amount of resistance across the
collector/emitter terminals (RCE)
when switched on, and therefore a
voltage drop. [The resistance is a
result of the PN junction doping
process during manufacturing.]
In addition to the transistor
voltage drop, the LED will also drop
somewhere between 1.2 and 3 volts
when it’s switched on (check the
datasheet under VF). Therefore, in
order to calculate the correct value
for resistor Rc, the voltage drop
across the collector/emitter (VCE(sat))
and the voltage drop across the LED
(V(LED)) must be included in the
formula. So, here’s the same Ohm’s
Law formula modified to account for
RC = Vcc – V(LED) – VCE(sat)
IC(MAX)
Now, let’s calculate the correct
resistance value for resistor Rc. Let
Vcc = 5V, V(LED) = 1.9V, VCE(sat) = .1V,
and IC(MAX) = 15 mA. The answer is:
Rc = Vcc – V(LED)– VCE(sat)
IC(MAX)
Rc = 5 – 1.9 – .1
.015
Rc = 200 ohms
The calculation shows that we
need a 200 ohm resistor for Rc in
order to limit the current through the
LED to a safe 15 mA. Notice, had we
used the basic Ohm’s Law formula
(Rc = Vcc / IC(MAX)), RC would
be 333.33 ohms. The real
problem with using a 333.33
ohm resistor for RC begins when you
actually breadboard the circuit, only
to find out the current you expected
through the LED is not the required
15 mA, but 9. 2 mA (a 39% loss).
Therefore, if you fail to add both the
LED and transistor voltage drops in
the calculation, your LED won’t be as
bright as expected.
Try and look at the LED and
transistor as small resistors. In a series
circuit, you would add all the resistor
values together to get the total
resistance, right? Well, all we’re doing
here is accounting for all the voltage
drops in a series circuit.
Figure 3 clearly shows what
happens to the collector current
(IC(MAX)) when you don’t include all
the voltage drops in the formula.
BASE TO CONTROL
The question now is how do you
control the transistor so it turns on
and off? Well, we have to do two
things: 1. Find the correct transistor
base current (IB) that will saturate the
transistor. 2. Calculate the resistance
value for the base resistor RB (see
Figure 1). The formula for finding the
base current is:
IB(EOS) = IC(MAX)
Beta (min)
Notice here, in order to find the
base current (IB), we divide the
maximum collector current (IC(MAX))
we want to go through the LED ( 15
mA) by the minimum Beta listed on
the datasheet (hFE). What is Beta?
Beta — also known as DC current
gain — is a ratio relating to how much
current gain you can expect through
a transistor’s collector terminal given
a certain amount of current going
into the base terminal. In other
words, the base current controls the
collector current. It’s kind of like a
small water valve controlling the flow
of water running through a large
pipe.
Having said all that — and this is
very important — Beta (gain) is only
used in amplifier design. When
you’re using a transistor as a switch
(digital mode), Beta has little effect or
meaning because the transistor is not
operating in the active region that
amplifiers work in. Once a transistor
switch is in saturation mode, there’s
no collector current gain beyond
saturation.
In other words, once a transistor
switch reaches the saturation point,
the gain formula IC = Beta x IB no
longer applies because the voltage
drop across the collector/emitter
terminals (VCE(sat)) has reached its
lowest saturation voltage of .1V.
When VCE(sat) reaches this voltage
May 2015 47
FIGURE 3.
IMPORTANT POINTS
1. The circuit designer (you) determines what the correct transistor collector
current (IC(sat)) should be by looking at the LED/transistor datasheets and verifying that
the current going through the transistor/LED circuit is below the maximum ratings for
both devices.
In other words, the saturation current (Ic(sat)) flowing through a transistor switch
is not determined by the transistor's internal electrical parameters, but rather by the
external components (resistor/LED) employed by the circuit designer.
2. Beta (DC gain) as listed in the datasheet has no meaning when a transistor is
used as a switch (saturation/cut-off). Only amplifier designers care about the various
levels of collector current (gain) in between saturation and cut-off. In other words, any
level of collector current in between the two operating states of "saturation" and "cut-off" (i.e., active region) is not important to the functioning of a transistor switch circuit.
3. "Saturation" in a transistor switch circuit is achieved when the voltage across
the collector/ emitter (VCE(sat)) is less than or equal to .1 to . 3 volts - depending on the
type of transistor. At that voltage point, the transistor appears to act like a simple SPST
mechanical switch that has been closed (On).
