Role of Op-Am in regulators-in-parellel circuit

[exleonhart]

As stated in the title. I dont understand the role of the LM308 Op-Am in the circuit. Someone enlighten me, please.
Here is the circuit:


**broken link removed**

 

[ericgibbs]

hi,
The 1K5 variable resistor sets the voltage output level, which is input to the NI of the OPA.
The junction of the two 5K0 senses the actual voltage level which is input to the INV of the OPA.

If the voltage at the INV input falls the OPA adjusts the current flowing thru the 2N2905 transistor in a sense that will correct the voltage output.

The OPA is the sensing/control element in the loop.

 

[MrAl]

Hi again,


To add to Eric's reply a little...


The op amp functions as a difference error amplifier, and
is responsible for all of the voltage regulation. Even though
the LM338's normally regulate voltage, they do not do that
here. The LM308 controls all three LM338's by sensing
the difference between pins 2 and 3 and generating an output
voltage at pin 6 that reflects that minor difference.
The voltage at pin 6 then changes the voltage at the emitter
of the transistor, then the ADJ pin which corrects the output
voltage.

For example, if pin 2 becomes lower than pin 3 the output
ramps higher, and if pin 2 becomes higher than pin 3 the
output ramps lower. This change of pin 6 varies the base
voltage of the 2N2905 and that being a voltage follower
the emitter voltage also changes. The change in emitter
voltage changes all the ADJ terminals, which changes the
output voltage. This is how regulation is achieved.
The current though the 1.5k pot is constant, so as
you adjust the pot value the voltage at pin 3 goes up
or down, which offers a new regulation point for pin 2
to have to follow. This is how the output voltage is
adjusted.

A typical scenario would go like this:

We turn the circuit on, and the output pin 6 ramps up
to regulate the voltage at say 10v because we have the
pot set to about 260 ohms. Note that pin 2 voltage is
roughly equal to pin 3.
Now we connect a load to the output, and the output drops
down a little because of internal resistances. This makes
pin 2 drop a little, which means now pin 3 is greater than
pin 2 so the output ramps up a little, which makes the
emitter ramp up a little, which makes the ADJ terminal
ramp up a little, which makes the output ramp up a little,
which puts the voltage back to 10v and makes pin 2 ramp
up a little, so pin 2 has become again equal to pin 3 and
equilibrium is again reached.

The opposite happens when we disconnect the load. The output
ramps up a little, pin 2 ramps up a little, the pin 6 ramps down
a little (because pin 2 is greater than pin 3), the emitter
ramps down a little, the ADJ pins ramp down a little,
the output ramps down a little back to 10v, and finally the
pin 2 ramps down a little and again becomes equal to pin
3 and the the circuit again is in equilibrium.


Note that all the ramping takes place pretty quickly so you dont
see too much difference at the output.

Also note that you can probably use a common LM358 for this app.
Regardless what op amp you use, you may need a 1k resistor in
series with the output of the op amp (to the input of the transistor).
This is just in case the op amp output draws current sooner than
the rest of the circuit can start up. What could possibly happen is
that the output of the op amp draws too much current through
the emitter of the transistor, which will latch the circuit up and
wont allow it to work properly. If it works without this extra
resistor then fine, but if not, insert the resistor and that should
take care of it.

One thing extra you might notice is that unlike many regulator circuits,
this one does not seem to have a reference voltage anywhere. It
really does however, as they are using the reference from the lower
LM338 to generate a reference in another part of the circuit.
I think this was pretty clever of the original designers.

 

[exleonhart]

Thanks : )