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

[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]

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

V_EBO

[Wimberleytech]

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?

In the context of your question, yes, BE acts like a diode. 
For a diode:  I = I₀(e^Vbe/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)

[petert]

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

V_EBO

Isn't that the negative of V_BE? In that case I still wouldn't know the maximum positive voltage allowed to flow from B to E.

[Wimberleytech]

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

V_EBO

Isn't that the negative of V_BE?...

Yes

[SilverSolder]

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

V_EBO

Isn't that the negative of V_BE? In that case I still wouldn't know the maximum positive voltage allowed to flow from B to E.

As @Wimberleytech said,  everything makes more sense if you think in terms of the current from B to E,  and think of the Vbe voltage as a kind of side effect of that.

As a first approximation, you can think of the CE current as the BE current times the gain of the transistor.

[GeorgeOfTheJungle]

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

V_EBO

Isn't that the negative of V_BE?...

Yes

Yes²  (Sorry!)

[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 Vb could tolerate?

In the context of your question, yes, BE acts like a diode. 
For a diode:  I = I₀e^Vbe/Vt - 1

For reference, to know what I₀ means: https://en.wikipedia.org/wiki/Shockley_diode_equation (where it is called I_s).
And this: https://en.wikipedia.org/wiki/Reverse_leakage_current
With the key information being:

For constant temperatures, the reverse current is almost constant although the applied reverse voltage is increased up to a certain limit. Hence, it is also called reverse saturation current.

Or in summary, the current through a diode is dependent on the voltage across a diode and the temperature, with an exponential relationship between voltage and current (matching the common transfer function graphs given for diodes). So far so good.

As @Wimberleytech said,  everything makes more sense if you think in terms of the current from B to E,  and think of the Vbe voltage as a kind of side effect of that.

As a first approximation, you can think of the CE current as the BE current times the gain of the transistor.

So given the maximum rating of the continuous collector current of I_C = 500mA, and assuming a gain range from 110 to 800, the maximum for I_B would be between 500mA/110 and 500mA/800, which is from 4.55 mA down to 0.625 mA.

What I conclude from this though is not that I_B couldn't be higher (no limit is specified in the datasheet and the base may be differently "fragile" from the collector, so a direct derivation based on gain might not apply), but that I_C would go beyond its maximum ratings if I_B exceeded the values computed above, and therefore it does not matter, much what the actual limit for I_B is.

Is that correct?

[Wimberleytech]

So given the maximum rating of the continuous collector current of I_C = 500mA, and assuming a gain range from 110 to 800, the maximum for I_B would be between 500mA/110 and 500mA/800, which is from 4.55 mA down to 0.625 mA.

What I conclude from this though is not that I_B couldn't be higher (no limit is specified in the datasheet and the base may be differently "fragile" from the collector, so a direct derivation based on gain might not apply), but that I_C would go beyond its maximum ratings if I_B exceeded the values computed above, and therefore it does not matter, much what the actual limit for I_B is.

Is that correct?

There are problems here.
1) hfe is a small-signal parameter and you are using it as a large-signal one.
2) hfe range given is for Ic = 2mA.  Thus your calculations do not apply.
3) dc current gain varies as a function of collector current.

You are getting a lot of good answers here in response to your questions.  But, I suspect we are not getting to the root of your problem.  It might be better if you present a circuit or problem you are trying to solve in the context of a BJT. 

[SilverSolder]

[...]

As @Wimberleytech said,  everything makes more sense if you think in terms of the current from B to E,  and think of the Vbe voltage as a kind of side effect of that.

As a first approximation, you can think of the CE current as the BE current times the gain of the transistor.

So given the maximum rating of the continuous collector current of I_C = 500mA, and assuming a gain range from 110 to 800, the maximum for I_B would be between 500mA/110 and 500mA/800, which is from 4.55 mA down to 0.625 mA.

That is good thinking, yes.

What I conclude from this though is not that I_B couldn't be higher (no limit is specified in the datasheet and the base may be differently "fragile" from the collector, so a direct derivation based on gain might not apply), but that I_C would go beyond its maximum ratings if I_B exceeded the values computed above, and therefore it does not matter, much what the actual limit for I_B is.

Is that correct?

Most datasheets do not specify the max base current, probably because it is assumed that you would not normally try to push more current than needed to achieve the target CE current - typically 10% of the collector current (depending on the min gain, e.g. your calculations above).   

There is one situation where it can make sense to "overdrive" the BE junction and push more current than that:  when the transistor is saturated.

For example, look at the attached spec for the common 2N3906 transistor, page 6, figure 14.   Here you can see how there is nothing much gained from increasing the base current beyond a certain point, once the transistor has saturated, but it might make sense to "squeeze it to the max" by driving it hard. (Another way of thinking of this, is that the gain drops quickly when saturation is reached, so you have to supply much more base current to get the last few drops out of the transistor.)

[petert]

I found some relevant posts:
https://electronics.stackexchange.com/questions/158522/bjt-absolute-maximum-base-current
https://www.rohm.com/products/faq-search/faqId/231
https://electronics.stackexchange.com/questions/353511/missing-bjt-parameter-i-b-max

[petert]

References regarding the original question:
https://electronics.stackexchange.com/questions/12477/how-do-bjt-transistors-work-in-a-saturated-state
https://electronics.stackexchange.com/questions/254391/terminology-for-bjt-regions-of-operation
https://electronics.stackexchange.com/questions/5336/i-dont-care-how-a-transistor-works-how-do-i-get-one-to-work
http://www.zen22142.zen.co.uk/Design/bjtsw.htm