QuickNote | Introduction | Mathematical Analysis | Applet Tutorial | Applet Worksheet | Quiz | SPICE/CAD | References | Feedback

Applet Worksheet

[Note to Instructors:  The Worksheet below is what we used in a Computer Lab Recitation for an Electrical Engineering junior-level course.  The students used this spreadsheet to report their work which the TA's graded near the end of the recitation session.  However, this could be used for homework assignment where the Students print and fill out the spreadsheet.]

Bipolar Junction Transistor Switching Circuit
 

Applet: BJT Switching Applet at  http://jas2.eng.buffalo.edu/applets/education/bjt/switching/index.html


II. Prob.4.97 of Sedra/Smith:  For the circuit shown in the applet, select a value for Rb so that the transistor saturates with an overdrive factor of 10.  The BJT is specified to have a minimum b of 30 and VCEsat = 0.2 V.  What is the value of forced b achieved ?
 

Applet Procedure:
1) Prepare the applet:
a) Click on the “Pause” button to stop the animation.  You can switch between Vi=Low and Vi=High by mouse cliking on the Vi area of the circuit.
b) Click the “Ckt Param” button.  Fill in High = 5.0V, Vcc = 5 V, Rc = 1E3.  Then click “OK”. Close the window.
c) Click the “BJT Param” button.  Fill in Beta=30, Vce,sat = 0.2V.  Then click “OK”. Close the window.
d) In the applet, four waveforms are displayed:  Two input waveforms, Vi(t) and iB(t); the total minority charge in the Base, Qb(t); and the Collector current, iC(t).  Note that the red, horizontal line in the Qb(t) waveform (the third waveform graph) is the Edge-Of-Saturation (eos) value.  This eos refers to the boundary between the forward active mode and the saturation mode.
e) After driving the BJT into cutoff (Vi=Low, 0.2V ==>  Hit “Resume” and then “Pause” after the Base minority charge has been completely flushed out and it displays “Cutoff”.), switch to Vi=High, 5V and run until the BJT enters saturation.
2)  Hand calculate and verify with applet numbers:


Hand Calculation
i)  Ic,sat = (Vcc – Vc,sat)/Rc = (     )
ii) Ib,eos = Ic,sat/b = (     )
iii) With Vi = High (5.0V), find Rb so that
    Ib = 10*Ib,eos (overdrive factor=10).
    Rb = (Vi – 0.7)/Ib = (    )

Read from Applet
 i) Ic,sat = (     )
ii) N/A
 iii) Set the  Rb that you calculated above in the applet.
        Run applet and read
        Ib = (         )
 
 

III. BJT TURN-ON AND TURN-OFF SWITCHING DELAYS

In this exercise, we shall learn about the causes of switching delays in BJT.  Faster circuits require small delay times.

1. Applet Instruction

1) Use the “Ckt Param” and “BJT Param” popup dialog boxes to set the following values and click “OK” button.
Rb = 1E4, Rc=1E3, High=5V, Low=0.2V, Vcc = 5.0V
Beta = 50, Vce,sat = 0.2V
Leave all other params at the default value.
2) Use mouse click on the Vi-area of the circuit diagram to switch Vi between High and Low.  Use “Pause/Resume” to run the applet (also,  the “StepForward” is sometimes useful to move the waveforms slowly).

3) To estimate the delay times, you may use the elapsed time value displayed in the circuit diagram or the vertical grid line spacing (the time axis values) in the waveform graph axes.

2. Turn-ON Delay Times
1) Prepare applet: Set Vi=Low  ==> run (click “Resume” button) until all the excess Minority charge in the Base is flushed out. ==>  Click the “Pause” button and Switch the input voltage Vi to High. ==>  Click “Resume” and when the BJT saturates (let it run a brief moment after it enters saturation), click “Pause”.

2) From  the applet, estimate the delay time between when Vi turns ON and when iC rises just above zero.
   First delay time* = (     ) ns.
* During this time, the Emitter junction of BJT (or the Emitter junction capacitance) gets charged to the Forward-biased condition.

3) From the applet, estimate the delay time between when iC has risen just above zero and when iC just reaches the saturation value (the Base profile display just changes from “Forward active” to “Saturation”.).
Second delay time** = (         ) ns
** During this time the Base is charged with the excess minority charge to the eos value.  This delay time is related to the Base transit time (tt = Wb2/ 2 Dn) by the minority carrier electrons.

4) Turn-OFF delay time is the sum of the above two delay times:
Turn-ON delay time = Emitter junction charging time + Base charging  time = (     ) ns.

5) Comment on Base Oversaturation+ (i.e., Base Overdrive):
Even after the iC saturates at the iC,sat, the Base minority charge continues to rise to Qb  >  Qb,eos.  In the Base minority carrier concentration profile, shown just above the BJT device cross-section, Qb > Qb,eos requires a forward-biased CBJ (Collector-Base Junction) as evidenced by the net positive excess electron Dn at CBJ.  Also, not that the minority profile maintains a constant slope or a constant dn/dx for Qb > Qb,eos.  Therefore, the Collector current is constant,  as can be seen from the formula iC = q Dn dn/dx, even while the total charge in the Base rises.
+ This Base oversaturation (or overdrive) stores unnecessary charge in the Base, which needs to be removed before the BJT is turned OFF, and adds to the Turn-OFF delay time.  But the Base current overdrive also allows to charge the junction capacitance and the Base region more quickly during the Turn-ON process.

6) Question: Which of the following controls the saturated Collector current, iC,sat ?
   a) The internal BJT device characteristics.
   b) The resistance, Rc.
Hint: Consider iC,sat = (Vcc – Vce,sat) / Rc;      iC = q Dn dn/dx.


3. Turn-OFF Delay Times

1) Prepare applet:  Set Vi=High (5V) ==> run the applet until the BJT is well into saturation  (as eeen by Qb > Qb,eos and the “Minority carrier profile in Base” displays ‘Saturation’] and click the “Pause” button to stop animation.

2) For the current applet state, hand calculate the NPN BJT terminal voltages, Vb and Vc using the iB and iC values read from the waveform graphs.
    Ve = GND = 0 V.
    Vb = High (5V) – iB * Rb = (        ) Volts
    Vc = Vcc – iC * Rc = (       ) Volts.
Are  the above voltages consistent with BJT in saturation++ mode ??      YES       NO
++ Hint: Saturation mode for BJT is both EBJ and CBJ are forward-biased.

3) Do the simulation:  Switch Vi to Low (0.2V) ==> Use the “Resume/Pause” button to measure the delay time and answer the following questions.

i) With the BJT in saturation (Qb > Qb,eos), verify the net Excess minority (electron) concentration at the junctions:
  Dn at EBJ = net positive or zero ?        (Circle one)
  Dn at CBJ = net positive or zero ?       (Circle one).

ii) Starting from the oversaturated state (Qb > Qb, eos), how much time elapses until the BJT reaches the Edge-Of-Saturation (i.e., Qb reaches Qb,eos, or in the Base profile, the display “Saturation” just changes to “Forward active”.)
       First delay time* = (         ) ns.
* This is the time required to remove the oversaturated Base minority charge.

i) At EOS, what is the condition of the excess minority concentration at the junctions ?
    Dn at EBJ = net positive or zero ?        (Circle one)
     Dn at CBJ = net positive or zero ?       (Circle one).

iv) Click “Resume” and then “Pause” so that the BJT changes from “Forward active” to “Cutoff”.  How much time is elapsed from the EOS to the Cutoff**  ?
  Second time delay = (       ) ns.
** Cutoff: iC = 0; Qb = 0.

v) At the Cutoff, what it the excess minority concentration at the Junctions ?
    Dn at EBJ = net positive or zero ?        (Circle one)
    Dn at CBJ = net positive or zero ?       (Circle one).

vi) Find the total Turn-OFF delay time (except for the Junction discharging time):
  Total Turn-OFF Delay Time = (         ) ns.
Which delay time seems to be dominant here ?  The first one or the second one ?

vii) Comment on the Turn-OFF process:  First, remove the oversaturated Base charge (usually the dominant delay); Second, remove the Base minority charge; Third, discharge the Junction capacitance (not shown in this applet.).
 


4. Improved Turn-OFF Switching Time by a Schottky-Diode Clamp at CBJ

Introduction:  In the modern BJT IC technology, a Schottky diode is connected in parallel to the CBJ PN junction in order to keep  the CBJ from becoming forward-biased, and thus to keep the BJT from becoming oversaturated. A Schottky diode is made of a Metal-Silicon junction, and functions similarly to the PN junction diode except for the following important difference:  Under the forward-bias, the forward voltage drop in the Schottky diode is typically 0.5Volts, whereas the typical forward voltage drop in a PN diode is 0.7 Volts at the working current level.  Therefore, if a Schottky diode is connected in parallel to the CBJ PN diode, then the CBJ PN junction can not be biased at more than 0.5V because the Schottky diode forward voltage drop will limit the maximum voltage that can appear across the CBJ PN junction.  Hence the CBJ is always kept below  0.7V and thus can not become forward-biased.  This way, the Schottky diode clamps  the CBJ at 0.5V or so, less than its needed voltage drop for a forward bias.
i) Applet observation: Check the “Schottky-clamp” ON and OFF a few times and observe the changes in the circuit diagram (i.e., the 'RTL circuit' in the applet) and in the BJT Planar Structure diagram (Note: the 'BJT Planar Structure' view can be changed to either a magnified view of the EBJ or to a panel which lists various parameter values by the Choicebox that you find in the top portion of the applet.).  Note the diode symbol (of the Schottky diode).  Note in the Planar structure that the Ohmic metal (Aluminium) contact to the p-Base widens to make contact to the n-Collector region when the Schottky-clamp is checked.  An Aluminium contact to n-Silicon makes the needed Schottky diode.
ii) Applet simulation: Check the “Schottky-clamp” ON ==> Drive the BJT into Cutoff mode (use Vi=Low) ==> Click “Pause” ==> Switch to Vi = High ==> Click “Resume”  ==> After a while, click “Pause”.
a) What is the excess minority charge at CBJ ?     (            )
b) Does the CBJ ever forward-biased with the Schottky-clamp placed ?     YES       NO
c) Measure the Delay Times:
 (c.1) Turn-ON Time = From Cutoff to EOS = (      ) ns.
 (c.2) Turn-OFF Time = From EOS saturation to Cutoff = (     )ns.
 (c.3) Which switching time is affected the most by the Schottky-clamp ?? Turn-ON  or Turn-OFF    (circle one).