NPN & PNP matched pairs at 10A

If you're really using mosfets, and not true bipolar transistors (there is a misuse/confusion of terminology in the question,) mosfets have a current sharing property whereas bipolar transistors have a current hogging characteristic. This means you might try to put two or more p-channel fets in parallel to equal the lower on resistance of the corresponding n-channel fets. This also furnishes two parallel thermal paths, further reducing the junction temperature rise.

The term "matched" is distinguished from the term "complementary"-- one  finds a matched differential pair of transistors at the input of an  opamp, but, a complementary pair of devices at the class-B output of a  power amplifier.The origin of class-B circuts is vacuum tubes, of course only available in one polarity. The matched pair of output tubes was not unlike the opamp's input matched pair. The tube circuit's center tapped primary winding allowed the sum of the fundamental (linear) part of the signal, but the difference of the second order distortion nonlinear response. This cancellation was not perfect, but certainly adequate to the first order. The availability of complementary transistors changed the usual class-B output stage topology forever.

Mosfets used as switches tend to exhibit resistive behavior, in contrast to bipolar transistors which show a voltage offset- Vcesat plus some low series resistance. In some circumstances a bipolar switch might have a lower Vcesat than the mosfet's I*R drop, giving the bipolar the dissipation edge. The problem of second breakdown in bipolar transistors is a legitimate concern, which is manageable, capable of being considered in a proper design.

The issue of matching is still not clear in my mind... the idea of equalizing the dissipation of the upper(p-channel) device with the lower(n-channel) device as was suggested in this group seems to be a red herring to me. As long as the device's ratings are observed, does the difference in Rdson cause any other problems? Either the same size device with higher Rdson runs at a higher (but tolerable) Tj, or, a beefier device with the same Rdson has the same dissipation with a likely lower Tj, since its J-C thermal resistance is lower in a larger device.

The concept of  a "balanced bridge" is the essential property used in wheatstone bridge measurements, but the term "bridge" describing the h-bridge implies the four branches, and not the balanced property of wheatstone bridges. When your h-bridge is operating properly, it departs from the balanced condition as much as possible.

A microstepping application of stepping motors requires a linear current drive, rather than a switch drive. Intermediate values of current in competing windings gives the effect of additional poles, if one considers the torque detent position possible in this situation. In this case, the output transistors are operating in a more linear region than switches, so, the on resistance or Vcesat is not as important.

Do I need to dig deeper into your application?

Jeff Furman

If you're really using mosfets, and not true bipolar transistors (there is a misuse/confusion of terminology in the question,) mosfets have a current sharing property whereas bipolar transistors have a current hogging characteristic. This means you might try to put two or more p-channel fets in parallel to equal the lower on resistance of the corresponding n-channel fets. This also furnishes two parallel thermal paths, further reducing the junction temperature rise.

The term "matched" is distinguished from the term "complementary"-- one  finds a matched differential pair of transistors at the input of an  opamp, but, a complementary pair of devices at the class-B output of a  power amplifier.The origin of class-B circuts is vacuum tubes, of course only available in one polarity. The matched pair of output tubes was not unlike the opamp's input matched pair. The tube circuit's center tapped primary winding allowed the sum of the fundamental (linear) part of the signal, but the difference of the second order distortion nonlinear response. This cancellation was not perfect, but certainly adequate to the first order. The availability of complementary transistors changed the usual class-B output stage topology forever.

Mosfets used as switches tend to exhibit resistive behavior, in contrast to bipolar transistors which show a voltage offset- Vcesat plus some low series resistance. In some circumstances a bipolar switch might have a lower Vcesat than the mosfet's I*R drop, giving the bipolar the dissipation edge. The problem of second breakdown in bipolar transistors is a legitimate concern, which is manageable, capable of being considered in a proper design.

The issue of matching is still not clear in my mind... the idea of equalizing the dissipation of the upper(p-channel) device with the lower(n-channel) device as was suggested in this group seems to be a red herring to me. As long as the device's ratings are observed, does the difference in Rdson cause any other problems? Either the same size device with higher Rdson runs at a higher (but tolerable) Tj, or, a beefier device with the same Rdson has the same dissipation with a likely lower Tj, since its J-C thermal resistance is lower in a larger device.

The concept of  a "balanced bridge" is the essential property used in wheatstone bridge measurements, but the term "bridge" describing the h-bridge implies the four branches, and not the balanced property of wheatstone bridges. When your h-bridge is operating properly, it departs from the balanced condition as much as possible.

A microstepping application of stepping motors requires a linear current drive, rather than a switch drive. Intermediate values of current in competing windings gives the effect of additional poles, if one considers the torque detent position possible in this situation. In this case, the output transistors are operating in a more linear region than switches, so, the on resistance or Vcesat is not as important.

Do I need to dig deeper into your application?

Jeff Furman

As Jeff said above there isn no need really to have upper halves and lower halves the same resistance as the total series resist ance will end up the same in either direction/bridge polarity. If for some odd reason you really do need equal resistances, you can use all n-channel devices and a gate driver IC for each device, this complicates the design and requires additional floating power supplies though.