Understanding Vce Voltage Drop of a BJT

[likephysics]

I don't understand the Vce voltage drop of a BJT.
If I try to think of the BJT as 2 diodes - Base emitter diode and base collector diode. Vce is the voltage across the cathode of the base-collector diode and cathode of Base-emitter diode.
How can it be 0.2V(when the BJT is saturated). I don't get it.

Can it be thought of in terms of resistance, like ON resistance in mosfets?

 

[Studiot]

The diode like connections are between
base and emitter
base and collector

The collector emitter connection is not diode like, as it contains an active component.

The saturated condition you are describing occurs in Common Emitter configuration.
Here the emitter is grounded or held at zero volts.
The active action of the transistor is such that the collector can be set or driven to any voltage between the supply rail and the saturation or botoming voltage (0.2volts in your case). This active action forces the appropriate current through any collector resistor to cause this.

And yes you can work in terms of CE resistance - the term transistor comes from 'transfer resistance' ie the CE resistance is controllable. Circuit analysis is not usually done this way because it requires knowledge of some hard to measure parameters.

This active action is, of course, known as transistor action and is precisely what distinguishes it from two back to back connected diodes. Diodes are passive components in that they cannot force the current or voltage in a circuit, their response depends upon the current and voltage instead.

 

[likephysics]

Thanks for the explanation.
Imagine a NPN transistor. I put the Multimeter probes at collector(close to base) and emitter(close to base). So Vce is the voltage here?
If so, why doesn't the Base-emitter diode drop show up here?
What is this CE resistance exactly. Is it the resistance of the base channel?
Why isn't this specified in the datasheets.
Can I use the CE resistance to compare the power consumption of BJT and Mosfet.