Hello, I am designing an inverter using ADUM6132 gate driver. I have two questions.
Data sheet does not say much about dead-time. Is it possible to create dead-time inside adum6132 or should I leave dead-time for signals going to the adum6132 as input?
My second question is that there seems to be two power supply needed for adum6132. 5V for one and 15V for another, nominally. Is it possible to set configuration with voltage divider? I am saying that I will supply +15V externally and use voltage divider to create +5V out of +15V and then use these two as input to adum6132. Would it work?
Thank you in advance,
The ADuM6132 has two independent inputs, VIA and VIB. Any deadtime needs to be introduced from the controller side. If VIA and VIB are both held high at the same time, both outputs will be high. There is no internal deadtime generator on the ADuM6132.
If you are looking to use resistors to create a voltage divider, the common issue you'll encounter is that as the 5 V load changes in impedance, the effective resistance on the bottom of the voltage divider will vary. To combat the voltage variance, you can either use a linear regulator, or a switched voltage converter. A linear regulator would be the lowest part count way to go, but the efficiency of a 15 V to 5 V linear regulator is 33%. If you can accept the power loss, it should be a fine way to do it.
Thank you for your in-depth respose to my previous question. It is crystal clear to me now.
I have one more question. I am trying to drive a mosfet with input capacitance of 5000pf. However, in datasheet of ADUM6132 pg:11, It says that ADUM6132 drives MOSFETS with typical input capacitance of 200pf. Datasheet also offers a way to drive transistors with higher input capacitances. Datasheet offers using current buffer for driving transistors associated with higher input capacitance levels. I have a question about this current buffer.
I know from datasheet that current can be max of 9mA (Iavail) (figure 16 from datasheet). Looking at the figure 1 below, using the formula VBB=I.Rg; Rg=15/9mA= 1667ohm. Moreover, for Cies to go from %10 to % 90, It takes Ton=2.2Rg*Cies. Then, Ton= 2.2*1667*5000pf = 18.3 us
For Cies to charge, it must not take that much of a long time. Either I am making a mistake in logic or ADUM6132 is not suitable to drive a transistor with 5000pf input capacitance. Could you clarify the issue?