It's a simple question. What's the turn-off time of this "diode" and does it have a recovery time when turning off?
Since the transistor does not get into saturation I think it should be the same as turning it off normally, so pretty fast. The transistor is D44H8.
Most transistors, and especially power transistors, exhibit long base-storage times that limit maximum frequency of operation in switching applications, unless clamped, such as in this circuit with Vce clamped by Vbe ( or with a Schottky diode B-C clamp or a Baker's dual diode clamp etc). In storage time tests Vbe is switched from Vbe .0.6 at some Vce(sat)/Ibe(sat)-10 to Vbe = -500 mV ( industry standard) whereas in this question Vbc is = 0 by design**
does it have recovery time when turning off?
Yes. But it will be less than the Recovery time is in the datasheet since the B-C diode capacitance reduces somewhat from 0.7V to 0V and pF reduces greatly for Vbc = - Vcc approx. with inverse voltage yet -500 mV (standard test) for Vbe to Vce storage test times.
Not that the D44 and D45 differ in \$C_{cb},f_T,t_D+t_R,t_F \text{ yet } t_S~\$ is not affected with both = 500 ns.
This is tested with Vbe= -0.5V after Ic/Ib=10 using Ib= 0.5A from the bottom of page 2 in the ON spec.
From nominal values in fig's 5, [6] and 7, [8] for D44, [D45] this means "Storage Time = 500ns" is measured at Ic=5A, Ib=500mA 25'C ;
Vce[sat]= 150, [275] mV
Vbe(sat) = 940, [960] mV
Yet when Vce=Vbe HFE is not 10 nor is current gain as high as hFE @ Vce=1000 mV (std test condition Vce=1V) but something in between so Ib is much lower and thus will have faster response time than Is @ 5A because Ib will be more like 2 to 3% of Ic rather than 10%. (Ic/Ib=10) yet the storage energy in the cloud capacitance layer E= 1/2CV^2 so the bulk capacitance in B-E will have the same voltage as Vce @ 5A but more Miller effect from Ccb @ hFE >>10.
If anyone has more to add or subtract. pls edit.
Edit (vangelo): I think it does deserve a direct comparison (signal x power BJT) to illustrate and I think it should not be posted as a separate answer:
This should actually be only a comment
CircuitLab transient analyzer shows that you are right. There's no such thing as storage time - not at least when the transistor is Cicuit Lab's version of 2N3904. Here's a circuit with 1N4148.
Signal V2 is bipolar square wave 5V peak:
The diode conducts to reverse direction about 10 nanoseconds. Transistor turns off substantially faster with the same signal and load:
The reverse conduction starts to decay as soon as the polarity is reversed, there's no full reverse conduction period (= no storage time). Schottky-diodes have the same property. Except they have about 50% lower voltage drop.
Transistor's BE junction unfortunately stands only a few volts of reverse voltage, so this may be not so good in high voltage circuits.
Not asked: The output voltage to the load seems to be about -5,7V when V2 jumps to -5V. That can look surprising. It's not. The carrier cloud in the semiconductor is like a charged capacitor which has minus pole towards the load. When V2 jumps to negative there's -0,7V in series with V2.
The peak which occurs when V2 jumps to +5V is the same to another direction. The semiconductor capacitances have got bigger voltages during the minus polarity period of V2. Transistor seems to have much bigger capacitance than diode 1N4148,
does it have a recovery time when turning off?
Yes. The transistor will not turn off until the concentration of minority carriers in the base drops significantly.
Since the transistor does not get into saturation I think it should [turn off] pretty fast.
Yes. Because the transistor is not in saturation, there will be less minority carriers in the base, and it will turn off quicker than a transistor in saturation. However, there is still a recovery time. It is just smaller than that of a saturated transistor.
