I have been looking for a simple h bridge for high power applications (say ~2kW total--whether that be 2kV at one amp or 2kA at 1V I really do not care. I figure I can always use a transformer to switch between the two). Anyway, the point is, I just want take a lot of DC power (two leads), and switch it across a load (two leads). In order to determine which way you want to switch it, you need two inputs (two leads). Here are just a few examples of full bridge h bridges with redundant leads and/or leads that should be tied together but are not:




Note the redundant leads on the first two. For example, in the second link, {13, 21}, {2, 19}, {10, 15}, and {4, 12, 14, 20} are clearly connected by wires. What is the point of having multiple pins if they are directly connected by wires? Furthermore, the last link, the only way I know how to use that design is with {G1, G4} on and {G2, G3} off or vica versa. Why not just tie gates together so that is the case? Then you have two inputs (each connected to two gates) two outputs, and two leads for power.

Although I would not mind some examples of what I am looking for, I am just as interested in knowing how to take full advantage of these seemingly incomplete or redundantly wired h bridges.


2 Answers 2


There are two functions that these "redundant" pins provide. Using your first datasheet as a reference, the grouping of pins helps to show their function.


Groups (9,10) (13,14) (15,16) and (11,12) are tied together to reduce resistance and inductance.

The pins directly connected to the source of each transistor (1,3,6,8) are lower resistance, lower inductance connections to the sources which is better for the gate drive circuit. Keep in mind that some gate drivers can drive over 9A peak gate current, so minimizing inductance is a must.

In terms of tying control pins together, that isn't possible with power devices due to dead time switching requirements and gate drive voltages.

  • \$\begingroup\$ As for reducing resistance and inductance, what is the point of reducing the resistance and inductance by doubling the number of wires if you are just going to solder them both to the same wire? Unless you are supposed to use twisted pairs... Also what do you mean you cannot tie control pins together? What is a "dead time switching requirements" and what do the gate voltages have to do with it. Please elaborate. \$\endgroup\$ Apr 23, 2012 at 22:34
  • \$\begingroup\$ @Feynman Paralleling the pins reduces the net resistance and inductance inside the package, which can get significant for switching converters and motor drivers. Dead time is the time when all devices are off to prevent current from shooting through from VDD to VSS. \$\endgroup\$
    – W5VO
    Apr 24, 2012 at 0:41
  • \$\begingroup\$ @Feynman You can't tie control pins together. For example, if you were using the module above for an H-bridge, you couldn't connect pins (7,4) or pins (2,5). This would cause the high-side device (gates 4 or 5) to blow when that side is turned off. (Vgate = 0V, Vgs = -VDD). \$\endgroup\$
    – W5VO
    Apr 24, 2012 at 0:46
  • \$\begingroup\$ So then could I connect them with two outward facing diodes? If I understand correctly, that should prevent them from being damaged when pulled low. \$\endgroup\$ Apr 28, 2012 at 19:43

In the case of doubling the power leads, the doubling changes current capacity, parasitics, frequency response, and thermal dissipation. The specs cited on the datasheet are tested with all the leads connected. To get those specs, guaranteed, one needs the same conditions.

In terms of the gates, separating the drives allows for custom deadband control, duty cycle control by shorting (0V) the output, and crossover timing.

Shorting the output (0V output) is achieved by having both high sides closed with bottoms open, or vice versa.

Deadband control allows for a briefly letting the diodes freewheel while switching the output to prevent load spikes by having both high and low side switches closed momentarily at the same time.

Duty cycle control allows modulation of the output voltage from zero to full scale.

One may also switch the output from zero to full at the moment of load current crossover for reliability or efficiency.

Not all applications are simply bang-bang full scale output.

Sometimes the precise timing of these things are the secret sauce in the application, and so are left to the a drive external to the power components.

  • \$\begingroup\$ As for the power leads, why not just double the thickness of one wire rather than having two. As for the deadband control, could you not get the same effect by connecting the two inputs in my hypothetical two input two output two power lead h bridge? \$\endgroup\$ Apr 23, 2012 at 22:26
  • 1
    \$\begingroup\$ @Feynman: In many applications, one will need to drive both sides of a motor low (dynamic braking). Further, in many applications which use PWM control, it will be better for various reasons to leave the low side on while modulating the high side, than to modulate both simultaneously (modulating both simultaneously will cause current stored within the motor's inductance to be sent back toward the supply; in some cases this may be good from an efficiency standpoint, but in other cases dumping current spikes into the supply could cause problems.) \$\endgroup\$
    – supercat
    Apr 23, 2012 at 23:08

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