# How are Transistors in the Cut-off Region in Double Reverse Bias?

To get an npn (say) BJT in the cut-off region, you set the base current to 0 and the voltage less than 0.7V. This means that $$\V_c > V_b\$$ and $$\V_e > V_b\$$. However, the emitter voltage is 0 (ground) normally, and the base voltage would be less than 0.7V, which is not less than 0, so $$\V_e < V_b\$$.

So when people say BJT's in cut-off have both junctions in reverse bias, what do they mean?

Here both BE and BC junctions have reverse external biasing voltage. There's no current except leakage as long as the breakdown voltages of the junctions are not exceeded.

Turn the polarity of the source in the left. Then the base current starts and the transistor can also sink some collector current from the source in the right. But now as drawn the transistor is in the cut-off state because there's no current except leakage.

The NPN transistor shown with its emitter at 0V, it is in the cut-off region when its base current is zero which is when the base is at 0V or is negative.

The base-emitter of the transistor has forward bias when its base is positive and has reverse bias when its base is negative.