2021-12-20 05:42 AM
Hi everyone,
I am using two mosfet as a switch (in order to open or close a contactor, turn-on or turn-off some PCBs, etc, not high frequencies are required), one high side mosfet and one low side mosfet. I want to control them with the L6395 driver.
Well, I have read about mosfet and its duty cycle but It is not clear for me:
AND THE IMPORTANT QUESTIONS:
I hope you can understand my english.
Thank you for you help
Solved! Go to Solution.
2021-12-21 02:44 AM
The control with the L6395 can be a bit confusing, but you have to clearly separate the two control modes: series connection and push-pull connection of the load, for which there are also two separate inputs LIN and HIN (see datasheet, table 2), which do not contain an interlocking function and can therefore be controlled completely independently.
Let's take a closer look at the series connection:
Fig. 6 of the datasheet shows the principle, fig. 5 a specific schematic in which the control is also more clearly visible: with a series connection, the current in the load can only flow if both output stages are switched synchronously, i.e. HVG and LVG are switched on or off at the same time.
In contrast to this, the L6395 in fig. 7 and fig. 4 works in push-pull mode, which is controlled by corresponding signals at the two inputs.
Now to the internal bootstrap circuit: since the bootstrap capacitor supplies the high-side driver and has to be charged periodically, the L6395 cannot, in principle, achieve a dute cycle of 100%, neither with the external diode nor with the internal charge pump.
However, there is probably a mistake in figures 5 and 6, because the bootstrap capacitor should be connected between the BOOT pin and the drain of the low-side MOSFET so that it is charged during the switch-on phase.
So to answer your important questions:
Regards
/Peter
2021-12-21 02:44 AM
The control with the L6395 can be a bit confusing, but you have to clearly separate the two control modes: series connection and push-pull connection of the load, for which there are also two separate inputs LIN and HIN (see datasheet, table 2), which do not contain an interlocking function and can therefore be controlled completely independently.
Let's take a closer look at the series connection:
Fig. 6 of the datasheet shows the principle, fig. 5 a specific schematic in which the control is also more clearly visible: with a series connection, the current in the load can only flow if both output stages are switched synchronously, i.e. HVG and LVG are switched on or off at the same time.
In contrast to this, the L6395 in fig. 7 and fig. 4 works in push-pull mode, which is controlled by corresponding signals at the two inputs.
Now to the internal bootstrap circuit: since the bootstrap capacitor supplies the high-side driver and has to be charged periodically, the L6395 cannot, in principle, achieve a dute cycle of 100%, neither with the external diode nor with the internal charge pump.
However, there is probably a mistake in figures 5 and 6, because the bootstrap capacitor should be connected between the BOOT pin and the drain of the low-side MOSFET so that it is charged during the switch-on phase.
So to answer your important questions:
Regards
/Peter