Post Go back to editing

Hello,

What is the continous Output current of this Driver?

It has an Output Peak current of 2.3 A, but how is the continous?

Is is possible to use it with 200mA Output current?

• Hello,

The ADuM4121-1 is intended to be used as a gate driver. The gates it is intended to drive are capacitive in nature, like MOSFET and IGBT gates, where once the gate charge is delivered, the gate driver doesn't need to provide more current. The peak current the ADuM4121-1 refers to for the 2.3 A typical is at a specific operating point. The ADuM4121-1 has a much higher saturation current, and if it is operated at higher output voltages and with a low external series resistor, it can easily reach above 7 A. This is the saturation current of the internal driving FETs, but in normal operation of a gate driver, it is much worse for power dissipation to enter the saturation region. Please keep in mind that although the driver is called a 2 A driver, this is not the absolute maximum spec for the output peak current.

For continuous current, the limit becomes one of power dissipation. The internal driving FETs of the ADuM4121-1 have an Rdson. The power dissipated within the ADuM4121-1 can be described as Iout^2 * Rdson if Iout were to be constant. In most gate driver situations, Iout is not constant, and is really a very short burst that lasts as long as the rise/fall time of the power switch being driven (often 10's of ns). In that case, the energy dissipated within the ADuM4121-1 on each edge is the integral of the Iout^2 * Rdson. This power should be multiplied by the Theta-Ja to get an approximate temperature rise of the ADuM4121-1 above ambient temperature.

Pdiss * Theta_JA + Temp_ambient = approximate junction temperature

A rough guess at the thermal rise due to 200 mA would be (conservative estimate):

(200 mA)^2 * 1.8Ω = 72 mW

72 mW * 104.2 ºC/W = 7.5ºC temperature rise

This looks acceptable. Is the application really looking to output 200 mA, or is that the average current for driving a switch gate?

RSchnell

• 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,