10.Transistor Inverter Applications II

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2 Νοε 2013 (πριν από 3 χρόνια και 8 μήνες)

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1

10.

Transistor Inverter Applications I
I


10.1
.

A Relay Driving Circuit

A driving ci
rcuit is required to operate the

relay used in a telephone
switching exchange. The coil of the relay has

an indu
ctance of 10
0mH
and an activation

current of 25
0mA
is required w
hen
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
V
OH min

= 0.8V
DD

and
which
operates
from a 5V supply.

The
deac
tivation time of the coil

(time for magnetic
field to release the contacts)

should not exceed 1ms. Transistors are
available which have a minimum β
F

= 50 and base
-
collector reverse
breakdown voltage of 80V.



Fig. 10
.1 Schem
atic Diagram of
the Relay Dr
iver



Step 1


Determine the collector current required in the transistor which is
simply the activation current for
the relay coil so that
:




O

V

V
i

= V
O
H min

T
2

coil

D

Z D

contacts

T
1


2

Step 2

Determine the
corresp
onding base current required to
b
ring the
transistor to the edge of saturation when turned ON by the gate input
as:




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 β
eta f
actors of two transistors
are

combined by using the emitter current of one t
ransistor as the input
base current of a second transistor as shown in Fig.

10.1. The effective
β
eta factor
is then the product of the β
eta factor of the two
individual
transistors. The minimum input base current required to the first
transistor can then b
e found as:




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:





Step 4


The base resistor can then be cal
culated

as before assuming

a
minimum input voltage
from the logic gate, V
iH min

and the fact that there
are now two transistors so that two V
BE

drops must be accounted for.
Then:




3

Step 5


A problem arises when turning the transistor
s

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 change
d and

can be quite large when the current is changed very quickly. If a
finite
turn
-
off time for the transistor
s

is taken as approximately 100ns,

t
he
value of the back
-
emf

can be estimated as:




If the 12V supply is assumed ideal, t
hen this voltage

remains

fixed, which
means that when the transistors are turned off the potential at the lo
wer

end of the coil rises towards
+
250kV

relative to ground potential
. In
reality it wil
l not go as high as this but will

be several hundred volts.
This
will bring the potential
at the
collector
s 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

V
D 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:




This mea
ns that to bring a current of 25
0mA to zero if it is assumed to
decay linearly requires a ti
me of:




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
colle
ctor
-
base reverse breakdown voltage of 80V, a back emf
approaching this could be allowed.


4


Fig. 10.2 Waveforms Showing Operation of the Relay Driver Circuit





V
D

V
ZD

V
CC

V
O

(t)

V
i

(t)

i
C

(t)

i
coil

(t)

diode

zener

diode

no clamp

d
iode

clamp

zener diode

clamp

t

t

t

t

250kV



5

This can be achieved by connecting

a Z
ener diode in series with

the
standard
diode
placed
across the relay co
il. To

obtain reasonable li
fetime
from the transistors, a Z
ener voltage of say 50V is reasonable.

Then the
revised rate of change of current can be found as:




This gives a revised turn
off time of:






Step
6


With a
n 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 diss
ipation 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, T
1

is driven by the logic gate
and providing it has sufficient base current it

will be driven into saturation
so that V
CE1

= V
CE sat
. The base of transistor T
2
, on the other hand, is fed
from the emitter of T
1
. In the first instance it might be assumed

that
provided this current was high enough, T
2

should also be driven into
saturat
ion. However, it can be seen from the circuit,

that

with the
collectors of the two transistors connected together
,

the base
-
collector
junction of transistor T
2

cannot become forward biased
.
In fact,
transistor T
1
, itself in saturation with V
CE1

= V
CE sat

,

maintains a minimum
reverse bias of this amount across the base
-
collector junction of T
2

, so
that the latter remains in the forward active mode with:




Most of the relay coil current flows into the collector of T
2

with only a
fraction

going into T
1
.

In this case the power dissipated in T
2

can be
estimated as:




This is
too high to use a transistor with a 250mW power rating and o
ne
with a 500mW rating will yield a much

a longer lifetime.