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I want to turn on and off both the mosfets at same time and I used IR2110 gate drivers for high and low MOSFETS.

When I try to simulate the circuit, after few seconds it shows error saying time step too small. What is the problem with circuit or simulations?

In the simulation graph,

Green represents the inductor current Blue represents gate voltage of M1 Pink represents gate voltage of M2

circuit diagram

error message simulation

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  • \$\begingroup\$ If you were building a half-bridge design, that is not the correct design. M1 and M2 should take turns being on then off. LTspice is warning you of a timing error you should not have. Please search "Half-bridge driver" on the web for crucial design information. \$\endgroup\$ – Sparky256 Feb 20 '18 at 21:21
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    \$\begingroup\$ @Sparky256 This isn't supposed to be a half bridge; this is a full bridge, but with two MOSFETs replaced by diodes. A perfectly fine arrangement if you only need unidirectional current trough the load, yet want to recover the energy stored in the inductance of the load. \$\endgroup\$ – jms Feb 20 '18 at 21:27
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    \$\begingroup\$ Are there any unusual currents or voltages being reported when LTSpice halts? \$\endgroup\$ – Spehro Pefhany Feb 20 '18 at 21:45
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    \$\begingroup\$ Between the H bridge output and those capacitors. Leave VB connected to the caps. Don't alter R3's position. It's just an experiment with what I have seen on previous problems with sims. \$\endgroup\$ – Andy aka Feb 21 '18 at 12:59
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    \$\begingroup\$ @lomeshpudipeddi then I would try to tackle that problem while using only one IR2110 because you dont know what's the connection inside this IC between VB and Ho. this might not be the solution but a better way to simulate your desired functionality. At the end of day, you are going to use only one IR2110, why not use it from very beginning ? \$\endgroup\$ – HerrderElektronik Feb 22 '18 at 8:54
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Not sure about the LTSpice error, but I do see one problem. In order for the high side IR2110 to work, the bootstrap capacitance (C1 and C3) must be charged so that 12 volts appears across it. This is done by pulling the VS pin to ground, which happens every time you switch the low side on a normally-configured half bridge. In your circuit, this capacitor will not ever be charged. C1 and C3 must be refreshed periodically by pulling VS to ground; the charge on the cap provides the current and voltage necessary to get the high side gate drive voltage to a voltage greater than your 12 V supply. If the charge on this cap is less than 8 volts or so, the high side switch will shut off. You could put in the low side FET instead of D2 and pulse it at the start and then again periodically.

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    \$\begingroup\$ But C1 and C3 do get pulled to ground (by the load inductance) every time M1 switches off while current is flowing trough the load. I know that to get the bootstrap caps charged for the first cycle (at startup) you do need a different method, but that isn't complicated: turn M2 on (while leaving M1 off), which after a while pulls the source of M1 low trough the load. \$\endgroup\$ – jms Feb 20 '18 at 21:40
  • \$\begingroup\$ Exactly correct. \$\endgroup\$ – John Birckhead Feb 21 '18 at 0:22
  • \$\begingroup\$ Thanks for the explanation @JohnBirckhead. Is there any other way without replacing D2 with FET and pulsing it periodically? Because I saw in some other models with different gate drivers, connected in the similar way work perfectly fine. Or would you give me an example of pulsing FET in my circuit? \$\endgroup\$ – lomesh pudipeddi Feb 21 '18 at 10:48
  • \$\begingroup\$ There is a good discussion at irf.com/technical-info/designtp/dt98-2.pdf \$\endgroup\$ – John Birckhead Feb 22 '18 at 14:18

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