Gate driving loss and gate driver for Buck

Question

I have the following question regarding gate driving loss in a synchronous Buck. When we want to turn on HS or LS we have to charge their input capacitance \$C_{\text{iss}}\$. This is done through a driver which is supplied by \$V_{\text{dd}}\$ and in it simplest form it can be considered a CMOS inverter. So suppose we are turning on a LS nMOS gate driving loss \$P_{\text{driving}}\$ is defined as

$$P_{\text{driving}}=C_{\text{iss}}V_{\text{dd}}V_{\text{batt}}f_{\text{sw}}$$

A certain charge \$Q_{\text{g}}=C_{\text{iss}}V_{\text{dd}}\$ is provided by the driver and I'm totally fine with it.

Now we are using the pMOS of our CMOS inverter to provide a path between \$V_{\text{dd}}\$ and gate switch. pMOS is operating in triode mode so there will be some heat dissipation or power loss since it behaves like a resistor. At this point I would write average power

$$P_{\text{driver}}=V_{\text{dd}}I_{\text{dd}}$$

where \$I_{\text{dd}}\$ is the current flowing from supply to driver to gate switch. Now if I want to calculate total power loss regarding driving is it right to sum up the two power losses? I've read several application notes but this thing isn't clear: some mention an IC power loss (which should be \$P_{\text{driver}}\$), others never talk about actual power dissipated by the driver. We need power to turn on the LS but we need power also for the driver. And gate driver power consumption impacts overall efficiency.

Answer 1

You are on the right track.

Your total loss is Vdd.Idd . On each turn-on cycle, a charge Qg is consumed from Vdd and delivered to the FET's gate. This (resistive via a FET) charging is precisely 50 % efficient -- half the energy consumed from Vdd is dissipated in the driver, and the other ½ is stored in the gate capacitance.

Note that gate capacitance isn't constant as the FET switches. it is better to characterize the device's switching gate charge which is well defined.

In the subsequent turn-off cycle, all this stored energy is dissipated in the driver's NMOS device.

So -- when charging, an energy E=Qg.Vdd is extracted from Vdd (this happens at a frequency fSW, so the power is Qg.Vdd.fSW). You end up with an energy stored in the gate of ½.C.Vdd^2, or ½.Qg.Vdd (since C = Q/V). The other half of the energy consumed is dissipated in the driver, adding the total Qg.Vdd.

Then, when discharging the FET, all the remaining stored energy (½.Qg.Vdd) is dissipated in the switch.

Therefore, in the end, the total power loss is Qg.Vdd.fSW, or Ciss.Vdd^2.Fsw when the driver is powered by Vdd. If it is powered (e.g. with a linear regulator), the regulator's efficiency changes the Vdd^2 term to Vdd.Vbatt

Answer 2

You should not attempt to calculate these kinds of losses using capacitance related to your MOSFET switch as this will yield inaccurate results.

The capacitances are time-varying and in very non-linear ways, so you have to use the integral of current with respect to time (charge) instead.

The power loss in your driver is best calculated by modeling your driver's high and low MOSFETs as resistances (especially if your high side driver FET is in the triode region) like so:

$$ P_{driver} = \frac{V_{gdrv}Q_{Gtot}f_{sw}}{2} x \left ( \frac{R_{GHI}}{R_{GHI}+R_{G}+R_{GI}} + \frac{R_{GLO}}{R_{GLO}+R_{G}+R_{GI}}\right )$$

Where:

• \$ V_{gdrv}= \$ Gate driver voltage (the power supply voltage)
• \$ Q_{Gtot}= \$ Total gate charge of the LS MOSFET
• \$ f_{sw}= \$ Switching frequency
• \$ R_{GHI}= \$ High side driver resistance
• \$ R_{GLO}= \$ Low side driver resistance
• \$ R_{GI}= \$ LS MOSFET gate resistance
• \$ R_{G}= \$ Series gate resistance

Or like this:

As you can see, you can very directly impact and reduce the driver power dissipation by increasing \$ R_{G} \$, but at the cost of increasing turn on and turn off times in the LS MOSFET. Depending on your specific specs and chosen MOSFET, adding or increasing the value of a gate resistor can actually result in higher total efficiency for the whole device. Or make it worse. It is definitely something you need to consider however when looking at the whole circuit efficiency.