*2) There is no pull down on the FET gate: This is the problem I first thought. This will reduce the efficiency of the circuit. The 330 Ohm (written on the board, but look, there is a 270 Ohm resistor installed) will discharge the gate in a given time. The IRFD has a 6nC gate charge. The 270 Ohm resistor will be able to discharge the gate in 6E-9 S * 270 = 1.62 uS. During this discharge, the drain current reduces linearly, meaning that the turn if is "smooth", and not sharp like we need. This is a issue particular to the 34063: it can only source current but not sink it.* The 270 ohm pull down resistor is the only change you can easily make to the existing design. At 1.62uS discharge time, that is more than 10% of the data sheet period with a 470pF timing capacitor: The transistor in switching regulators is meant to operate either ON or OFF, never in linear mode. Turning the transistor ON to charge the inductor is easy as the regulator output transistor is capable of delivering 1.5A but that resistor can't turn the transistor OFF. Add to this the fact that when the controller wants to turn off the FET with 12V on the gate, the gate has to fall about 6V before the transistor really begins to turn off which throws off the timing of the control loop. One last thing that happens is that when the transistor does start to turn off, the fast rising drain voltage is coupled into the gate through the drain-to-gate capacitance trying to turn it back on which results in a period of unhealthy oscillation.
Since the gate driver is capable of 1.5A of output current, you can lower the pull down resistor pretty dramatically without breaking anything except the low load efficiency (Which can't be very good with the transistor getting "Hot"!). You will have to scale up the power rating of the resistor by replacing it with a larger, perhaps axial part, but this should improve your transistor heating. A 100 ohm gate pull down resistor will improve your turn off time by a factor of 3 but at the cost of about 1.25W of dissipation (Use a 2W resistor). Another way to speed this up is to use an active pull down circuit on the gate in place of the passive pull down resistor. Using a current sink instead of a resistor has the added benefit of drawing the same current regardless of the applied voltage vs the RC discharge curve and it is less susceptible to the drain-to-gate coupling. Where the discharge current of the 270 ohm resistor is 44mA at the start of the gate discharge, it falls to only about 15mA at the critical 4V gate threshold voltage. A current sink like that shown below pulls the same current regardless of the applied voltage down to it's dropout voltage of 1V which is well below the 2V minimum gate threshold voltage of the MOSFET. The current sink at 4V looks like an equivalent 80 ohm resistor and at 2V a 40 ohm resistor, i.e. 3 to 7 times faster turn off for the same power dissipation of the 270 ohm pull down. The 20ohm resistor does not have to dissipate a lot of power as this task is shifted to the transistor. Increasing the current is now just a function of adjusting the 6.2K resistor. -- You received this message because you are subscribed to the Google Groups "neonixie-l" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send an email to [email protected]. To view this discussion on the web, visit https://groups.google.com/d/msgid/neonixie-l/c4d0ab65-a138-4e4b-ba30-cf9f99a2dca6%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
