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Before talking about driving high-side switches, let us first review how the usual low-side switches work. Whenever a low-side switch (MOSFET) has to be turned on, 5V or 10V is applied to the gate of the MOSFET, thus turning it ON. In order to turn it off, 0V is applied to the gate. Now let us take the case of an NMOS used as a high side switch. Let us first drive it in exactly the same way as the low-side switch is driven: 10V at the gate to turn it on and 0V to turn it off http://everycircuit.com/circuit/5593154097774592 . What do you see in the linked simulation? You see that the load recieves 6-7V and the MOSFET drops 3V across its drain-source path. This is because the source of the MOSFET is connected to the load and not to ground. When 10V is applied to the gate, the MOSFET begins to turn on. Current starts flowing through the MOSFET and the load and the voltage across the load starts rising until the point where it reaches 6-7V. At this point, the GATE TO SOURCE voltage of the MOSFET is around 3V, which is not enough to keep it in its ohmic region. And any further increase in the voltage across the load decreases the gate to source voltage, which tends to turn off the MOSFET. Due to this, the voltage across the load doesn't rise above 7V and the MOSFET, operating in its saturation region, drops 3V. In order to turn the MOSFET ON properly, we need atleast 5V across the gate to source, which mean we need to supply the gate driver with atleast 10V + 5V = 15V. But how can that be done with a 10V input voltage?
To overcome this problem, a diode and a capacitor are used in order to derive a supply voltage for the gate that is higher that the available supply voltage. This is called bootstrapping. Look at how the 100nF capacitor and diode are connected. Initially, when the MOSFET is off, current flows from the 10V supply through the diode, the capacitor and the 2ohm load until the capacitor charges to around 9V (It doesn't charge up to 10V because of the diode drop). When the MOSFET turns on, initially, the gate only recieves 9V from the capacitor. Now this seems like a step backwards, but watch what happenes next. As the load voltage starts rising, the voltage at the source of the MOSFET starts rising. This now adds onto the 9V that was stored in the capacitor due to the presence of the diode, which is now reverse biased. So as the voltage at the source increases from 0V to 10V, the voltage at the gate can rise up to 10V + 9V = 19V! This means that the gate to source voltage is now 19V - 10V = 9V! This is higher than the 3V that was previously available. Due to this, the MOSFET is kept ON in it's ohmic region and drops very little voltage, allowing the entire supply voltage to be available to the load. Thus, the MOSFET has been successfully and properly turned on.
Now the next step is to supply an appropriate voltage to the bases to the BJTs in order to turn them on. They too need a higher voltage, just like the MOSFET, and cannot be driven directly from the signal source. Thus, a level-shifter circuit is needed in order to shift the signal's voltage level up in order to drive the BJTs properly. This is achieved using the auxiliary MOSFET (a small signal one) and some resistors. You can analyse this circuit on your own. To see how the absence of a level shifting circuit make the bootstrap circuit ineffective, see http://everycircuit.com/circuit/6414760470642688 .
N-Channel MOSFETs are easily driven when used as a low-side switch, but in circuits the use MOSFETs in topologies such as the Half-Bridge or H-Bridge, N-Channel MOSFETs are needed for high side switching and are driven this way.
MOSFETs used: IRF540N as the main switching transistor and BS170 as the auxiliary transistor for level shifting.
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