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What is this circuit called?

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It seems like that would be called, "A linear voltage regulator loaded by a constant current." At the top, you have a linear regulator with a PMOS pass-FET which regulates VOUT to equal VIN, and at the bottom you have a current mirror which draws current out of VOUT. Even though current is drawn out of VOUT, the op-amp at the top still maintains the VOUT=VIN relationship.
 

The circuit is called "Regulated drain current Mirror". It increases the output impedance and any variation makes output voltage constant.
 

The circuit is called "Regulated drain current Mirror". It increases the output impedance and any variation makes output voltage constant.

Dear varunkant2k,
In this file shown below, page 18 we can see the structure of drain regulated current mirror.
**broken link removed**

A important thing is making VDS1 = VDS3. However, from my figure, drain voltage of MN1 not equal to MN2.

Anyone has another answer?
 

drain voltage of MN1 not equal to MN2.
Need not to be , the advantage of the circuit is Drain voltage at the output stage becomes DC constant. And provide very high impedance at that node. Such a node generally helpful for baising the sensitive nodes.
 

It's just a class A voltage follower.
You can just call A1 the first stage amplifier, and MP1/MN2 a class A output stage.
 
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    ZekeR

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varunkant2k, this circuit is not useful as a regulated drain current mirror. A current mirror circuit ultimately outputs a current, which this circuit does not do.

I think checkmate hit it on the head. This can best be described as a buffer with class-A output stage.
 

There may be an application specific purpose of the class-A output stage but it isn't obvious. Possibly an asymmetrical current limiting action is intended.
 
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    ZekeR

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Is there a difference if the Op amp has negative feedback instead of positive feedback? In other words if you have Vin on the (+) input and the feedback on the (-) pin?
 

The circuit has negative feedback, not positive feedback. The output of the opamp goes through MP!, which inverts the signal, before being fed back to the non-inverting (+) input.
In other words if you have Vin on the (+) input and the feedback on the (-) pin?
If you do that the output voltage will either jump up to VDD and stay there or jump down to ground and stay there.
 

varunkant2k, this circuit is not useful as a regulated drain current mirror. A current mirror circuit ultimately outputs a current, which this circuit does not do.
Hi Zeker , you are right. Because the circuit itself is bit confusing /and incomplete.
OPAMp input is given VIN. that makes circuit no use other than operating pass transistor MP1 as active (triode) load.
Generally such circuit used in delay blocks.
 

Hi Zeker , you are right. Because the circuit itself is bit confusing /and incomplete.

Yes, circuits are generally confusing; truly understanding them usually takes years of study and practice, coupled with the right mentality.

OPAMp input is given VIN. that makes circuit no use other than operating pass transistor MP1 as active (triode) load.

As long as VIN is less than VDD by Vov (~0.3V or thereabouts), MP1 will be in saturation.
 
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    FvM

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As long as VIN is less than VDD by Vov (~0.3V or thereabouts), MP1 will be in saturation.
Actually benefit of the circuit comes when Pass transitor works in the deep triode region.(why?).
For higher swing, you need to operate PM1 in deep triode region and thats why high gain used.
 

Actually benefit of the circuit comes when Pass transitor works in the deep triode region.(why?).
For higher swing, you need to operate PM1 in deep triode region and thats why high gain used.
Where do you get this insight? You are also apparently assuming a specfic purpose of this circuit (otherwise there won't be a "benefit"). What is it? I didn't hear but guesses about it, including my own.
 

Where do you get this insight? You are also apparently assuming a specfic purpose of this circuit (otherwise there won't be a "benefit"). What is it? I didn't hear but guesses about it, including my own.
Hi FVM, sorry if i was not clear. This circuit is frequently used in ring oscillator to servo the on-resistance of differential stage active load.This is also called replica circuit.
I will explain the deep triode region benifit:
Tras. MP1 , overdrive voltage will be constant = VDD-VIN. If gate voltage of MP1 goes below VIN-vth, transitor MP1 will be in triode region. and it will vary its resistance to make drain voltage constant equal to VIN.
More detail you can find out in "design of Analog CMOS Integrated Circuits" by B. Razavi. page no514.
 

I see what you mean.



Of cause this specific application can't be guessed from the circuit shown in the initial post. It's different at least by this details
- exposing a different output node
- using a variable Vin voltage rather than Vref
nevertheless it's the same topology.

The deep triode versus saturation point is a matter of transistor parameters and intended circuit behaviour and can't be derived from the basic circuit as such, I think.
 

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