The assumption is completely wrong for power MOSFET. Maximum Vgd can be calculated by combining Vds and Vgs ratings.since a MOSFET drain and source is interchangeable, therefore I should also assume that the max Vgs also implies the max of Vgd
The assumption is completely wrong for power MOSFET. Maximum Vgd can be calculated by combining Vds and Vgs ratings.
It's not unusual to have input or bus voltage higher than maximum Vgs and in all these cases, the gate driver needs to implement some kind of Vgs voltage limiting. A simple driver would use a PMOSFET as assumed in your circuit, a zener diode resistor circuit between gate and source and a transistor current source to span the voltage difference. If fast gate drive is an objective, a push-pull driver stacked below Vin can be used.
An alternative is a NMOSFET with a standard level shifting high-side driver. But it needs an auxilary circuit to charge the bootstrap capacitor.
it would be better to design with N-CH mosfet like irf540 that is robust and very cheap.
P-CH 100v buck design is not a good and up to date solution.
For driving high side n_ch mos just search the IRF.com and you will find excellent drivers.
besides you can go with synchronous buck with two irf540 and one half bridge driver from IR with minimum components.
Check this out:
**broken link removed**
There are many new 200v N mosfets with very low RDS . IRF540(not 450!) is fine for under 5A application.
IR2183 has more powerful output in comparison with IR2117 and it will be important for driving larger mosfets at higher frequencies.
IR2117 is fine for IGBT driving .
But a bootstrapping driver is problematic with single switch buck converters, since the bootstrap capacitor may not necessarily be charged prior to the start of operation. It can be done, but not without careful consideration to circuit startup.As previously mentioned, a bootstrap driver for a single transistor buck converter needs an auxilary circuit to charge the bootstrap capacitor. IR20153 is an example of a driver integrating this feature.
But a bootstrapping driver is problematic with single switch buck converters, since the bootstrap capacitor may not necessarily be charged prior to the start of operation. It can be done, but not without careful consideration to circuit startup.
Yes but doing this safely is difficult, given that the common mode voltage of the bootstrap capacitor will depend on exactly how you power up the circuit, what the load is doing, etc... the simplest way I can see to do it is to actually have a small FET in parallel with the catch diode. Before starting the converter, you turn it on briefly, which will charge the bootstrap cap, and then immediately start running the converter before it discharges. But you'd have to make sure the small startup FET and the main buck converter FET never operate at the same time.To solve this, could I take out the bootstrap capacitor, charge it, and release it back to the circuit?
Yes but doing this safely is difficult, given that the common mode voltage of the bootstrap capacitor will depend on exactly how you power up the circuit, what the load is doing, etc... the simplest way I can see to do it is to actually have a small FET in parallel with the catch diode. Before starting the converter, you turn it on briefly, which will charge the bootstrap cap, and then immediately start running the converter before it discharges. But you'd have to make sure the small startup FET and the main buck converter FET never operate at the same time.
What is your input/output voltage range? This may be an application where a gate drive transformer is the best solution.
A gate drive transformer is probably easiest for you then, assuming it's correctly designed and you're not trying to operate at very high frequencies.Hi mtwieg,
Thanks for your reply. My input voltage spans from 110v max to 70v min, but nominal at 100v. Output fixed at 28v constant no matter what.
Hi intelligent minds,
I've chosen to substitute the IR2117 with an optocoupler as my gate driver now. Have been in some research and have found out that the HCPNW-3120 is quite suitable for such high freq. operation (100kHz for my case).
Basically 3 main questions to address:
- Could I connect the bootstrap circuit (see attached) for the optocoupler using the orginal designed circuitry for the IR2117 as attached in earlier post?
- How would the adjacent diode of the free-wheeling as suggested in most bootstrap guides succeeds in preventing the Ve PIN not falling lower than the potential of (GND-Vf of the free-wheeling diode). Based on the fact that both Schottky are the same. Could it be because the free-wheeling diode (on the right) conducts more current than its neighbor diode (on the left) and hence the forward voltage drop of one diode would be higher than the other in nature?
- Based on the datasheet of the optocoupler: https://www.farnell.com/datasheets/652282.pdf, should I add a resistor between point A & B (see attached) to 1) limit the input current and 2) lower the voltage drop to the recommended range of 1.5V?
Thanks in advance.
The circuit involves a risk to fail in recharging the bootstrap capacitor in some load and PWM duty cycle situations. The problem has been particularly addresses by mtwieg.
I wonder, where the circuit with gate resistor connected to Ve comes from. I don't see it e.g. in the Avagotech datasheet. I don't feel a need for an additional protection diode.
The optocoupler input current should be set to the specfied value, the maximum ratings have to be kept. In most cases, a current limiting resistor will serve this purpose.
There are many new 200v N mosfets with very low RDS . IRF540(not 450!) is fine for under 5A application.
IR2183 has more powerful output in comparison with IR2117 and it will be important for driving larger mosfets at higher frequencies.
IR2117 is fine for IGBT driving .
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