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I have a H-bridge that is being supplied by 12V and using around 10A. I am turning it on/off with a microcontroller that has an output of 0V-5V. The switching frequency is "super low" so the gate charge of the MOSFET is not a problem. I am trying to "convert" this 5V logic to 12V logic (0V-12V), and thought of something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Is there any redundacy, any problem or anything missing from this design? I can see this work, but this will be in "somewhat of a commercial product", and would like it to be as robust as possible.

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  • \$\begingroup\$ What manner of part is "OA1"? \$\endgroup\$ – Spehro Pefhany Feb 2 '15 at 16:03
  • \$\begingroup\$ @SpehroPefhany Its a TL072: ti.com/lit/ds/symlink/tl074.pdf \$\endgroup\$ – Golaž Feb 2 '15 at 16:08
  • \$\begingroup\$ @Tut Schematic modified. I dont have any rail-to-rail at hand... \$\endgroup\$ – Golaž Feb 2 '15 at 16:38
  • \$\begingroup\$ Most H-Bridges accept logic-level inputs; are you sure yours doesn't? \$\endgroup\$ – Nick Johnson Feb 2 '15 at 16:46
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    \$\begingroup\$ Take note of what Spehro touched on at the end of his answer. During the transition, both the high-side and low-side MOSFETs will be on (shoot-through) causing a current spike. This is a problem when the gates are tied together. Better to drive them separately with some short dead-time when they are both off. \$\endgroup\$ – Tut Feb 2 '15 at 18:09
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I suggest something more like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Drive into a 10nF load looks like this:

enter image description here

Edit: The violet trace is your circuit- as you can see, very very slow and it does not have enough swing to turn the MOSFETs off reliably- Vgs(th) can be as bad as 2V (time scale changed to show response of slow circuit).

enter image description here

You really want this circuit to switch rapidly or to insert some dead time- both MOSFETs will be one for a time when it is switching which causes shoot-through current.

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  • \$\begingroup\$ Wont R2 cause any problems? I though push-pull stages mustn't have a base resistor? Are there any drawbacks? \$\endgroup\$ – Golaž Feb 2 '15 at 16:50
  • \$\begingroup\$ R2 is for level shifting. The only problem it causes is to make the rise time slower than the fall time- a trade off between current consumption and speed. \$\endgroup\$ – Spehro Pefhany Feb 2 '15 at 16:59
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I first saw the following circuit in a TI app note many years ago. It works well and is robust. I have used it or variations of it in several projects (and products).

schematic

simulate this circuit – Schematic created using CircuitLab

There are a couple of things to consider:

1) This circuit inverts the drive signal.

2) R3 is added to ensure the FET drive remains OFF while the controller is in reset (i/o pin is input instead of output).

3) Value of R1 may need to drop if PWM frequency is high or if FET has large gate capacitance.

4) Add low-value resistor in series with FET gate as close to the FET as possible. That's why it's not shown on the schematic - this gate resistor is part of the FET circuit rather than the driver. The purpose of the resistor is to reduce or eliminate parasitic oscillation that can occur of the FET is located some distance away from the driver circuit. This resistor value is usually fairly low - 22R to 100R. I use 47R in most of my low-frequency (1KHz or less) PWM circuits.

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  • \$\begingroup\$ There wouldn't be a problem with voltage not swinging close to GND, when using MOSFETs with -2V and 1.7V thresholds? \$\endgroup\$ – Golaž Feb 2 '15 at 16:43
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    \$\begingroup\$ Do you have a reference to the app note or title or subject ? \$\endgroup\$ – efox29 Feb 2 '15 at 17:09
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    \$\begingroup\$ @Golaž It doesn't look like that should be a problem. The diode has only 1V forward voltage and the typical Vbe for a 2N4401 would be 0.7V when on so in both cases, there's a fair amount of margin. \$\endgroup\$ – horta Feb 3 '15 at 1:13

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