Question about the switching behavior of a transistor as a switch

Question

I was doing some testing and my intent was to run a transistor as a switch. I was providing an input square wave from a source, and was looking at the fall time characteristics.

Basically this is the circuit I was using to operate it, with the square wave being provided to the base through the resistor.

However while observing the fall time characteristics i.e when the square wave is turning off. I was getting a signal plotted on the scope below.

Basically the green signal is the plot for Vce and the yellow one(falling) is Vbe. Now I would expect to see Vce rise after Vbe falls down, but I was observing it rising a little before Vbe and there is a small kink where the left cursor is, where it gets triggered. Can someone suggest reasons to why it would cause the Vce trigger to soon when its higher than the base voltage to keep it ON ? Thank you for looking.

Answer 1

A run through SPICE shows that the delay is real. Delay depends upon the transistor characteristics. The transistor does turn off well after the base voltage drops quickly to zero. In your circuit, the collector current would be of interest, since LED brightness is directly proportional to current flowing through. A plot of collector current does show that collector current falls toward zero about 0.6 uS after Vin drops to zero, in the SPICE simulation:

Notice the small step in base voltage at the 1.0 uS mark, when the transistor is told to turn off (just like your measurement). At this same moment, collector current pulses up above the 12 mA mark, because of base-to-collector capacitance. This short pulse is difficult to see in the blue plot of Ic.
MOSfet switches don't have this storage delay, but their very large capacitances must be charged and discharged - requiring large pulse currents.

Answer 2

The basic problem is your base drive. Assuming that your yellow is actually Vbe (and not the input to the 1k resistor), two things become apparent. First, Vbe is not being measured accurately, since your trace seems to indicate about a 1 volt swing, instead of about 0.7 volts. Second, you'll notice that Vbe starts falling at the cursor, just at the same time Vce starts rising. If you go back to basic theory, you'll find that the current through the base is exponential with respect to the voltage, so that "gentle" drop in voltage is actually producing a rather large drop in base current. As a result, Vce is rising.

I suggest that, instead of looking at Vbe, you monitor the input to the 1k resistor. I'm pretty sure you'll find that it drops over a period of more than a microsecond, and probably pretty linearly. As it does, the base current will drop about the same amount, although this won't have a big effect until the drive voltage gets down around a volt or two. But it does have an effect which you can see.

Answer 3

A lot of things floated through my mind when reading your question. (I upped you because you took the trouble to actually look at things and come up with such a question -- too rare here, I think.) One of them was charge storage in the BJT bulk and transit times. Another was to wonder how you'd set things up for measurement. Another was that you should check out your observations by also seeing if a spice simulator also would show the effect.

But as all these flew by, another much more obvious one arrived that probably explains the hogs share of what you are seeing and stopped me from wasting time on the rest. It's that a mere \$60\:\textrm{mV}\$ change in the base-emitter voltage would lead to a 10 fold change in the collector current.

Think about that for a moment. Say the LED current is \$20\:\textrm{mA}\$ at 100%. To go from 100% to 90%, the base voltage would change by perhaps \$2.5\:\textrm{mV}\$. With \$1\:\textrm{V}\$ per division, and noise, there's no way you'd see that. To go from 90% to 9% (nearly OFF), the voltage would drop by another \$60\:\textrm{mV}\$. Merely 6% of one division on your scope. And the LED is nearly OFF and therefore the \$V_{CE}\$ has risen to nearly it's maximum value now.

Of course the display would look like that. And I don't need to invoke charge storage and transit times, worry about your wiring or your hookups, or anything else. It's the expected behavior without such complications. Those complications may be there, too. But they aren't needed to explain your observation. I'm sure spice will confirm this.

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You didn't set things up on the scope well enough to drill in deeper. The explanation above is enough to explain things at your current resolution. But, if you set things up to provide better precision (time and voltage) to narrow in further, it will turn out that my explanation above is no longer enough and you may be able to start pointing to further details which remain unexplained by this single principle alone.

Of course.

At that point, you are starting to dig into why BJT models are in fact more detailed and more complex and why there are so many added terms.

But at the level your display currently shows? That one basic principle is sufficient.