Download 10 Transistor Inverter Applications II

10. Transistor Inverter Applications II
10.1. A Relay Driving Circuit
A driving circuit is required to operate the relay used in a
telephone switching exchange. The coil of the relay has an inductance
of 100mH and an activation current of 250mA is required when
operating from a 12V supply. The driving circuit is to be controlled
from the output of a CMOS logic gate, which can source a maximum
current of 1mA with a minimum output high logic voltage VOH min =
0.8VDD and which operates from a 5V supply. The deactivation time of
the coil (time for magnetic field to release the contacts) should not
exceed 1ms. Transistors are available which have a minimum βF = 50
and base-collector reverse breakdown voltage of 80V.
VCC
coil
ZD
contacts
D
IC
IB
T1
T2
RB
VO
Vi = VOH min
Fig. 10.1
Schematic Diagram of the Relay Driver
Step 1
Determine the collector current required in the transistor which
is simply the activation current for the relay coil so that:
IC  250mA
1
Step 2
Determine the corresponding base current required to bring the
transistor to the edge of saturation when turned ON by the gate input
as:
IBEOS 
IC 250 mA

 5 mA
βF
50
If a safety overdrive factor is allowed, this will be even higher and is
well above the maximum current that the CMOS gate can provide. The
current required from the gate output must be reduced somehow.
Step 3
A solution can be found in the form of the Darlington transistor
configuration. In this configuration, the βeta factors of two transistors
are combined by using the emitter current of one transistor as the
input base current of a second transistor as shown in Fig. 10.1. The
effective βeta factor is then the product of the βeta factor of the two
individual transistors. The minimum input base current required to the
first transistor can then be found as:
IBEOS 
IC comb
I  IC2
I
250 mA
 C1
 C2 
 0.1 mA  100A
βF comb
βF1βF2
2500
βF
If a reasonable overdrive factor of 3 is allowed to cater for variations
in beta factors and temperature, the actual base current is given as:
IB  L IBEOS  3 x 100 A  300 A
Step 4
The base resistor can then be calculated as before assuming a
minimum input voltage from the logic gate, ViH min and the fact that
there are now two transistors so that two VBE drops must be accounted
for. Then:
RB 
VRB
IB

VOH min  2VBE SAT 0.8 x 5V  2 x 0.8V
2.4V


 8 kΩ
IB
300 A
0.3 mA
2
Step 5
A problem arises when turning the transistors off to deactivate
the relay. Since the coil is inductive, a back emf will be developed
across it when the current through it is altered. When the current in
the coil is forced to zero by turning off the transistors, a back-emf will
be generated across the coil which tends to oppose this change. This
back-emf will therefore be in the opposite direction to the voltage drop
which prevailed across the coil when the relay was activated. The
magnitude of the back-emf depends on the rate at which the current is
changed and can be quite large when the current is changed very
quickly. If a finite turn-off time for the transistors is taken as
approximately 100ns, the value of the back-emf can be estimated as:
VLB  L
di
- 250 x 10 -3
 100 x 10 - 3 x
 -250 x 10 3  -250kV
-9
dt
100 x 10
If the 12V supply is assumed ideal, then this voltage remains fixed,
which means that when the transistors are turned off the potential at
the lower end of the coil rises towards +250kV relative to ground
potential. In reality it will not go as high as this but will be several
hundred volts. This will bring the potential at the collectors of the
transistors well above the reverse breakdown voltage.
A common solution to the back emf problem is to connect a diode
in reverse across the coil. This limits the back emf to –VD ON = -0.5V
and thereby protects the transistors. The problem with this is that the
rate at which energy can now be removed from the coil is limited as:
di VLB
- 0.5


 -5 As -1
-3
dt
L
100 x 10
This means that to bring a current of 250mA to zero if it is assumed to
decay linearly requires a time of:
Tturn off 
 250mA
0.25

 0.05s  50ms
5
 5As 1
This is much longer than the limit given in the specification which is
1ms. The deactivation time can be reduced by allowing a higher back
emf to develop across the coil when deactivating. Since the transistors
have a collector-base reverse breakdown voltage of 80V, a back emf
approaching this could be allowed.
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Vi (t)
t
iC (t)
t
diode
zener
diode
icoil (t)
t
250kV
no clamp
zener diode clamp
VZD
diode clamp
VCC
VD
VO (t)
t
Fig. 10.2
Waveforms Showing Operation of the Relay Driver Circuit
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This can be achieved by connecting a Zener diode in series with the
standard diode placed across the relay coil. To obtain reasonable
lifetime from the transistors, a Zener voltage of say 50V is reasonable.
Then the revised rate of change of current can be found as:
di VLB
- 50.5
50


 -500 As -1
-3
dt
L
0.1
100 x 10
This gives a revised turn off time of:
Tturn off 
 250mA
0.25

 0.0005s  0.5ms
1
500
 500As
Step 6
With an active current level as high as 0.25A, it is wise to check
any relevant power ratings. Assuming that the given relay has been
chosen correctly by the user, the only thing of concern is the power
dissipation in the transistors. Before this is done, the mechanism of
operation of the transistors should be looked at a little more closely.
In the Darlington configuration used, T1 is driven by the logic
gate and providing it has sufficient base current it will be driven into
saturation so that VCE1 = VCE sat. The base of transistor T2, on the other
hand, is fed from the emitter of T1. In the first instance it might be
assumed that provided this current was high enough, T2 should also be
driven into saturation. However, it can be seen from the circuit, that
with the collectors of the two transistors connected together, the
base-collector junction of transistor T2 cannot become forward biased.
In fact, transistor T1, itself in saturation with VCE1 = VCE sat , maintains a
minimum reverse bias of this amount across the base-collector
junction of T2 , so that the latter remains in the forward active mode
with:
VCE2  VCE1  VBE2  VCE sat  VBE on  0.2V  0.7V  0.9V
Most of the relay coil current flows into the collector of T 2 with only a
fraction going into T1. In this case the power dissipated in T2 can be
estimated as:
PT2  VCE 2 x IC  0.9V x 250 mA  225 mW
This is too high to use a transistor with a 250mW power rating and one
with a 500mW rating will yield a much a longer lifetime.
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