ADuM7234 for an H-Bridge questions.

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,

  • 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

  • Hello Ryan, thanks for your reply:

    Now taking a look into the data sheet of the ADuM3223, the section that you mentioned it´s well explained,

    clears the doubts about the purpose of serie resistors on the MOSFET gates and why to reduce the inductances as much as possible.

    To clarify, the 10Ω load that I mentioned previously is the external load that will be driven by the H-Bridge, actually it will be working as an square wave inverter. Do you see any trouble about power disipation if the load is pure resistive? In that case how can I take it into account for the design?

    You got right the question about VDDB, I will follow your indications.

    Thanks.

    Jean.

  • Hello,

    I made the calculation of the current, following the equation: I = (C * V * F)  and it´s around 32mA.

    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.

    I´m going for these transistors http://www.irf.com/product-info/datasheets/data/irfp250n.pdf , if I´m not wrong the input capacitance it´s around 2.2 nF, in fact it would be interesting to know the upper limit to be prepared.

    Thank you for your time.

    Jean.

  • 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

  • The Theta-ja for the package is 76ºC/W. This means to keep the part below 125ºC if the ambient temperature is 70ºC (a rough estimate for some motor applications, your case can easily vary between 25-100ºC), the gate driver needs to dissipate less than:

    125-70 = 55ºC

    55ºC / 76ºC/W = 0.724 W.

    This is the total power dissipation with both channels and quiescent current.

    The power dissipation will be shared between the external gate resistor and the IC in a ratio:

    Pdiss_ic = Rdson / (Rdson + Rext)

    The internal RDson of the ADuM7234 is probably around 2-3Ω, so if a 2-3Ω external resistor is used, half the power dissipation will happen off chip.

    The Ciss number quoted in MOSFET datasheets is a good starting point, but it often less than the effective capacitance seen by the gate driver. It is good practice to multiply the Ciss by about 3 to get a good safety margin. I believe running at 1MHz may produce a lot of self heating, and depending on the ambient temperature of your application, it might be close to the limit.

    I hope this helps.

    Ryan