BJT transistor storage time

Thread Starter

cmartinez

Joined Jan 17, 2007
7,084
What does "storage time" in a bjt transistor mean?

For example, in the 2N3906 datasheet it says:

Capture.JPG
I understand delay, rise and fall times... but is storage time the "delay time" that it takes for the transistor's current to start falling after it has been switched off?
 
Last edited:

bertus

Joined Apr 5, 2008
20,768
Hello,

From the wiki:

Turn-on, turn-off, and storage delay
The Bipolar transistor exhibits a few delay characteristics when turning on and off. Most transistors, and especially power transistors, exhibit long base-storage times that limit maximum frequency of operation in switching applications. One method for reducing this storage time is by using a Baker clamp.
https://en.wikipedia.org/wiki/Bipolar_junction_transistor

Bertus
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,084

Jony130

Joined Feb 17, 2009
5,183
Storage time (ts) is the time required for the BJT to come out of saturation. This is the time required for the VC to reach 10% of its high-state value (Vcc)
I do some real world measurements of this circuit
New1.PNG

With anti-saturation diode (I do not have any Shockley diode).
New2.PNG

But speed-up capacitor will also help.

new3.PNG

959849.png
 

#12

Joined Nov 30, 2010
18,210
Darn you, @cmartinez
Your questions are getting sophisticated enough that I can't answer them without looking up the material, and people with first hand experience still have more to contribute than I could translate out of a book for you.

Congratulations on your progress, sorry you have passed this teacher, I don't like you any less, but I can't function at the level you need.

apologies and reassurances.
 

crutschow

Joined Mar 14, 2008
25,423
A baker clamp is the one using diodes, but what about the one using an RC at the transistor's gate? does it have a name too?
Speedup capacitor.

Note that some BJTs are designed to have a low saturation delay, expressly for their use as a switch. Specific impurities are diffused into the base region to increase the carrier recombination rate, thus reducing storage time.
The 2N2369, for example, has a saturation delay of only 15ns, a factor of ten or more improvement over most general purpose BJTs.

Besides the Schottky Baker clamp, which is used in TTL "S" devices to reduce saturation delay, the emitter-coupled-logic (ECL) is designed so that the transistor is always in the active region, thus saturation delay cannot occur.

FET transistors have no intrinsic turn-off delay which is one of the reasons they excel as switches.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,084
Exploring google/Wikipedia is left as an exercise for the student.
Thanks, Ian... I did explore the wikipedia/google ... but I'm not sure I understand it thoroughly...
The wikipedia says, in the baker clamp section (which, BTW I had already looked up before you mentioned it, see post #3):

Clamp circuits were also used to speed up cutoff transitions. When the transistor is cutoff, the output is similar to an RC circuit that exponentially decays to its final value. As the circuit gets closer to its final value, there is less current available to charge the capacitor, so the rate of approach lessens. To reach 90 percent of the final value takes about 2.3 time constants. Cutoff clamping reduces the output voltage swing but makes the transition faster. Clamping the collector voltage to 63 percent of the final value allows a factor of two speed increase.

I'm guessing that the transistor's base behaves like a capacitor then (of course it does... the base must be charged to a certain potential before it starts conducting... right?) but I'm still at a loss understanding why an RC arrangement at the base (see bottom figure in post #4) speeds the transistor's turn-off time... is it because the capacitor is helping discharge the base when the source signal is pulled down?

@#12, thanks for the compliment... but it's not that my questions are getting fancier... it's more like I'm a pesky questioner :D

EDIT: changed the word gate to to base in the previous paragraphs, to make it consistent with transistor terminology
 
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ian field

Joined Oct 27, 2012
6,539
Speedup capacitor.

Note that some BJTs are designed to have a low saturation delay, expressly for their use as a switch. Specific impurities are diffused into the base region to increase the carrier recombination rate, thus reducing storage time.
The 2N2369, for example, has a saturation delay of only 15ns, a factor of ten or more improvement over most general purpose BJTs.

.
The 2369 is sometimes chosen for avalanche transistor ultra-short pulse generators.

Its not "up there" with devices specifically made for the job, but if you select devices you can get pretty impressive results.

Not really up to making a multi-GHz sampling scope, but its top notch for most amateur applications.
 

crutschow

Joined Mar 14, 2008
25,423
.......................
I'm guessing that the transistor's gate behaves like a capacitor then (of course it does... the gate must be charged to a certain potential before it starts conducting... right?) but I'm still at a loss understanding why an RC arrangement at the gate (see bottom figure in post #4) speeds the transistor's turn-off time... is it because the capacitor is helping discharge the gate when the source signal is pulled down?
....................
It's more than just the gate-source capacitance, it's the carriers generated in the base region that must be removed or recombined before the transistor turns off.
When the input signal goes to 0V the capacitor generates a momentary negative voltage at the base-emitter junction which pulls out the excess charge and speeds the turn off.

Say you have a 0V-5V-0V control signal. When it's at 5V the capacitor charges to 5V on the signal side and about 0.7V on the transistor base side.
When the signal goes from 5V to zero, the output of the capacitor drops from 0.7V to about -4.3V. That is what pulls the excess charge out of the base.
 

AnalogKid

Joined Aug 1, 2013
8,596
The speedup capacitor is charged with one polarity when the voltage driving the base is greater than the voltage at the base due to the IxR voltage drop across the base resistor. When the input signal goes to zero, or anything below the base voltage, the capacitor is reverse-biased, and literally sucks charge out of the base region and the two capacitors equalize the available charge between them. Another way to look at it is that at high frequencies (like those in a fast switching signal), the cap's impedance is much lower than that of the base resistor. So it dissipates zero power at steady state, but looks like a short circuit during the switching edge for a higher base current. It works to improve both the on and off times for BJT's, FET's signal diodes, whatever.

ak
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,084
... looks like a short circuit during the switching edge for a higher base current. It works to improve both the on and off times for BJT's, FET's signal diodes, whatever.
ak
I think I get it now... kind of clicked when you mentioned it looks like a (very brief) short circuit...
But what's that last comment about FETs? Is placing an RC arrangement at the gate advisable too? I thought that a FET's gate capacitance was one of those issues one had to deal with when driving it... so adding a cap at the gate makes it kind of confusing for me...
 

ian field

Joined Oct 27, 2012
6,539
I think I get it now... kind of clicked when you mentioned it looks like a (very brief) short circuit...
But what's that last comment about FETs? Is placing an RC arrangement at the gate advisable too? I thought that a FET's gate capacitance was one of those issues one had to deal with when driving it... so adding a cap at the gate makes it kind of confusing for me...
When I serviced PC monitors, I often found the driver chip (UC3842 was the most common) drives the MOSFET gate via a forward diode, that allows the inclusion of an emitter follower to give more current drive while discharging the gate capacitance.

You could try including UC3842 in the search string for a few well known brands of CRT monitor schematic/service manual and hope you get lucky with one that shows that arrangement.
 

AnalogKid

Joined Aug 1, 2013
8,596
MOSFETs, and particularly power MOSFETs, are famous for having large gate capacitances relative to BJT's. But the idea is the same and so are the consequences and solutions. There usually isn't a series resistor to a gate (as opposed to a base), and you can't have only the cap in series because the gate needs to see a DC level to function, so the solution is gate driver chips and circuits. You'll see gate driver chips with 9 A capability in a tiny SO package. They can't make 9 A for long, just long enough to charge or discharge the gate.

ak
 
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