Hello,
I have seen several circuits similar to the attached one. I can understand how the upper part works (Voltage regulator ICs...), but I find some difficulties in the lower part. Here are some questions:
1) Why the op-amp IC is fed through a 100 resistor and from the adjust's of the voltage regulator ICs?
2) What does the connection between pin 7 and the output of U3 do?
3) The feed back through the 5k resistor, the transistor connection... why are they connected like that??

 

I think you forgot the attachment.

 

Your image link did not work.
Try posting a link to the web page instead, and indicate which schematic it is that you are talking about.

 

I edited the post. You can see the circuit now.

 

1) Why the op-amp IC is fed through a 100 resistor and from the adjust's of the voltage regulator ICs?
2) What does the connection between pin 7 and the output of U3 do?
3) The feed back through the 5k resistor, the transistor connection... why are they connected like that?

1: The op-amp is fed from the output of the regulators. The 100 Ohm resistor is a feed to the adjust pins from the output voltage. As the voltage between the out and adjust of the regulators is fixed (that's what the regulator IC is trying to regulate), it's also a constant current source through the transistor.

2: see 1.

3: The two 5K resistors form a divider, so the input to the opamp is at half the regulator output voltage.

The voltage at pin 3 of the opamp will be set by the value of the 1k5 pot + 150 ohm resistor. Due to the constant current source feeding the transistor, this is forming an adjustable voltage reference.

The opamp compares that reference to the half output voltage on pin 2, and drives the transistor to control the voltage on the reg adjust pins so the opamp inputs have a matching voltage.

I don't know how well it would work in practice without some capacitors added to stabilise the regulators and control the feedback.

 

1) Why the op-amp IC is fed through a 100 resistor and from the adjust's of the voltage regulator ICs?

The opamp is being powered by the output of regulator U3, not the 100 Ohm resistor.
U3 will attempt to keep a nominal 1.25 difference between the OUT and ADJ terminals by sourcing current from the OUT terminal. The 100 Ohm resistor is effectively connected between the OUT and ADJ terminals. Therefore, the current through the 100 Ohm resistor will be approximately 1.25v/100=12.5mA.

2) What does the connection between pin 7 and the output of U3 do?

That is how the opamp is supplied with it's regulated Vcc/+V.

3) The feed back through the 5k resistor, the transistor connection... why are they connected like that??

The 5k resistor on the output and the 5k resistor from pin 2 to ground form a voltage divider. This gives the opamp feedback at a level of 1/2 the output voltage. The opamp attempts to keep the inverting input's voltage the same as the noninverting input by changing the voltage on the output (pin 6). The transistor is used as a voltage follower; the gain of the transistor multiplies the current that can be sunk by the output of the opamp. The more current sunk by the transistor, the higher the voltage on the noninverting input of the opamp. The output voltage can thereby be adjusted by the 1.5k pot. Note that 1.5k is not a standard value for pots; they are typically available in values like 500 Ohms, 1k, 2k, 5k, 10k, etc.

 

Well, dang. Too slow, get back.

 

Thank you both.
Now I see how much my questions are simple and easy.. very easy.. and especially the questions 2) and 1) and also 3). What you clarified are obvious in the circuit. But still I have a question, why is the op-amp fed form the output of U3 and not from the final output?

Here is another op-amp based voltage regulator/power supply:

In general, I think it's better (simplest, more efficient...) to use this circuit as a variable power supply, replacing the npn transistor by a power MOSFET one, instead of the first circuit. Am I right? The second circuit with mosfet/mosfets is better than the first?

 

Thank you both.
Now I see how much my questions are simple and easy.. very easy.. and especially the questions 2) and 1) and also 3). What you clarified are obvious in the circuit. But still I have a question, why is the op-amp fed form the output of U3 and not from the final output?

At that point, it doesn't make a heck of a lot of difference, however under heavy load the output of the regulator will be slightly higher than the actual output.

Here is another op-amp based voltage regulator/power supply:

In general, I think it's better (simplest, more efficient...) to use this circuit as a variable power supply, replacing the npn transistor by a power MOSFET one, instead of the first circuit. Am I right? The second circuit with mosfet/mosfets is better than the first?

You really can't say "efficient" and "linear regulator" in the same sentence.

In the 2nd circuit, the output will be limited by the gain of the transistor vs the current source ability of the opamp.

Replacing the transistor with a power MOSFET would minimize the current necessary from the output of the OPAMP (seeing as they are voltage controlled rather than current controlled), but instead of a somewhat fixed 0.7v difference between the base and emitter, you would have the threshold voltage of the MOSFET vs drain current to contend with.

A much bigger problem will be what to do with the concentration of heat from all the power that a single MOSFET will be dissipating?

The first circuit posted would likely stand several times the current that a single MOSFET would, when used as a linear regulator.

However, if you used the MOSFET as a switching regulator instead, your power dissipation would be far lower.

 

Right. Thank you for the information .