there is no relationship between the source of the MOS tube and the negative end of the DC input.

乱世煮酒论天下

The driver ground and the negative pole of the main circuit are not the same. How to ensure that the driver voltage loads the correct turn-on and turn-off voltage between the gate and the source? I can think of several ways, such as adding an isolated power supply. The isolated power supply and the main circuit share a common ground. Since isolation can be added, there are optical isolation and magnetic isolation. Other isolations have not been thought of yet.

Or this topology uses bootstrapping when driving the upper and lower bridges of the inverter, which requires a special high common-mode driver chip.

se7ens

Indeed, current drivers all use chip solutions, and rarely use separate devices to build

Generally, there is a bootstrap circuit or isolation circuit inside the chip to output the drive signal, which means that the drive signal is actually floating.

乱世煮酒论天下

I think this question can be answered with the following circuit diagram. Although this circuit diagram has many problems and is made by me at random and cannot be used at all, it can help us understand the ground supply problem between the driving circuit and the main circuit of the switching power supply.

R5, R2, and R3 are the output voltage detection circuit. R3's voltage divider is fed back to the control loop. From the above circuit, it can be obtained that the ground of R3 is the ground of the input power supply. At the same time, the voltage collected by R3 is input to the amplifier. Here, an op amp and a triangle wave comparison wave are used to generate a PWM wave. Because the collected voltage is to be sent to the op amp, the ground of the op amp power supply and the ground of the negative pole of the main circuit power supply are the same ground. The first question is, how is the power supply of this op amp generated? Does it have to be grounded with the main circuit?

The output of the op amp drives the power control loop. A push-pull circuit is used here to drive the MOS tube. Similarly, the output of the op amp is also based on the ground of the op amp power supply. At the same time, as the output drives the push-pull circuit, the ground of the push-pull circuit is also the ground of the power supply. The second question arises: is this feasible in the actual circuit?

The last step is to drive the MOS tube with a push-pull circuit. The gate drive of the push-pull output is based on the magnitude of the gate voltage to the source voltage. Here, the ground of the power tube MOS tube is the source. However, in the previous question, the ground of the push-pull circuit is the ground of the power supply of the operational amplifier, which is also the negative pole of the main circuit. Will this have an impact? How to solve this problem?

乱世煮酒论天下

I did another simple simulation. This time I should understand it. The basic simulation has no control. The circuit is as follows:

Figure 1 shows the output voltage amplitude under different driving voltages, and Figure 2 shows the output voltage amplitude under different driving switching frequencies. When the switch tube is turned on, the ground of the driving gate is the negative pole of the main circuit. When the switch tube is turned off, the diode is turned on for freewheeling, and the ground of the driving gate is one diode conduction voltage drop lower than the negative pole of the power supply. In this way, it is sufficient to set the power supply voltage of the driving circuit near this ground.

It can also be seen that increasing the switching frequency does not have a significant effect on the overall amplitude of the output voltage. It can only greatly reduce the volume of the inductor, but it will increase the EMC problem.