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Will a BJT transistor still be in saturation mode when exceeding 0.7V?

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petert:
Hi,

I am trying to understand the details of transistors's operation modes, following the tutorial from Sparkfun: https://learn.sparkfun.com/tutorials/transistors/operation-modes
Please don't beat me haha, I am trying to get a basic understanding, before I dig into even more advanced descriptions to avoid drowning in countless relationships and graphs.

As an example I took a general purpose NPN BJT: BC547 (datasheet).

For saturation mode, Vbe(on) is specified to be between 0.58V and 0.70V (or 0.77V for higher current).
So when reaching this level for Vbe (say 0.6V), and ensuring Vc is lower than Vb (such that Vbc > 0), the transistor should be in saturation mode, right?
Keeping Vce greater than Vce(sat) is just optional for entering saturation mode, but not exceeding Vce(sat) would prevent any current to conduct between C and E?

What if I increase Vbe even more (and keep Vce unchanged), would it sill work? I couldn't find an absolute maximum rating for Vbe, or is that the 0.7V/0.77V?

petert:
Ok, I got confused. Since BE acts like a diode, the voltage drop will be between the min and max ratings given in the datasheet (so between 0.58 and 0.7/0.77). So Vbe(on) is analogous to the forward voltage ratings of a diode.

Any remaining voltage will simply be dropped at any component(s) following the emitter of the transistor.

Yet, diodes fail when a certain voltage is exceeded, so what would be the maximum voltage for the BC547, that the base could tolerate?

T3sl4co1l:
Vbc actually has very little to do with it.  Most saturate at positive Vbc, some at negative (particularly power transistors, particularly high voltage ones).

Saturation might better be defined by the increase in h_oe, or the decrease in h_fe, following some threshold (e.g. a change of 2x or 10x or whatever).

(Look up the definitions for h-parameters, if needed.  They're ratios of device input and output voltages and currents, specifically, the changes in them around a center bias level.)

This is reflected by the design process, where we might use a BJT as a saturated switch at some forced hFE(sat).  That is, the base is driven with Ib, the collector has a load current less than Ib * hFE(sat), and the collector voltage will pull down to whatever it saturates to.

The fact that the collector voltage has a low impedance, is another way of saying "large h_oe".  That is, the output conductance, it looks conductive, like, y'know, a switch -- in contrast to the off state (no current flow, or dynamic resistance), or the linear range (current flow, but high dynamic resistance).

Tim

GeorgeOfTheJungle:

--- Quote from: petert on February 05, 2020, 01:44:43 pm ---Yet, diodes fail when a certain voltage is exceeded, so what would be the maximum voltage for the BC547 that Vb could tolerate?

--- End quote ---

VEBO

Wimberleytech:

--- Quote from: petert on February 05, 2020, 01:44:43 pm ---Ok, I got confused. Since BE acts like a diode, the voltage drop will be between the min and max ratings given in the datasheet (so between 0.58 and 0.7/0.77). So Vbe(on) is analogous to the forward voltage ratings of a diode.

Any remaining voltage will simply be dropped at any component(s) following the emitter of the transistor.

Yet, diodes fail when a certain voltage is exceeded, so what would be the maximum voltage for the BC547 that Vb could tolerate?

--- End quote ---
In the context of your question, yes, BE acts like a diode. 
For a diode:  I = I0(eVbe/Vt - 1)                  <<<corrected...I left out some important parentheses!!

Diodes are specified in terms of maximum current, not maximum Vbe.  You need to think in terms of current.

BJTs are current-controlled devices.  You should not think in terms of a voltage source across BE and dialing up that voltage to some particular Vbe to achieve some collector current or collector voltage.

Editing this post to add a chart from a commonly used diode(s)

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