R2 forms an important part of the current limiting circuit. When current travels through this
resistor a voltage develops across it. When the voltage approaches 630mV, ie. 630mA
through a 1R resistor, the internal electronics in the LM723 starts to pull down the base
voltage on Q1. This effectively lowers the output voltage to make sure the current doesn’t
climb much above that value.
The value of R2 (and also R3) will determine the actual value of the current limit. For
example, a 1R resistor will set the current limit to be around 630mA and a 0.82R resistor will
make it around 770mA. The resistor value is chosen so that it will develop around 630mV
across it when your chosen current limit would be going through it.
I
limit
= 0.63V/R
limit
I
limit
is the current limit value in amperes and the R
limit
is the value in ohms of the resistor.
However, there are several things to be taken into consideration should you decide to increase
the current limit above the values I have chosen.
Firstly, you need to ensure your transformer, internal or external, will be able to handle the
additional current you are asking from it. The simple rule of thumb is:
I
ac
= 1.8 x I
dc
.
Where I
ac
is the steady state AC current in the transformer's secondary coil, and I
dc
is the
current taken by the load connected to the +15V and -15V rails. For example: The Yamaha
PA-20 can supply 0.94A of alternating current into a simple resistive load from each of its two
outputs. But by the time the output of the PSU has been rectified and smoothed the DC
current taken should not exceed 0.52A on a continuous basis.
Secondly, the amount of current needed to drive the base of the pass transistors, Q1 and Q2,
is directly related to the amount of current draw from the supply lines. Thus to ask for more
from the supply would be to ask more from the 723 on the positive rail and the op-amp on the
negative rail. The actual relationship between load current and drive current is related to hfe or
the gain of the pass transistor. Choosing a pass transistor with too little gain will mean that
you cannot obtain your required current. The recommended TIP35C and TIP145 devices have
been tested to work with a mains transformer that can supply up to 1.1A into the output load.
In this case the current limiting resistors, R2 and R3, would both be 0.56R.
Thirdly, the printed circuit board's copper tracks are not thick enough to pass huge amounts of
current. The more current a track has to carry the greater the unwanted voltage drop across it
and the hotter it becomes.
Fourthly, you need to consider how you are going to get rid of the extra heat the power
devices will develop. The actual amount of heat dissipated by the power devices will increase
with the more current you take from the supply. It is not quite as simple as a doubling of
current will give twice the amount of heat given off, but it is not far off it. The amount of heat
dissipated (in watts) will be equal to the voltage drop between the collector and emitter
multiplied by the current running through it (the DC load). The temperature will rise with
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