Everyone knows that IGBT single tube is quite fragile, the same current capacity IGBT single tube, more than the same current capacity, MOSFET is fragile, that is, in the inverting H bridge, there is no problem, but IGBT, may The boot is blown up. This makes many people estimate. At that time, I saw the buddy of the fish machine with FGH25N120and, which reflected it was easy to burn, and did not think about it.
Only when I was in my work, I found IGBT, I found me failed. I really thought that an IRFP460, 20A / 500V MOSFET, I use a SGH40N60UFD40A / 600V IGBT? It will not be blown, but the actual situation is, after loading, suddenly add the load and revoke the load, it is blown several times, I thought it was not welded, and then changed, it was blown away, so white white Waste a lot of IGBT.
Later, some laws were found, that is, the measures to use peak current protection can make IGBT will not blow, I will tell these things in detail, I don't want to forgive.
We see this question several parts to solve:
1, drive circuit;
2, current collecting current;
3, protection mechanism;
First, drive circuit
The IGBT used in this time is ixys, IXGH48N60B3D1, detailed specifications are as follows: IXGH48N60B3D1
The drive circuit is as follows:
This is a very typical application circuit, which can be used in IGBT or MOSFET, but there are somewhat different.
1, there is a negative pressure generating circuit,
2, isolated drive,
3. Power supply separately.
First let's take a look, this circuit is not protected, it is used in the inverter 100%, but we can clear the essence of this circuit.
Let's talk about it first:
1: Drive resistor R2, this is very important in the driving, and the D1 is closed, and the IGBT's CGE is quickly discharged. It is actually needed, and this D1 can not, or in the D1 loop. The resistor is a gate resistance when the 0FF is closed.
Several waveform photos, different gate resistors, and high-voltage HV + 400V, the actual situation of the upper and lower IGBT gates.
The above figure is that when the negative pressure is canceled, the gate waveform between the upper and lower 2 tubes is in the case of 10R.
The above figure is a 2-tube G polar waveform without adding DC400V, and the figure below is the gate waveform of the 2 tube in the case of DC400V.
Why will there be a spike? This should be said from the internal situation of IGBT, simply, there is a parasitic capacitor on the IGBT GE, which combines a parasitic capacity of the other CGC to form a pool, which is QG, in fact, this and MOSFET is also very like .
So let's see why 400V plus it, you will generate spikes at G level on the tube. Take a flower to show a Buddha, catch a picture to explain:
As shown in the figure above, when the upper official opened, it was closed. Because of the opening of the official, this time to introduce the concept of DV / DT, this is abstract, no matter it, simple and popular saying is to open the management At the time, the taller is equivalent to pass through, + DC400V voltage immediately added to the Class C below the tube, so high voltage immediately by generating an induction current, this induction current is calculated on the formula, this Under the common action of the RG resistance and the driving internal resistance, a spike voltage is formed on the gate of the lower tube, as shown in the screenshot of the oscilloscope above. So far, there is no concept of Miller capacitance, understanding these, and then seeing the specification, what is the Miller capacitor, what is impact on the circuit, it is easy to understand.
This spike has many bad sites. From the above oscilloscope screenshot, it can be seen in the spike, the tube has actually reached the 7V voltage, that is, in this time period of the spike, the upper and lower two tubes are commonly turned on. The tube is shorted, but because there is TON time, this current is not too big. The pipe does not blow, but it will heat it. With the greater the power of transmission, this situation will be more serious and greatly affecting efficiency.
Originally, it is necessary to issue a wave-like photo after the addition of negative pressure, and the negative pressure can make this spike in a safe level range. The oscilloscope needs a U disk to guide the graph, which is clear, today is latched and not touched the instrument, then go up again.
Second, current collecting circuit
Speaking of this step, it is not far from protection. My experience is that the current collecting speed is very fast, so that I can quickly tell the circuit after crossing or short circuit - ", there is a problem here. Let IGBT close up quickly and safely.
How can this circuit be implemented? For inverter circuits, we can use the resistance to sample directly, or the VCE tube pressure drop detection method can also be used. There are many discussions in this forum, but there is no truly used, truly actually applied, testing circuits (special drive chip exceptions), because each actual application is not the same For example, the IGBT parameters are different, there are many parameters that need to be adjusted, and certain experience is needed to adjust.
We can start from the simplest way, using resistance to detect this current, short circuit, can generate pressure drop, comparator and this voltage, resulting in the ultimate flow or short circuit signal.
You can use this picture, because the principle is very simple, it will be very easy to achieve it, you can adjust it without how many parameters can be adjusted.
The figure above is the current of the sampling h bridge, for example: if the IGBT is 40A, we can take the peak current, which is 80A, corresponding to the above, RS 0.01R, if the flow exceeding 80A pulse current is then A 0.01R * 80A = 0.8V voltage is generated on this resistor. After R11 and C11 blame to the + terminal of the comparator, compared to the reference voltage from the--terminal, the--end reference resistor is not correct, actually In addition, in this example, this example can be used to be rotate to 0.81V to-end, and if the voltage on the sampling resistor RS exceeds 0.8V, the comparator is turned to the SD 5V level. In the external circuit. The level signal of this change is the signal we need to use short-circuit overcurrent.
With this signal, how do we turn off IGBT? We can see if the situation takes soft shutdown or take direct hard shutdown.
Take soft shutdown, it can effectively prevent the case where the voltage is increased in a closed moment, the closing property is very soft, very gentle, very suitable for high-voltage high-power drive circuits.
If the hard shutdown may cause the voltage overshoot on the high pressure DC, such as the DC400V high pressure in the first figure may become instantaneous to become a DC600V. At that time, I saw some information on the record, it was very difficult to understand: The relationship is closed, is there a high pressure? The actual situation is true.
If you are difficult to understand, you can do a test, there is a water tower in the family, the clearel, the water tower is upstairs, the faucet is on the first floor, open the faucet, the water is left, then close this faucet with very fast speed, you Will hear the water pipe has a sound, and the water pipe will have to vibrate (do not know the right to say, please "introduce the faucet, this example, I have to thank my teacher, the teacher see us is too stupid, speak the triode The metaphor of the characteristics of the characteristics, I would like to thank him here), IGBT is the same in the bridge circuit. When IGBT is severely short, if the horses are hard to close the IGBT, it will only cause the induced voltage on the busbar (as for why you can check the relevant information, many information are said), the pipe can resist the past, such as You have a very good absorptive capacitor on the DC high-pressure bus line, there are multiple absorption circuits, etc. ...
If the tube is turned off, it will be invalid. It is also useless. IGBT will still be shorted by overflow pressures, and this short circuit is no way to recover, it will immediately damage a lot of circuits. Sometimes there is no overvoltage can also cause this phenomenon. The principle of this failure is unknown, but it can be imagined that it may be due to other parasitic capacitances and Miller capacitors associated with the tube, or due to overcurrent When the short-circuit signal occurs, the IGBT has undergone the engine effect, and the relationship is not dead.
There is also the third way, called: Secondary closure, this way is simple, it is detected short circuit, overcurrent signal, PWM this pulse does not plan soft shutdown or shut down, but immediately The corresponding VGE drive pulse voltage is reduced to about 8V to determine whether it is still overcurrent or short-circuit area, if still, continue to use this 8V driver, until the set time, such as multiple US US, or this is the case Off, if yes, PWM will return to normal. This method is generally different, so we don't do in-depth research.
Understand these, we can see the situation to specifically adopt those ways, I think therein in the 2kw level, DC380V, directly take hard shutdow, can meet the requirements, only need a good capacitor in the H-bridge in parallel. You can use 600V IGBT.
The key point is to detect fast, after testing, it is necessary to turn off fast, only it is fast, IGBT will not burn.
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