Linear Regulator Circuit: Understand +Vcc, V_load, T1/T2, V_out & R1

[foobag]

http://img715.imageshack.us/img715/2721/capturehoq.png

I am having a bit of trouble understanding this circuit. My textbook is pretty vague on the topic of three terminal voltage regulators, really only emphasizing on the LM317. For this circuit, is it necessary to know +Vcc , or V_load?

Furthermore, are T1 &T2 transistors behaving as a p-n-p current mirror?

I understand the Zener diode as acting as the reference input for the op-amp. However I am unable to understand how to calculate V_out and how R1 plays a role here.

Any insight would be deeply appreciated!

 

[berkeman]

The main loop will force the Vout node to the same voltage as the ZD voltage, because of the gain of the opamp, right?

Then you are right, T5-->T4 is a current mirror, and T1-->T2 is a 2nd mirror for that current. The T5 current is determined by the Vout and R1 (notice how the input to the mirror ends up looking just like a diode to ground below R1...?

But at first glance, I can see at least one reason why they are using this double-mirror to source current into the ZD, instead of just using a resistor. Can you think of what advantage(s) there might be to using this double current mirror instead of just a resistor to the ZD from Vcc?

 

[berkeman]

And a Quiz Question -- What can you add to the current mirrors to make them much better (more accurate) mirrors?

 

[foobag]

The main loop will force the Vout node to the same voltage as the ZD voltage, because of the gain of the opamp, right?

Then you are right, T5-->T4 is a current mirror, and T1-->T2 is a 2nd mirror for that current. The T5 current is determined by the Vout and R1 (notice how the input to the mirror ends up looking just like a diode to ground below R1...?

But at first glance, I can see at least one reason why they are using this double-mirror to source current into the ZD, instead of just using a resistor. Can you think of what advantage(s) there might be to using this double current mirror instead of just a resistor to the ZD from Vcc?

Ok that does make sense, so Vout would be the ZD voltage of 5volts. With the 2 current mirrors, its better to use over a resistor since they do not dissipate as much heat, also have high impedance and are generally small and cheap.

so does this mean R1 = Vout / I(T5), where current of T5 is influenced by the ZD current of 1mA?

So R1 = 5V/1mA = 5 kiloohm?

 

[foobag]

And a Quiz Question -- What can you add to the current mirrors to make them much better (more accurate) mirrors?

i am not entirely sure to the answer to this question

 

[berkeman]

Ok that does make sense, so Vout would be the ZD voltage of 5volts. With the 2 current mirrors, its better to use over a resistor since they do not dissipate as much heat, also have high impedance and are generally small and cheap.

so does this mean R1 = Vout / I(T5), where current of T5 is influenced by the ZD current of 1mA?

So R1 = 5V/1mA = 5 kiloohm?

You are getting close...

First, the voltage across R1 is less than 5V (but not by a lot). Why?

And you are setting the desired current of 1mA for the ZD. Maybe that's what you meant anyway.

BTW, even if you don't want to answer my Quiz Question about improving the mirrors, you should definitely try to figure out why they are using the double current mirror for setting the ZD current, instead of just a resistor from Vcc. That's a key part of this circuit.

 

[foobag]

is the voltage across R1 less than 5volts because of the ZD voltage drop of 0.7 volts?

so they are using current mirrors here instead of resistors to produce a constant steady current with extremely high impedance, and no affect on loading? is this why they have implemented the transistors in favor of resistors?

 

[berkeman]

is the voltage across R1 less than 5volts because of the ZD voltage drop of 0.7 volts?

Yes, good.

so they are using current mirrors here instead of resistors to produce a constant steady current with extremely high impedance, and no affect on loading? is this why they have implemented the transistors in favor of resistors?

It's not so much a loading issue. You're going to load Vcc with about 1mA either way (current mirror source for I(ZD) or resistor source for I(ZD).

Hint -- what happens with the resistor source option for I(ZD) when Vcc varies? They haven't given you a spec for Vcc in the problem, have they? What if it varies by 2:1 worst case?

 

[foobag]

ah so if Vcc varies a lot, than the current I(ZD) through the resistor source won't be as steady as using the current mirror source for I(ZD), basically because of Ohm's Law (R = V/I)?

the extra info they gave about the T4&T5 having Vbe (base-emitter) as 0.7Volts and beta >100, do they have any influence on the calculations, or just standard values?

 

[berkeman]

ah so if Vcc varies a lot, than the current I(ZD) through the resistor source won't be as steady as using the current mirror source for I(ZD), basically because of Ohm's Law (R = V/I)?

the extra info they gave about the T4&T5 having Vbe (base-emitter) as 0.7Volts and beta >100, do they have any influence on the calculations, or just standard values?

Good. And no, if they are assuming ideal (and ideally matched) transistors, then the gain and Vbe don't mean much for the mirror. BUT, with real transistors, high gain is useful, and the matching of the Vbes within the mirror is important. That's what I was getting to with my Quiz Question. There is a simple improvement to the basic current mirror circuit you were given, which helps a lot in improving the mirror action in the face of slightly different Vbes, which you get in the real world.

Just for fun you might try Google Images or just a search at wikipedia.org, to see what the modification is, and to understand why it helps.