Active load vs active diode

Question

It makes sense that when VGS = VDS, the channel resistance (lower slope) is large compared to the resistance in ohmic region (steeper slope,) so a MOSFET can be used as a resistor.

From a previous chapter I know that a BJT acts as a diode when the base is connected to the collector.

As I'm new to MOSFET circuits, I always compare MOSFETs to BJTs to understand better. Almost all the BJT circuits work just fine (logic/functionality wise) when we replace the BJT with a MOSFET.

Now it is confusing because with a BJT, the arrangement gives a diode fixing the base emitter drop to around 0.7 V.
Sith MOSFET, the arrangement gives just a resistor, not a diode.

Gate-drain shorted MOSFET gives a resistor, but base-collector shorted BJT gives a diode. Why?

Answer 1

Now it is confusing because with bjt, the arrangement gives a diode fixing the base emitter drop to around 0.7V. But with mosfet, the arrangement gives just a resistor, not a diode.

It's all about the scale of things.

An NPN BJT with the base shorted to the collector looks like a diode with approximately 0.7 volt drop because; the forward-biased base-emitter region is basically a diode (a PN junction) with 0.7 volt forward volt-drop. Sure there are a few intricacies involved but, the bottom line is this.

For a MOSFET, the gate-source region is high-impedance and thus the gate-source voltage can rise (within reason) to cause drain-to-source conduction without any current flowing into the gate. Hence the volt-drop of an N channel MOSFET when gate connects to the drain is like a diode but scaled much bigger that a BJT.

The upshot of this is that the MOSFET appears more linear and like a resistor but, if you pushed the applied voltage a bit further it would start to behave non-linearly and more like a diode but, a diode with a much bigger forward volt-drop than that associated with a normal diode or a BJT.