NEED CLARIFICATION

[techguru]

An additional LM395(Q2) is driven from the negative power supply lead of the LM308 to provide some output current sink capability
(2A) so the supply can be quickly programmed even with large capacitive loads? wHY DO WE WANT Q2 AND WHY CURRENT SINK IS NEEDE IN POWER SUPPLY?

[hugo]

Hi Guru,

Let's say we have a lab power supply without a load. You turn the voltage adjust pot clockwise/anticlowise the output voltage increases/decreases accordingly.
With large capacitive loads the output voltage will not follow exactly the voltage adjust pot movement, so to fix the issue that current sink stage is needed. 

[techguru]

Hi hugo,

          can you explain with a situation ? it will be helpfull.

[nicalitz]

The larger the capacitance of your load, the slower the rate of discharge (generally speaking). If you want the output voltage of your power supply to respond to adjustments in a timely manner, you need to enable said capacitive load to discharge and quickly and efficiently as possible. This is why you need Q2

[techguru]

hi nicalitz,
              Then you mean to say that Q2 transistor act as a resistor and Load capacitor tends to discharge through this resistor. Is it correct???

[Damianos]

A transistor, usually, acts as a ... transistor! It is a device that the conductivity, between collector and emitter, depend on the bias of the base.
The circuit is designed like an asymmetrical class B amplifier. When the output has to be reduced (for whatever reason) the op-amp tries to do this by pulling current through R11 and R14 to the R13 that biases the Q2 and it "helps the situation"...

[Raj]

guru baba shishaya
 explaination with water anology.
Think of the capacitor as a water tank and power supply is feeding water into that tank

the voltage is water level...

you have currently set the pump (power supply) to mantain water (voltage) to 20cm (20 volts) high...

later you push a button to bring it down to 10cm (10 volts)

without opening a in the tank drain (artificial load q 2)  the water will remain 20cm high (20 volts potent) until the water evaporates (the ineffitiencies of capacitor emptys it)