0
\$\begingroup\$

I've done a good bit of experimentation and learned quite a bit from other members of the site here. I've built several circuits that drive an N-channel MOSFET on the "low side". But to build an H-bridge inverter I need to drive MOSFETs on both the "high side" and the "low side".

Although I have the components to assemble the necessary driver circuit, I am tending to agree with other contributors here that I should just buy a "driver" chip. From what I understand most of these are just a charge pump internally with some required external components.

I think I understand the usage of such circuits, but have a few questions. Most of my knowledge comes from reading the datasheet of the IR2125 at this point.

I found it here: http://www.irf.com/product-info/datasheets/data/ir2125.pdf

  1. All driver chips list a value like V-offset. Is this the maximum voltage they can produce above ground in the circuit? For example, if I have a 170 VDC current source, the gate of the high side for most MOSFETS needs to be around 180 VDC with respect to ground. This would give me a Gate-Source voltage of approximately 10 volts.

  2. Most drivers have a IN or SIGNAL pin. This is what controls whether the driver is charging the gate or discharging the gate. I'm assuming my signal will always be 5 volts coming from a microcontroller. If I connect the V-B (High Side Floating Output Voltage) of the driver to the source of the high side MOSFET, do I need an optocoupler to drive the driver circuit?

  3. Drivers list a T-on and a T-off. When I consider the maximum switching frequency of a circuit, do I just need to sum those values with switching speed of the MOSFET I am using?

  4. I do not understand how the "current limiting feature" of the IR2125 is meant to work. It shows an additional connection to the MOSFET I do not fully understand. Is this driver meant to work with MOSFETs like the IRCZ44? Can I just ignore the current sense pins on the driver, or is there a cheaper driver I can select?

  5. The companion to the IR2125 is the IR2121 for the "low side". Are these chips just meant to be a cheap way to get gate drive voltages from a +5 VDC power supply?

\$\endgroup\$
  • \$\begingroup\$ What is the input DC voltage to your half-bridge? What currents are you going to switch by MOSFETs? \$\endgroup\$ – Master May 14 '16 at 17:20
  • \$\begingroup\$ There are reliable Half Bridge driver ICs. They receive 2 logical signals and drive 2 MOSFETs: high side one and low side one. However, they operate typically up to 80 V DC input. \$\endgroup\$ – Master May 14 '16 at 17:24
  • \$\begingroup\$ However, if your switching voltage is high (200 V or above), you have to use isolated drivers. And the choice of the right driver depends on the power you switch. \$\endgroup\$ – Master May 14 '16 at 17:26
1
\$\begingroup\$
  1. I am not sure the term v-offset has universal meaning. The best way to tell is to look at the Recommended Operating Conditions (on page 2 of the datasheet), it listed Vs (a specific pin label) can operate up to 500V max. In conjunction with a note above that all voltages are referenced to COM, it is a positive indication that the driver is designed to operate up to 500V.

  2. On page 3 of the datasheet, there are specifications of the "IN" pin -- VIH and VIL indicate 5V TTL levels; I-in+ and I-in- indicate low current requirements meaning it can be driven by any logic output.

  3. t-on and t-off are propagation delays and have some bearings on the maximum frequency, they are also important when you have to time a bridge to minimize shoot-thru. But for the overall maximum switching speed of the MOSFET, the rise and fall time (t-r, t-f) play bigger roles. In most cases, you don't want to operate near the "maximum" switching rate because it would be too inefficient, and also too "non-linear" if that is important.

  4. The current limit feature operate by sensing the voltage difference between the CS and Vs pins. The threshold values are given on page 3 as V-CSTH+ and V-CSTH-, with the typical value being 0.23V. As an example, connect a 0.1 ohm resistor from CS to Vs as in the diagram on the first page. When the current through the resistor gets to 0.23V/0.1ohm = 2.3A, the current limit would kick in and turn off the HO pin.

  5. It is a cheaper circuit for low side. The gate drive (and supply) voltage has to match that of the MOSFET and is not necessarily 5V.

\$\endgroup\$
1
\$\begingroup\$
  1. The gate can be driven as high as the offset voltage plus 18V (Vb).

  2. You need an isolator if you want to isolate the microcontroller from the supply bus. If the bus is connected to the mains it could be a requirement for safety reasons (or perhaps the safety isolation could be provided elsewhere and the micro would be directly connected to the mains). Certainly for development you'd want isolation. Don't forget that you have to switch the input continuously for it to work.

  3. It's a bit more complex than that if you want it to work well, but ton+tr+toff+tfall would be the fastest it could (more or less) fully switch the gates.

  4. The current limit works by sensing a voltage across a very low value resistor. When it reaches 230mV nominal it is sensed as overcurrent. There is probably some kind of blanking logic in there for when it switches to create a short delay to account for the gate driver current and stray inductances. The schematic shows a device with an extra sensing lead- one trade name is SENSEFET. The extra lead splits off a fraction of the current passing through the main source lead. You can think of it as a small MOSFET in parallel with a much larger MOSFET, and the little buddy MOSFET does a small share of the work. For example, the NTMFS4833NS has a sense ratio of 387:1, so a 100A drain current would result in a 258mA sense current, and a resistor of 0.9\$\Omega\$ would trip the overcurrent at about 100A.

  5. Not just that, they drive significant current into the gates to get them to switch promptly. If you don't drive enough current the MOSFETs will switch sluggishly and you'll get a lot of extra heating, especially at higher switching frequencies. So it's level shifting and a fairly high current push-pull output stage. Plus UVLO, which is an important system self-preservation circuit.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.