2024-07-12
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Table of contents
1.2. Four commonly used isolation solutions
2. Pulse transformer isolation
2.1 Working principle of pulse transformer
2.2. Impact of bleeder resistance on switching circuit
3. Optocoupler isolation drive
3.1、Optical coupler isolation driving principle
3.2 Analysis of the power supply driven by optocoupler isolation
4.1、TI's UCC27200 is a typical bootstrap boost driver chip
4.2. Bootstrap boost driving principle
4.3. P-type tube driving circuit
4.4. Taking Buck circuit as an example, PMOS is used as the high-side switch
4.4.1 PMOS BUCK simulation waveform
Be sure to watch the original video, this note is only for the convenience of knowledge review!
Video link: (text comes from the txt in the link)
Because the control potential of the switch may be high voltage
Let's look at such an H-bridge circuit. The voltage at point A is uncertain. If the switch below is turned on and connected to the ground, it is 0V. If T1 is turned on and connected to PVCC, it is 200V. If T1 and T3 are not turned on and are completely symmetrical, then I think this point divides the voltage equally, 100V. Of course, it may be another voltage value. Then if I want to turn on T1, how many volts should the control potential of gate point B be? This is also floating, so we need to isolate the driver.
Pulse transformer isolation, optocoupler isolation, bootstrap boost and P-type tube. The first two are genuine isolation circuits and the last two are two workarounds.
We know that transformers can isolate all potentials and only transmit the potential difference to themselves. The origin of pulse transformers is because they are suitable for high frequencies because your pulse control signal is a square wave. Pulse square waves contain high frequencies and cannot use industrial frequency transformers. Its waveform is also asymmetrical, which is different from the main high-frequency transformer of general switching power supplies. In addition, pulse transformers are generally not available for purchase and need to be customized or made by yourself. And the turns ratio is mostly step-down type.
Let's look at an H-bridge circuit.
Only the high-side switches T1 and T2 need to be isolated and driven because the potential of these two points is floating, while the two tubes of the bridge arm on the low-voltage side do not need it. O14 represents the control signal of the two diagonal switches 1 and 4. We turn on the diagonal alternately. Then O23 represents the control signal of the two diagonal tubes. Then 1, 4 and 2, 3 are transformed through an inverter, that is, they are complementary alternating conduction forms.
When T5 is turned on, the power current flows from VCC to the ground through the primary of the pulse transformer, and the secondary of the pulse transformer flows through the current driving the gate of the switch. When T5 is disconnected, if you want to turn off this switch, the gate will discharge through R2.
Let's see that when the bleeder resistor is very large, 100kΩ, and can hardly discharge electricity, we find that the waveform of the output voltage should be almost a 200V square wave, but now it is only 20V, which means that the switch is not turned on correctly.
Why? Because our driver does not only charge the parasitic circuit of the gate, charging is turning on and discharging is turning off. If the discharge resistance in the discharge link is so large, you cannot discharge the electricity and it is actually never turned off, and the drive is not successful at all. Then reducing the discharge resistance to 1kΩ is successful, but there is a large delay. The switching delay is serious, and the discharge is too slow. When we reduced it to 100Ω, the delay is still acceptable, but satisfactory. After reducing it to 10Ω, it is a perfect square wave.
Then to improve the pulse transformer drive we useTotem Pole Structure To drive the pulse transformer:
Both charging and discharging are active charging and discharging with large current. So now the output waveform is very perfect 200V square wave
(1) What is the floating phenomenon of drive level?
Taking the H bridge as an example, the voltage level at point A is uncertain. When the lower switch is on, it is 0V, and when the upper switch is on, it is 200V. So if I turn on the high-voltage bridge arm, I don't know how much potential my gate B should have.
(2) Principle of pulse transformer isolation
The current flowing through the primary of the transformer can transfer energy to the secondary. The secondary of the transformer is connected between the gate and the source. No matter how much potential there is, I can always load a voltage between the gate and the source to control the conduction of the switch.
(3) The significance of totem pole drive
If the totem pole drive is not used, the gate discharge resistance is very large and there is no reliable drive at all. Simply reducing the discharge resistance will bring great power consumption. Therefore, we should use totem pole drive at this time. That is to say, whether it is charging or discharging, switches are used to complete the charging current and discharging current of the gate. Large current is used for totem pole drive, and a perfect square wave is obtained.
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In the figure, VCC_T1 and GND_T1 represent the power supply for the secondary side of an optocoupler. The independent power supply has nothing to do with the power supply of the primary side of the optocoupler. These two GNDs are not the same ground.
How many isolated ionization sources are needed to drive an H-bridge:
First look Control circuit and low-side switch driver power supply VDD That is to say, the power supply for the control signal is shared with the main circuit to supply the low-side switch. There are two optocouplers on the high side that need to be powered: VCC_T1, GND_T1, VCC_T2, GND_T2.
We cannot draw so many power supplies here, so we will not isolate the low side if possible. The high side switch driver power supplies VCC_T1, VCC_T2, GND_T1, and GND_T2 are completely independent.
Use two voltmeters to observe the voltage difference between the ground of the isolated power supply and the ground of my circuit:
Simulate the circuit. This is the square wave obtained by the inverter bridge H bridge. You can see that its rising edge isThe current rising part is not perfect. Why? Because we don't use a totem pole here. This driver wants to turn on the gate. VCC flows through the resistor and then flows to the gate. So it doesn't turn on quickly, but it turns off quickly because it turns off directly. This is a strong zero weak one circuit.
If we want to achieve a good driving effect, we need to add a totem pole driver after the optocoupler. Let's look at the control voltage signal. Our control signal only gives 10V, but the voltage difference between each ground, that is, the ground of the isolated power supply and the GND of my entire board is 200V, and the voltage difference is fluctuating. This is why my optocoupler isolation needs an isolated power supply.
The essence of optocoupler drive The light is only responsible for isolating signals, transmitting signals and providing driving energy for isolated power supplies. I drew batteries in the picture for these isolated power supplies, but in reality we still use transformers to get them from the AC power supply, which is the essence of optocoupler isolation. You still have to use transformers and use independent power transformers.
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Since the bootstrap capacitor must be charged frequently, it is not a real battery, so it is only suitable for situations where the high-side and low-side switches are alternately turned on.
The fast recovery diode is integrated inside. It is used to control the upper and lower bridge arms of a half-bridge. The low-side switches T1 and T2 must be turned on first. When T2 is turned on, the 12V power supply charges the bootstrap boost capacitor C through the diode to charge 12V. Then T2 is disconnected.It must be that T2 is disconnected before you want to turn on T1。
After T2 is disconnected, the potential of the capacitor is not connected to the ground but to LOAD. The 12V power of this capacitor will power the high-side DRIVE HI module to drive T1. This capacitor now acts as the driving power supply for the high-side switch. So what is the voltage at this point? No matter how much it is, it will raise the source voltage by 12V when it is connected to the load.。
This is a step-down circuit, so its maximum voltage is 20V. However, the source voltage VF1 of this NMOS tube is floating. When the switch is turned on, it is connected to 20V. This voltage is 20V. When the diode is turned on, it is connected to the ground, which is close to 0V. So although its voltage is not high, But it is still difficult to control the floating gate due to the source voltage At this time, we can use PMOS instead of NMOS as the switch tube
At this time, for PMOS, its source voltage is fixed at 20V. When my gate voltage is 20V, it is turned off. When it is lower than 20V enough to reach the threshold voltage, it is turned on. We can also add a switch T1 to formInverter Circuit this is okayOur control signal does not need to switch between 0V and 20V It can achieve this with a TTL level output voltage of 0V and 20V
The level of VF1 is 19.8 close to 20V when the floating switch is turned on.
Why is SD1 -256mV when it is turned on? When the diode is on, this is 0V. The diode is on, the voltage drop is so the voltage is slightly negative.
The control signal is 5V TTL level 5V, 50% duty cycle output voltage 50% duty cycle, 20V Buck circuit output voltage 10V in line with the theoretical value
The gate control voltage floats between 0V and 20V because I used an inverter which is a Buck circuit made of PMOS. PMOS can also be used for bridge circuits. For bridge circuits with a total voltage below 200V We can also use P-type tubes to replace N-type tubes to drive
The high side is replaced with PMOS. We must pay special attention to the withstand voltage of switches T5 and T6 that constitute the inverter. You must also meet the PVCC voltage level.
(1) Principle of bootstrap boost drive
For the half-bridge circuit, if I open the low-voltage bridge arm first, the 12V power supply can charge the bootstrap boost capacitor C. Then when T2 is disconnected, the 12V level charged on C will automatically float upward. In short, it is a 12V power supply to power the power module of the high-voltage bridge arm. This is the principle of bootstrap boost drive.
(2) P-type tube driving principle
Even for circuits with low voltage, such as Buck circuit, the source VF1 voltage of its switch is floating, so it is difficult for us to give the gate a suitable potential for driving. In this case, we can use PMOS to replace NMOS. The source potential of PMOS is fixed at 20V. We use an inverter to make a 0V and 20V control signal to achieve reliable on and off control of PMOS. This is the driving principle of PMOS tube replacing NMOS.
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