# What is the purpose of D1 on the gate of the MOSFET? [NCP3065]

I've built this DC-DC LED Driver, and don't quite understand the purpose of D1. In particular it works with a 1N4007, but I wanted to replace it with a Schottky type for faster switching since the 4148 is much faster than a 1N4007, but the circuit fails to power up properly.

I understand that diodes conduct in one direction, but there is no need to prevent reverse flow of current in this situation. The only thing I can think of is maybe for it to turn off faster in the sense of having a constantly present ~0.7V on the diode to accelerate the turn off?

Or possibly is this diode creating a lower voltage essentially like a bootstrap device since Q1 will be turned OFF when the SWC is open to ground?

-

The short answer- it's there to shunt current away from the base of the MMBT3904 BJT and to allow SWC to pull the gate down.

SWC on the NCP3065 is the collector of a darlington. When it is "off" (SWC is pulled high by R2), the 1K resistor R2 and Q1 act together to suck the gate charge out of Q2 quickly. The transistor allows a relatively high value (1K) resistor to be used, since the base current is multiplied by the $h_{FE}$ of the transistor at the collector, so it's like using perhaps a 30 ohm resistor, without the horrible power dissipation and wasted current a 30 ohm resistor would have when SWC goes low.

When SWC goes low (the darlington turns on), it directly pumps the gate charge into Q2. Just before it turns on the gate is close to Vin, so the gate-source voltage is close to zero and Q2 is off. As it turned on, it pulls the base of Q2 below the emitter, so it's well off, as it drops 0.6V further, the diode becomes forward biased and it limits the $V_{EB}$ to no worse than -1V, well within its -6V rating. It then conducts the gate current from Q2, charging the gate and turning it on (as well as conducting the current from R2). Thus Q1 is off and D1 allows SWC to pull the gate of Q2 down directly.

It may help to visualize the current flow if you imagine a capacitor from gate to source of Q2. Part of the capacitance is gate-source capacitance and part is actually Miller effect from gate-drain capacitance.

The combination of Q1, Q2, D1, and R2, together with an N-channel or PNP switch to ground (inside the chip, in this case) is a useful building block where a relatively fast high-side switch is required (for example, for a buck regulator as here).

-
If you don't mind, would you please explain in little detail on how making the SWC low pumps the gate charge into Q2 and turns the Q2 on? How is diode forward biased in this situation? –  dr3patel Mar 13 '14 at 4:05
I added a bit more detail in an edit –  Spehro Pefhany Mar 13 '14 at 4:22
Alright, this makes more sense. However, do you have any theories why using an SB240 might not work vs using a 1N4007 does? In particular the output current massively drops, and it appears the frequency jumps up, and gate oscillations occur. Would this be some self resonant frequency? The capacitance of the SB240 is almost 200pF, wheras the 1N4007 is 15pF. Scope Traces Upper trace is gate, lower is drain. –  SubZero Mar 13 '14 at 5:26
Suspiciously close period to 1K * 200 pF. 1N4148 is only 4pF. Hmmm –  Spehro Pefhany Mar 13 '14 at 5:54
I think I might have found out why, however I do not fully understand why which will probably be my next topic to understand. I was told that a faster diode will result in more oscillations. I tried bypassing the diode with a 0.1uF capacitor and viola it's back in business although with a 24% drop in efficiency. 87% -> 63%. –  SubZero Mar 13 '14 at 5:58