Phy 3311 Electronics Lab #10 Transistors
In this lab you’ll be introduced to the transistor.
Part I: Bipolar Junction Transistor testing and identification
As we discussed in class, bipolar junction transistors (BJTs) come in two varieties, NPN and PNP, which
have equivalent behavior if one reverses the polarities of all voltages and currents. As their names
imply, they consist of three adjoined pieces of doped semiconducting material, creating two pn
junctions. The “end” pieces (which have the same type of doping) are referred to as the emitter and the
collector. The “center” piece, with the opposite type of doping, is referred to as the base.
With resistors, inductors and most capacitors, the two terminals were interchangeable. With diodes
and electrolytic capacitors, the two terminals were not interchangeable (+ and -, or cathode and anode),
but clearly identified as such. Unfortunately, there is no industry-wide standard for labeling or lead
placement on transistors, so we often have to figure it out for ourselves. When you buy transistors,
they frequently come in a package which identifies the lead placement, but if you don’t have the
package, you’re on your own. However, even if we know the type and lead placement, we may still
want to check to see if a transistor is functioning properly.
Select an NPN and a PNP transistor from the drawers. Consider the
figure at right, which gives three ways of viewing NPN (upper) and
PNP (lower) transistors: (left) as a physical component consisting of
three doped pieces of semiconductor, (center) as a schematic
symbol, and (right) an analogous circuit (a reasonable approximation
when only two of the three leads are connected) consisting of backto-back diodes. Mnemonic: the arrows for NPN transistors Never
Point iN, and the arrows for PNP transistors Point iN Proudly.
Carefully draw a picture of the transistor, and label the leads 1, 2 and 3. With your drawing you should
be able to unambiguously identify which lead is which. Fortunately, the drawers in our electronics
cabinet have little pictures that identify the leads, but we should still verify it for ourselves.
While some fancy multimeters have transistor-testing capabilities, in which you can plug all three leads,
and the pinout and type are displayed on screen, ours don’t. So set your meter to the Diode setting, and
measure the voltage between each combination of leads, in both polarities. You should have six
measurements in all, which will either be of the form “reverse bias” (reading OL on the meter), or
“forward bias” (typically a “diode forward voltage” around 0.6 V; record this voltage). Make a table of
your six readings, and follow the steps below to identify the type and lead configuration of your
transistor. (If you should find yourself in a situation without a meter with a Diode setting, you can also
use your ohmmeter as we did two weeks ago, although interpreting the results is a little more involved.)
When measuring between the collector and emitter, you should read “reverse bias” in both directions,
since one of the “diodes” will always be in reverse bias. Determine which two leads measure in reverse
bias either way; those are the collector and emitter (you’ll figure out which is which in a bit). The
remaining lead, by process of elimination, is the base, which you can now identify on your diagram.
When measuring between the base and either of the other two leads, you should get “reverse bias” in
one direction, and “forward bias” in the other. Referring to the above “diode equivalent” pictures, if the
“forward bias” condition happens when the red lead is on the base, you have an NPN transistor. If the
“forward bias” condition happens when the black lead is on the base, you have a PNP transistor.
Now, look at the values of the voltages recorded between the base and the other leads in forward bias.
Generally, the emitter is more heavily doped than the collector, so the base-emitter junction will often
have a slightly higher forward voltage drop than the base-collector junction. (If you use the ohmmeter
setting, the base-emitter junction will have a slightly higher resistance than the base-collector junction.)
So you should be able to figure out which lead is the collector and which is the emitter.
If you cannot tell unambiguously which lead is which, if you can determine the manufacturer or model
number of the transistor, you may be able to dig up literature on it (the “specification sheet”) which
identifies the type and lead placement. The spec sheet is actually quite important anyway, because it
also lists other parameters, such as voltage, current and power limits, which you’ll need to know if
you’re going to do any serious work. (In our case, as long as the identification of the base lead is
consistent with what the picture on the drawer says, we’ll use its identification of the collector and the
emitter, if we can’t tell the difference between the base-emitter and base-collector voltages.)
Finally, if your six readings make no sense at all – if any leads read as a short, or read “forward bias” in
both directions, or all three leads record “reverse bias” in both directions – your transistor may be
faulty. Or, it may not be a BJT at all, but some other component (such as a FET) with three leads.
Now that we’ve determined that we have two good transistors, have identified the type (NPN or PNP)
and determined the lead placement, we’re ready to do some work with them!
II. The Transistor Switch
Assemble the circuit shown at right, using RB = 2.2 kΩ, your NPN transistor,
and one of the switches you soldered leads to in a previous lab. Use RC ~ 150
Ω to limit the diode current; you should now know how to wire the LED so it
is in forward bias. For the supply VCC, use the ‘A’ output of your DC power
supply, and set VCC = 5 V.
In this circuit the switch controls a larger output current being delivered to
the load (diode plus RC). We consider two cases: “on,” corresponding to the
switch being closed, and “off,” corresponding to the switch being opened.
When the switch is open, the base current IB is obviously 0, but when the switch is closed, the resistor RB
controls IB. Note that the emitter is grounded, so the base is at a voltage VBE, where VBE is a typical
“forward biased diode” voltage ~ 0.5 - 0.7 V. So,
IB = (VCC – VBE) / RB
The voltage source VCC is what actually delivers power to the load (diode and resistor RC). In particular,
IC = (VCC – VCE – VD) / RC
Note that VD will have some nonzero value, typically around the VK you found in Part II. When IB
changes, VCE also changes (through “transistor action”), which causes the collector current IC to vary. In
the cutoff regime, IB = IC = 0, and VCE will rise as high as is required (VCC – VD) in order to stop current flow
in the collector circuit. In the linear regime, IC = βIB, and that equality is maintained by adjustment of
VCE. In saturation, VCE = 0 and IC is controlled only by parameters external to the transistor.
What happens as you turn the switch on and off? Does the LED turn on and off? (If not, you’ve
probably wired something incorrectly.)
Using your multimeter, measure the voltages VRB, VD (the voltage across the LED), VRC (the voltage across
the LED’s current-limiting resistor), VBE and VCE. From VRB and VBE, you can compute IB, and from VRC, VCE
and VD, you can compute IC. Do this both for the cases that the switch is open, and the switch is closed.
What is the value of VBE? How does it compare with what you expect, given that the base-emitter
junction is a forward-biased diode with IB = 0 when the switch is open, and IB = some finite value when
the switch is closed?
What are the values of IC and VCE? Based on what we learned in class, what mode (cutoff, saturation,
linear) is the transistor operating in when the switch is open? When the switch is closed?
Based on your values of IC and VCE, would you say that this transistor works well as a switch?
Finally, what is the effective current gain of this circuit? That is, what is the ratio of IC to IB? (Note that
since this transistor is not necessarily in the linear mode, that IC/IB does not equal β, but instead is some
other value which is a function of the things to which the transistor is attached.)
III. Basic transistor characteristic curves
Assemble the circuit shown at right, using the ‘A’ and ‘B’
outputs of your BK Precision power supply as the base bias
and collector bias voltages VBB and VCC. Use R1 = 100 kΩ (yes!), RL = 1 kΩ, and your NPN transistor.
Using Kirchhoff’s laws, we can write equations describing the relationships between the voltages and
currents in the circuit:
(left loop)
(right loop)
(node at transistor)
VBB – IBR1 – VBE = 0
VCC – ICRL – VCE = 0
IE = IB + IC
(Note that while we may be tempted to say that IC = βIB, that is only true – and only approximately so! in the linear operating regime of the transistor. We do not know if that is true for this circuit!)
A. Base characteristic curve
First, with the collector side of the circuit disconnected (that is, the collector is physically disconnected
from everything else), make measurements of VBB and VR1 as VBB is slowly turned up from 0 to 5 V (again,
start with coarse steps, and fill in afterwards). From VBB and VR1, you can calculate VBE and IB:
VBE = VBB – VR1
IB = VR1/R1
Make a plot of IB versus VBE. Since the isolated base-emitter junction is simply a silicon pn diode, you
should see an exponential dependence of IB versus VBE, with a knee voltage VK ≈ 0.6-0.7 V. How does
this VK compare with the base-emitter forward bias voltage you measured in Part I? You can try fitting a
Shockley-type exponential curve to this data, but be aware that the “base-emitter” characteristic curve
does not necessarily follow the Shockley model when IC is nonzero!
B. Base/collector gain curve
Next, connect the collector (“output”) circuit with VCC = 5 V and measure VBB, VR1 and VRL, as VBB is
increased from 0 to 5 V. From VR1 and VRL you should be able to calculate the four most important
parameters that describe the transistor’s operation – VBE, IB, IC and VCE:
VBE = VBB – VR1
IB = VR1/R1
IC = VRL/RL
VCE = VCC - VRL
Make a plot of IC (on the y-axis) versus IB (on the x-axis).
On your plot, identify the regions
corresponding to “saturation,” “cutoff” and the linear operating regime.
We expect to see a linear relationship between IC and IB in the linear regime, with
IC = βIB
Perform a linear fit in the linear regime to determine the value of β for your transistor. Typically, β is on
the order of 100, although values as low as 25 and as high as a few hundred aren’t uncommon.
We expect that for increasingly large values of IB, the collector current should increase to the point of
saturation and can increase no more; this occurs when VCE → 0 (and VL approaches VCC), so that
ICsat = VCC/RL
How does this expected saturation current compare to what you observe? What happens to VCE as IC
increases? Does it approach zero as IC approaches ICsat?
C. Collector characteristic curves (as time permits)
Finally, we will determine the characteristic curves of the collector.
Set VBB = 1 V and measure VR1; this corresponds to a (relatively) fixed value of IB = VR1 / R1.
Now adjust the collector supply voltage VCC gradually upwards from 0 V in approximately 1 V steps up to
10 V, and measure VRL at each step (you can go back and fill in gaps, or take higher voltages, later). From
VRL you will be able to calculate the collector current IC and collector-emitter voltage VCE, as
IC = IL = VRL/RL
VCE = VCC - VRL
For small values of VCC you should see VRL (and IL = IC = VRL/RL) increase linearly with VCC. This is because
VCE is very nearly zero, meaning that the current IC is limited by the value of VCC; as VCC grows so does IC.
Although this may be counterintuitive, this corresponds to saturation, since in this region IC is as large as
it can get due to the VCC/RL limit, and is also only rather weakly dependent on IB.
As VCC continues to increase, VRL should nearly level off (corresponding to a relatively constant value of
IC), although you will see that IC does weakly increase with increasing VCC (and hence VCE). This occurs as
the transistor regulates the current flow by varying the value of VCE in order to maintain IC ≈ βIB, and
corresponds to the linear operating regime.
Eventually, VRL (and hence IC) will start rising very rapidly with increasing VCC (while VCE stays relatively
constant), as the transistor enters breakdown mode. Take one or two points in this breakdown regime,
and then stop; this corresponds to the reverse voltage limit on the reverse-biased collector-base diode
being exceeded, and you don’t want to go too far beyond this limit! In the breakdown regime, the
reverse-biased collector-base diode, which normally acts as the current-limiting element that regulates
IC, fails to do so, and the transistor ceases to operate as it should.
Once you’ve collected data covering the saturation, operating and breakdown regimes, graph your
values of VCE (on the x-axis) versus IC (on the y-axis). You should see something that looks rather similar
to what we studied in class yesterday – first, a rapid increase of IC with VCE (saturation), then a relatively
constant IC (linear regime), then IC increasing rapidly with VCE (breakdown).
Once you have done this, repeat your measurements of VCC and RL for a few different values of VBB (and
hence IB). Once you have an idea of what the “landscape” looks like, this should go more quickly, as you
will only need to take a half-dozen or so data points for each VBB value – two or three in the saturation
regime, another two or three in the linear regime, and one or two more in the breakdown regime.
Plot these curves (corresponding to different values of IB) on the same set of axes, to produce a set of
collector characteristic curves for your transistor.
D. Epilogue
At this point you should have three characteristic curves for your transistor – the base characteristic IB
vs. VBE, the gain characteristic IC vs. IB, and the collector characteristics IC vs. VCE as a function of IB. With
these three sets of curves it is possible to design good switching and amplifier circuits by selecting the
appropriate bias voltages for the input and output side.
Finally, note that the circuit we used is actually terrible to use as an amplifier, since its gain is highly βdependent! (It’s not too bad as a switch, as long as we take care to ensure VB cannot “float.”)
However, we used it because it is relatively simple to analyze.