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On and off at the switching frequency and that's no problem. Okay, some more things we need in this circuit. Amplifiers to make them turn on and off in this way. Likewise, when the logic input is low, we want Q1 off and Q2 on. Okay, this logic signal typically is referred to this ground, and so it will go high and low with respect to ground. And the, this upper block must also then perform the function of level shifting.
It has to level shift this signal up to this node voltage to provide the proper signals at the input of this buffer. Something else we have to do, is we have to make sure that we never turn both Q1 and Q2 on at the same time.
If we ever do that, then they'll short out our input voltage, what we've been calling bg, and even if they're on for a few nanoseconds at the same time, during those few nanoseconds we can get a large current spike that flows through the two transistors, that can be many, many amps, and that current spike can damage the transistors and make the circuit fail. Or at the very least it will cause a large amount of power loss. So what we do is this logic circuitry, these boxes also implement what is called break before make operation or non-overlapping conduction circuitry.
Which make sure that first first you turn the on or conducting transistor off. We wait some amount of time called a dead time, and then we turn the other opposite device on. So, both devices are off for some short amount of time That is long enough to make sure that, when we account for switching times and driver delays and things, we never have the case where both transistors can be on at the same time.
Okay, let's consider what happens, let's suppose we had Q2 on and we want to turn it off, so our logic signal will start out low. Q2 is on, then we, we go to switch the logic signal high. There is some dead time. And then we turn Q1 on. So one question is, what happens during the dead time? Which our device conducts if any. Okay, well the answer is during the dead time the inductor current continues to flow and we can't interrupt the inductor current.
We're, let's assume here that the inductor current is positive and flowing to the output. If we have a positive input and a positive output voltage then the inductor current will flow in this direction. If both transistors are off the inductor current has to flow somewhere, well the only place it can flow then is through D2, through this body diode. The inductor current will forward bias the body diode, and it will conduct during the dead time and the switch volt no voltage VS will remain low with D2 conducting.
Then right here where we turn Q1 on the, the upper gate driver will turn Q1 on. And what it does then is we have current flowing this way that will both make D2 go through a reverse recovery, and cause switching loss, and also it will supply the inductor current. After the reverse recovery of D2 is done, D2 will be off, and then Q1 will conduct and supply the inductor current.
So we have switching loss that happens, even with synchronous rectifiers because of the need to have dead times, we get switching loss caused by the body diodes. Let's consider some further details about the switching times, and exactly how these devices switch. Where we explicitly showed the MOSFET with its body diode, and with a three capacitances, gate drain and gate resource capacitance, and the output or drain resource capacitance. So here's our, our driver's symbol and what I'm going to do is represent the driver output terminals here, with a simple Thevenin-equivalent model.
Here, in which we model the output terminals with some Thevenin-equivalent voltage source, and some Thevenin-equivalent resistance. Okay, we know the Thevenin-equivalent voltage is simply the open circuit output voltage, which is something that will switch between zero and in this case 12 volts.
And r Thevenin is the Thevenin-equivalent output resistance of the driver. Now, real drivers are more complicated than this, but this is actually a reasonably good first order model that explains all of the things we want to talk about today. And in a typical driver, we can think of this Thevenin-equivalent resistance as being the on resistance of the MOSFETs that are inside the driver. That are driving the output terminal of the, of the of the chip.
And so if we buy a driver that's rated at 12 volts and one amp for example then this thevenin equivalent resistance would be 12 volts divided by the one amp or 12 ohms. So what I've done here is to insert the gate driver, Thevenin-equivalent model in, in place of the low side driver, and I've replaced the MOSFET with an equivalent circuit model that contains the capacitances, the body diode.
And here we have a dependant current source that models how the drain current depends on the gate voltage. So we're going to draw some waveforms. So, it's 12 volts, and under these conditions The gate to source voltage of Q2, this voltage here, is is high also, and is equal to 12 volts. The switch node voltage vs of t is equal to ground because the lower MOSFET is on, and at this point what we'll do This will make our driver switch low.
So the Thevenin-voltage goes low. We have an RC circuit where the Thevenin resistance is connected across Vgs and it will pull current out of the, this gate. And discharge these capacitors. So they'll be some kind of RC type decay, and voltage will go low.
Recall the body diode is still on, to cnoduct the inductor current, and that will keep this vs switch node voltage low. Okay, and it will stay low for the dead time, so here is the dead time. Or perhaps even, we should call this whole time here the dead time. Okay, at this point, the upper gate driver turns Q1 on. Now, as we just mentioned a few minutes ago, this makes the body diode of Q2 undergo a reverse recovery, and so it takes some time for that reverse recovery to happen before this voltage at this node can start to rise, but eventually, it does rise.
You can see that this node here is all of a sudden going high. We're trying to hold VGS low. So that means the voltage across this capacidence has to increase along with the drain voltage. So we have current flowing in this direction through CGD.
Now, the big question is, once that current gets to the node here at the gate, Where does it go from there? It has to keep, it has to complete the loop an keep flowing someplace. Okay, now where can it flow? Can it flow this way? Well, we hope it does.
But the trouble is, we initially, the voltage here is zero, so we have zero volts here, and we have zero volts here, coming out of this the driver. Or V thevanin is zero. So you can see right there. So there's no voltage across our R thevanin. And if there is no voltage across this resistor, then by ohms law there is no current through the resistor. So, we have zero amps of current flowing through the resistor.
What that means is that the only place that this current can go is this way. Through CGS. And when it turns on again, then it tries to pull this node back down. And what happens is we get oscillations. Where this transistor will turn on and off many times. And we get all kinds of switching loss in the process. So, the gate to drain capacitance causes problems. And it's a common problem to, that we have to battle when we're designing these kind of circuits.
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Igbt gate drive basics of investing archie s place menu for diabetics
Lec 22: Introduction to Gate DriversPREDICTED ETHEREUM DROP
The difference between Gate to Emitter voltage is called as Vge and the voltage difference between collector to emitter is called as Vce. As the current flow is relatively same in both collector and emitter, the Vce is very low. IGBT applications are vast in electronics field. Due to low on resistance, Very high current rating, high switching speed, zero gate drive, IGBTs are used in High power motor control, Inverters, switched mode power supply with high frequency converting areas.
In the above image, basic switching application is shown using IGBT. The voltage difference across the load is denoted as VRL. The load can be inductive also. And on the right side a different circuit is shown. The load is connected across collector where as a current protection Resistor is connected across the emitter.
The current will flow from collector to emitter in both cases. On the right side circuit, the current flowing through the load is depends on the voltage divided by the RS value. It is an ideal switch device to replace GTR, which is widely used at present with its ability to turn off and is also widely used in all kinds of solid-state power supply.
And it requires a reasonable drive circuit, but its improper control may cause damage, such as IGBT damage due to overcurrent, and affects the performance of the whole machine. In a word, the drive circuit is very important to IGBT. So this paper mainly discusses the driving and short-circuit protection of IGBT, based on the analysis of its working principle, then designs and simulates the overcurrent protection of the drive circuit.
To make IGBT turn on and off safely and reliably, the driving circuit must meet the following conditions. To increase the switching speed, it is necessary to have a suitable gate bias voltage and gate series resistance. This value is sufficient to allow IGBT to reach saturation and then getting conduction, which can minimize the conduction loss. The effect of RG on switching loss is shown in Fig. To decrease the steepness of the front and rear edges of the control pulse, prevent oscillation and reduce the voltage tip pulse with a large IGBT collector, it is necessary to add a gate series resistor RG.
Thus, according to the current capacity and voltage rating, and switching frequency of IGBT, it is necessary to select a suitable RG, usually from dozens of ohms to hundreds of ohms. To get a more specific value of RG, it is suggested to refer to the device manual.
When the upper and lower arms conducting, the power supply voltage is almost all added to the two ends of the switch, at this time, the larger the short circuit current is, the smaller the saturation voltage drop will be, during this time, the device would be damaged due to the large current. The main drive circuit adopts push-pull output mode, which effectively reduces the output impedance of the drive circuit, improves the driving ability, and makes it suitable for the drive of high power IGBT.
The over-current protection circuit uses the principle of desaturation of the collector. When an over-current occurs, the IGBT will be turn off.
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