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I recently discovered the Bias Resistor Transistors (BRT) / Digital Transistors as commercial solution for low power, cheap solid state switches.

This IC is offered by multiple suppliers (Rohm, NXP, Infineon, Onsemi...) and all of them use transistors of BJT type. enter image description here

Why there's not a mosfet integrated solution for BRTs?

My first guess would be that since BJTs have higher transconductance, can deliver current faster and have lower capacitance, so they might be more useful to be used as simple gate drivers for higher power transistors.

However, I have found circuits where a low side digital transistor is used to drive a high side, low power PMOS transistor. Is that only for redundancy purposes? Just 'cause BRTs are very cheap? Wouldn't just adding a series resistor to the mosfet's gate a cheaper solution?

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  • \$\begingroup\$ Did U not appreciate my wisdom? \$\endgroup\$ Jan 22 at 14:29

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Why there's not a mosfet integrated solution for BRTs?

  • A BJT needs a series resistor to make it's input compatible with a logic level voltage source whereas, a MOSFET is voltage driven to the gate hence it doesn't require a series resistor to make its input compatible with most common logic level voltage drives.
  • A BJT has much much lower input capacitance compared to a MOSFET hence, the pull-down resistor is less problematic for high-speed switching. Compare this with a MOSFET; it has quite a high gate-source capacitance and, this can vary from a few tens of pico farads to several nano farads hence, there is a less clear-cut value of pull-down resistor that fits many user requirements and, inevitably, the user would likely have to fit an external pull-down resistor to suit the speed of their application.
  • A BJT is guaranteed to activate (to within a defined specification) with logic levels using the series resistor and pull-down resistor whereas a MOSFET has quite a variation in gate-source voltages to suit the drain current requirements of the user. This means that adding a series resistor and pull-down resistor is unattractive to most pro users.
  • MOSFET circuits sometimes do need a series gate resistor but this is usually in the region of a few Ω to tens of Ω. Due to the nature of most MOSFET switching circuits, the power dissipated by this resistor means it might have to be a bigger package compared to what some internal resistor might be.
  • Some MOSFET applications will suffer if there is an inbuilt series resistor hence it doesn't make sense to load this component into a MOSFET package.

BTW your diagram is misleading as it shows a PNP transistor with emitter connected to ground (a highly unlikely scenario in most applications): -

enter image description here

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    \$\begingroup\$ Hi Andy, thanks for your explanation. Your reasoning makes sense. I got this symbol from an Onsemi's datasheet of a digital transistor. \$\endgroup\$ Jan 21 at 13:07
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schematic

simulate this circuit – Schematic created using CircuitLab

Driver switches with open Collector or Drain always invert.
So a high side FET switch must use a low side driver which must be NPN.

The example above shows "Pre-biased Bipolar Junction Transistor (BJT)" It typically uses input logic level to the pre-biased NPN to a high voltage Pch switch. A uC's IO ports use push-pull CMOS drivers or as high impedance input, which cannot exceed their own supply. Thus an inverting NPN makes perfect sense.

, I have found circuits where a low side digital transistor is used to drive a high side, low power PMOS transistor. Is that only for redundancy purposes? Just 'cause BRTs are very cheap? Wouldn't just adding a series resistor to the mosfet's gate a cheaper solution?

Yes they are cheap solutions, but it is necessary because the load supply voltage is often greater than the logic levels.

BJT's do have a C-B Miller Capacitance but it is much smaller than the D-S FET output capacitance spec'd as Coss.

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