Reply 20 of 25, by mkarcher
Your board has the block of four jumpers to the right of the 3.3V/4V select jumper. Boards with a jumper block like this typically do not support auto-voltage using the VOLTDET pin.
What you call "VRM" is in fact just a single NPN transistor. It's one of the important parts of the regulator, though. The middle pin ("collector") is connected to the 5V rail of your power supply, the right pin is the output ("emitter") and connected to the Vcc pins of the processor. The left pin is the control pin and is typically 0.5 to 0.7 volts above the output voltage when the VRM is operating.
It looks indeed like the transistor is bad. The left pin (near the ISA slot) should never be more than 0.8V higher than the right pin (near the 3.3V/4V jumper), except if an exceptionally high current is going to the left pin. I am very confident that the circuit on this board doesn't let high currents get near the left pin, and even if it did, the pin would likely be at a slightly lower voltage than the central pin. If the transistor is just overloaded, I would expect 2.8 to 3V on the left pin when the right pin is at 2.0 to 2.2V.
On the other hand, if I misunderstood what you said about left and right pins, and the pin in the middle and the pin near the 3.3V/4V jumper are both at +5V, whereas the pin near the ISA slots is at 2.2V, this would indicate that 5V is getting to Vcc on a different path than through the transistor (and thus "overrides" the VRM), and the regulator circuit in response pulls the control pin down to 2V, because the output voltage is way higher than the requested voltage of 3.3V or 4V. Two possible sources of "shorting out" the regulator are other broken components on the board, or accidentally not putting the 3.3V/5V-block into the 3.3V position.