drive%20mosfet

# As described under section 12 a mosfetigbt driver

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he output N -Channel and P-Channel MOSFETs, practically eliminating cross conduction. As described under section 1.2, a MOSFET/IGBT driver incurs losses. Let us derive formulae to compute this power loss in a driver: PD(on) = D x ROH x Vcc x Qg x fsw Eq. 5.2 ROH + RGext + RGint PD(off) =(1-D)xROLxVccxQgxfsw Eq.5.3 ROL + RGext+ RGint where: ROH = Output resistance of driver @ output High ROL = Output resistance of driver @ output Low fsw = Switching frequency RGext = resistance kept externally in series with Gate of MOSFET/IGBT RGint = Internal mesh resistance of MOSFET/ IGBT D = Duty Cycle (value between 0.0 to 1.0) Qg= Gate Charge of MOSFET/IGBT Total loss PD = PD(on) + PD(off) Note also that in general, RGint is small and can be neglected and that ROH = ROL for all IXD_ drivers. Consequently, if the external turn-on and turn-off gate resistors are identical, the total driver power dissipation formula simplifies to: PD=PD(on)+P D(off)= ROHxV cc xQgxfsw Eq.5.4 ROL + RGext Let us review with some examples: Assume that we are driving an IXFN200N07 for a Telecom power supply application or for a UPS/Inverter application at a switching frequency of 20 kHz. Furthermore, RGext = 4.7 Ohms and gate supply voltage is 15V. On page two of the IXD_409 Data sheet, we read the value of ROH = 1.5 Ohms ( Maximum). For Qg, refer to Data Sheet of the IXFN200N07 and go to Gate Charge vs. VGS curve and look for value of Qg at Vcc = 15 V. You can read it as 640 nC. Substituting these values into Eq. 5.4 yields: PD = 1.5 x 15 x 640 x 20,000 x 10-9 1.5 + 4.7 PD = 46.45 mW Assuming an ambient of 50 o C in the vicinity of IXD_409PI, the power dissipation capability of IXD_409PI must be derated by 7.6mW/oC, which works out to be 190 mW. The maximum allowable power dissipation at this temperature becomes 975190=785 mW. However, as calculated above, we will be dissipating only 46.45 mW so we are well within the dissipation limit of 785 mW. If one increases fsw to 500 kHz for a DC-to-DC Converter appl...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online