Boost 5V to 18V: charge pump + LDO or conventional boost converter?

Question

I am trying to build a H-bridge driver using discrete parts. Now I have a problem regarding generating the 10+V gate voltage for the NMOSFETs I used.

The H-bridge and its controls is like this:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

My question is, how do I generate this +18V rail? DC-DC Boost converter, or charge pump to ~20V and LDO it down a bit?

Don't worry about my 5V rail, it is capable of 30+ amperes.

Can you use PMOS on top? Then you would not need to generate a special voltage to turn them on. — user57037, Dec 25, 2014 at 3:09
PMOS have a lower current rating and larger Rds(on). This will cause problems. — Maxthon Chan, Dec 25, 2014 at 3:22
I searched on digikey. I found many parts that have Rds < 44 mOhm and Id larger than 33A. I didn't find anything in same TO220 package, though. What is your max current anyway? And will you be in volume production? — user57037, Dec 25, 2014 at 4:44

Spehro Pefhany · Accepted Answer · 2014-12-24 23:17:07Z

0

A simple boost converter would probably be the easiest for the currents involved.

Those drivers are not going to be the best- fast turn on + slow turn off = lots of shoot-through.

answered Dec 24, 2014 at 23:17

Spehro Pefhany

384k21 gold badges329 silver badges862 bronze badges

\$\begingroup\$ Then how to reduce the shoot through then? I can reduce the gate pull-up resistors. \$\endgroup\$
– Maxthon Chan
Dec 24, 2014 at 23:50
\$\begingroup\$ The easy and expensive way is to use gate driver ICs such as those from IRF that include dead time as well as push-pull drive. If you don't want to do that, use a push-pull driver (two MOSFETs rather than a MOSFET and a resistor) or a similar circuit with two BJTs. Either will be better than what you have. Consider inserting small delays (100 ns or whatever) to allow the opposing power MOSFETs to switch off. \$\endgroup\$
– Spehro Pefhany
Dec 24, 2014 at 23:56
\$\begingroup\$ And the IC way is what I cannot afford to do. I know very well about those ICs when going through IRF portfolio searching for power MOSFETs that fits my need but those are just too expensive. About push/pull driver my microcontroller only have 5V output, how to make that work? \$\endgroup\$
– Maxthon Chan
Dec 25, 2014 at 1:34
\$\begingroup\$ Add another transistor instead of the resistor- three transistors per driver. One (plus a resistor) for the level shifter and two complementary transistors to drive the gate. \$\endgroup\$
– Spehro Pefhany
Dec 25, 2014 at 3:16
\$\begingroup\$ Rephrase this, open drain inverter circuit for level shifting, push-pull inverter circuit for driving the gate, right? \$\endgroup\$
– Maxthon Chan
Dec 25, 2014 at 3:21

| Show 1 more comment

1 Answer 1