ADuM7234 for an H-Bridge questions.

Question

Hello, 
 I´m designing an application that has to control a resistive load with an H-Brige activated by two ADuM7234: 
 - Maximum voltage 200V. 
 - Frequency around 1Mhz 
 - Resistive load 10 Ohm´s 
 I´m following the document http://www.analog.com/static/imported-files/tech_articles/Inside_iCoupler_Technology_Driving_an_H_Bridge.pdf 
 And I have some questions that I would appreciate if someone can help me to clear it out: 
 - For small loads I red that the optional resistor at the gates of the MOSFET could help to avoid ringing, which value do you suggest? 
 - The Vgs voltage that is set at VDDB(in the pointed document 15V) how much current could need, any idea how to calculate that? 
 - I´m planning to have the system separated in two boards, one the H Brige and the other with the gate drivers and the components required(SMD), 
 could it be a problem to extend the MOSFET gate inputs over a BUS cable to the SMD board? 
 Any remark would be really appreciated. 
 Regards,

EZAdministrator · Accepted Answer

This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

Thank you,
EZ Admin

RSchnell · Answer

Hello, 
 The series resistance mentioned in the datasheet is used both to reduce overshoot seen at the gate being driven, as well as offset power dissipation between inside the chip and the external resistor. For lighter loads, the external resistor is mainly used to dampen the response. There is a section in the ADuM3223 datashet that talks about this. It is titled "Output Load Characteristics", and it's contents can be applied to the ADuM7234 as well. Basically, the more inductance between the part and the gate, the more external resistance is generally required. This value is often in the 3-10Ω range for a 2nF load with up to 2 inches of trace between the driver and the MOSFET. 
 You mentioned a 10Ω load. Is this the external resistance? Or is this a load that will be present from the gate driver output to the gate driver output's ground? If the load is always present, we need to do some calculations about the power dissipation. These gate drivers generally run a capacitive load, so if the output is held high, the power driven into the gate eventually goes to zero. 
 A BUS cable between the driver and the MOSFET can cause excessive inductance, leading to poor performance. One of the benefits of using a driver is cleaning up the input signal for the FET. The closer the driver can be to the MOSFET, the better the situation. 
 I don't totally understand your question about the VDDB = 15V. The choice of the VDDB voltage is usually driven by the required voltage for the MOSFET being driven, choosing a value well above the device threshold, while at or below the recommended gate voltage. The current the part will require will be related to the power the driver outputs. Each charging and discharging cycle of the MOSFET will take approximately: 
 C x V^2 
 energy, where C is the input capacitance of the FET, and V is the peak voltage (VDDB). Multiply this by the operating frequency to get power. Divide the power by VDDB to get current seen in VDDB. you should also add the quiescent current to get a more accurate value. 
 Please let me know if you need any more clarification on these points, or if I misunderstood your questions. 
 Ryan

RSchnell · Answer

Hello Jean, 
 The actual load on the H-bridge is usually not the most important factor for the gate driver. The load will help decide what MOSFET to use, and the gate capacitances of the MOSFETs will lead to the choices for the gate driver. 
 The power delivered to the load will be a function of the load impedance, duty-cycle of the converter, and the bus voltage. 
 One issue that might come up is the power dissipation in the chip. You mentioned you will be switching at 1MHz. The 2MHz maximum frequency mentioned in the datasheet of the ADuM7234 is a max determined for a very light capacitive load at the MOSFET gate. in most cases, 2 MHz operation isn't actually feasible with a normally sized MOSFET. If you have a MOSFET in mind, we can go through the calculations to see the dissipation that will happen in chip. I believe the 2MHz limit on the ADuM7234 assumes a 1nF equivalent capacitance. 
 Ryan

Partner Forums

Partner Libraries

Products Mentioned

Associated Drivers

ADuM7234 for an H-Bridge questions.