Download 9 Transistor Inverter Applications I

9. Transistor Inverter Applications I
9.1.
Introduction
The simple transistor inverter circuit finds many applications,
particularly in discreet form. It is very often used as a current buffer
where only a limited amount of current might be available, from the
output of a logic gate, for example, and more than this current is
required to drive some other component or element. It is also common
in level shifting circuits to convert the logic levels from one system to
another. It is commonly used as a buffer to drive LEDs, buzzers, relays,
solenoids and even small loudspeakers. Some of these applications will
now be examined as worked examples.
9.2.
A LED Buffer
A buffer is required to drive a red LED having the properties given
in the graphs shown below in Fig. 9.2 and Fig. 9.3. A brightness of
100mcd is required from the LED when illuminated. The buffer is to be
driven by a logic gate having a VOH min = 0.8VCC operating from a 5V
supply. A bipolar transistor is available which has a minimum βF of 50,
VBE sat = 0.8V and VCE sat = 0.1V.
VCC
IC
RC
IB
T
RB
VO VCE sat
Vi = VOH min
Fig. 9.1
Schematic Diagram of LED Buffer
1
Brightness
1000
luminous
intensity
I V mcd
100
10
1
0.1
1
Fig. 9.2
10
forward current
I F mA
100
A Plot of the Luminous Intansity Properties of the LED
I vs V Characteristic
150
IF
140
mA
130
120
110
100
90
80
70
60
50
40
30
20
10
0
0
0.5
1
1.5
2
VD
Fig. 9.3
V
Electrical Characteristics of the Light Emitting Diode
2
A schematic diagram of the proposed buffer is shown in Fig. 9.1.
This differs from the single transistor inverter circuit only in that a LED
is included in the collector load as well as the resistor, RC. The collector
resistor serves to determine the level of current through the LED when
the transistor is ON.
Step 1
Determine the forward current needed in the LED to provide the
brightness level required. Fig. 9.2 shows the luminous intensity, which
can be interpreted as brightness, as a function of the forward current
in the diode. The coordinates of this curve for 100mcd brightness are:
IV = 100mcd, IF = 30mA
Step 2
Determine the forward voltage drop across the diode for the
above value of current. Fig. 9.3 shows the electrical current-voltage
relationship from which this can be found for a current of 30mA. The
coordinates are:
IF = 30mA,
VD = 1.5V
Step 3
Calculate the voltage drop across the collector resistor, RC, which
is given as:
VR C  VCC - VD - VCE sat  5V - 1.5V - 0.2V  3.3V
Step 4
The value of the collector resistor can then be found as:
RC 
VRC
IC

VRC
IF

3.3 V
 110 
30 mA
Step 5
Check the power rating of the resistor required to ensure that it
does not overheat. The power dissipated in RC is given as:
PRC  VRC x IC  3.3 V x 30 mA  100 mW
3
A resistor with a 125mW rating could theoretically be used. However,
it is normal practice not to use circuit components at more than 75 80% of their power rating in order to reduce failure rates and extend
their lifetime. Hence, a resistor with a power rating of 250mW would
be used to avoid being at this limit. This is the standard power rating
for most resistors in low voltage circuits. The power rating of the
transistor is not critical as there is only a low voltage of VCE sat = 0.2V
across it.
Step 6
The base current required to drive the transistor should now be
determined. This is primarily influenced by the βF of the transistor. To
bring the transistor to the edge of saturation when driving the LED
requires a base current given as:
IBEOS 
IC
I
30 mA
 F 
 0.6 mA
βF βF
50
Since the minimum value of βF has been used, there is no need to allow
for variations in this. Moreover, the load conditions are fixed as only a
LED is to be driven. This means that really only temperature variations
are present. This can be adequately allowed for by using a base
overdrive factor of 2. Then:
IB   IBEOS  
IF
30mA
 2x
 1.2 mA
βF
50
Step 7
Finally the value of the base resistor, RB , can be calculated as:
RB 
VRB
IB

VOHmin  VBE sat
IB

0.8 x 5V  0.8V
3.2V

 2.7 kΩ
1.2 mA
1.2 mA
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9.3 A Level Shifting Circuit
A circuit is required to convert the HI and LO output levels of the
comparator in a threshold detector which operates from a ±10V power
supply to the logic levels of a standard TTL family of gates operating
from a single 5V supply. Each TTL gate input can be treated as having
an equivalent resistive load of 5kΩ connected between the supply rail
and the input. The level shifting circuit must provide a fan out of 10. A
transistor is available which has a typical βF of 100 and a base-emitter
reverse breakdown voltage of 5V.
V
CC
RC
= 5V
IC
+10V
10 stages
+
RB
_
V
Vo
IB
i
-10V
Fig. 9.4
A Schematic Diagram of the Level Shifting Circuit
Step 1
Determine the load current required to drive 10 gates forming a
load as outlined in the specification. This can be done assuming a
perfect logic LO output from the level shifting circuit. Then:
IL 
VCC  VO
5V

 10mA
R L/N
5k /10
5
Step 2
Determine the total collector current needed in the transistor.
This can be done by making an allowance for current through a
collector resistor RC under unloaded conditions. A value of RC of 10kΩ
will allow a collector current of 0.5mA which is sufficient. This means
that:
ICTOT  IC  IL 
VCC  VO
 IL  0.5  10  10.5mA
RC
Step 3
The base current can now be calculated. To ensure that the
transistor is at the edge of saturation requires a base current of:
IC 10.5mA

 105A
βF
100
The value of βF used is a typical value so this is subject to
manufacturing process and temperature variations. The worst-case
load conditions have already been accounted for in dealing with the
fan-out requirements and so a safety loaded overdrive factor of σL = 3
should suffice so that:
IB  σL
IC
 3 x 105A  315A
βF
Step 4
Finally, the value of the base resistor can be determined. This is
chosen to allow the above value of base current to flow through it
when being driven by a HI level input from the comparator. Many
comparators do not provide an output voltage that goes to either
supply rail. Those using bipolar technology tend to give an output
voltage which is about 1 to 1.5V away from either supply rail voltage
and this should be allowed for in calculating the value of the base
resistor, RB. Then:
RB
VOHmin VBE sat
IB

VCC (comp) 1.5VVBE sat
IB

10V 1.5V0.8V
7.7V

24.4kΩ
315 μA
315 μA
In this event, the nearest value of 22kΩ in a 5% manufacturing
tolerance series would be used.
6
Step 5
There is just one remaining design detail that must be addressed.
The specification states the base-emitter reverse breakdown voltage is
5V. This means that the maximum reverse bias voltage which the base
emitter junction can tolerate is 5V (i.e. max -ive VBE = -5V). As the
comparator is fed from a bipolar ±10V supply, its LO output voltage
will be of the order of -8.5V which will reverse bias the base-emitter
junction. Since negligible current flows through the base resistor, RB,
under this condition there is no voltage dropped across it and the
entire -8.5V appears as a reverse bias across the base emitter
junction. This will exceed the maximum reverse bias allowed. The
problem can be solved by connecting a diode across the base emitter
junction in the opposite direction to the junction itself as shown in Fig.
9.4. When the comparator output is HI this diode will itself be reverse
biased (but will be chosen to have a higher rating than the transistor),
the transistor turns ON and the level shifter output goes LO as
required. This can be inverted by the first logic gate which is driven by
the circuit. When the comparator output goes LO, the transistor turns
OFF and the level shifter output goes HI as required. The diode
conducts under this condition and limits the reverse bias across the
base-emitter junction to -VD = -0.5V.
There is, in fact, a comparator on the market which has the
intrinsic ability to do level shifting. The LM311 marketed by National
Semiconductor Inc. provides an open-collector output stage. This is an
output transistor that has no load connected to the collector which is
simply brought to an output terminal on the chip so that the user can
connect their own load. Likewise, the emitter of the transistor is
connected to a pin on the chip so that users can choose their own
common or ground connection. This is shown in Fig. 9.5.
+VS
+VCC
+
VCC
REXT
0V
_
-VS
Fig 9.5
The LM311
7