Exact working of Widlar Current Mirror .

[brainbaby]

As involvement of emitter resistor in current mirrors has certain drawbacks like restricting the output voltage range to a lower value..thus in order to cancel the draw back of emitter resistors they say to use a
Widlar current mirror …Widlar current mirror is designed to provide very small output currents without requiring very high resistance values(Re) or drastically restricting the output voltage range…..how exactly thus Widlar current mirror does it…yet it involves two resistors( R and Re)..??

 

[Jony130]

Ic1 is equal to Ic2 only if Vbe1 = Vbe2 (I assume perfectly matched transistors)
But in this case we have Vbe1 = Vbe2 + Ic2*Re. So as you can see Vbe1 is now divided into Vbe2 and VRe.
Because now Vbe1 is no longer equal to Vbe2. We clearly see that Vbe2 < Vbe1. So Ic2 < Ic1.
http://forum.allaboutcircuits.com/threads/current-mirror-explanation-misleading-diode-equivalent.92655/page-2#post-678790

 

[meBigGuy]

see 11.8 in this chapter
http://wiki.analog.com/university/courses/electronics/text/chapter-11

 

[brainbaby]

Ic1 is equal to Ic2 only if Vbe1 = Vbe2 (I assume perfectly matched transistors)
But in this case we have Vbe1 = Vbe2 + Ic2*Re. So as you can see Vbe1 is now divided into Vbe2 and VRe.
Because now Vbe1 is no longer equal to Vbe2. We clearly see that Vbe2 < Vbe1. So Ic2 < Ic1.
http://forum.allaboutcircuits.com/threads/current-mirror-explanation-misleading-diode-equivalent.92655/page-2#post-678790

But if Ic1 is not equal to Ic2 because Vbe2 < Vbe1 ...so then how can we say that the circuit behaves as a current mirror...i mean for a current mirror both Ic1 and Ic2 should be equal...

 

[brainbaby]

see 11.8 in this chapter
http://wiki.analog.com/university/courses/electronics/text/chapter-11

please explain this phrase of the text you mention:
"Figure 11.11 is an example Widlar current source using bipolar transistors, where the emitter resistor R2 is connected in series with the emitter of output transistor Q2, and has the effect of reducing the current in Q2 relative to Q1"
Since by adding emitter resitor in Q2 it has an effect of reducing collector current in Q2 relative to Q1...then why we call it a current mirror..since ideally a current mirror both Ic1=Ic2..but in this case Ic1 is not equal to Ic2..so is it reasonable to call the circuit a current mirror...??

 

[Jony130]

But who said that mirror current must match ?

 

[brainbaby]

But who said that mirror current must match ?

so what could be the definition of current mirror then...?

 

[The Electrician]

.so is it reasonable to call the circuit a current mirror...??

Not everybody calls it a current mirror: http://en.wikipedia.org/wiki/Widlar_current_source

 

[brainbaby]

Not everybody calls it a current mirror: http://en.wikipedia.org/wiki/Widlar_current_source

rather than its a modification of a current mirror into a current source ...better to call it a current source...
but what does this means---> enabling the current source to generate low currents using only moderate resistor values...
they are talking about which resistor R1 or R2..

 

[Jony130]

R2 of course. And this circuit has been used only in IC. But today no one use this circuit anymore. Because value of the resistor is proportional to its length on the chip. So resistors with large values will take up an inconveniently large area on the chip. And this is why transistors replace the resistors in todays IC chip (transistor require less space on the chip).

 

[analogdesign]

R2 of course. And this circuit has been used only in IC. But today no one use this circuit anymore. Because value of the resistor is proportional to its length on the chip. So resistors with large values will take up an inconveniently large area on the chip. And this is why transistors replace the resistors in todays IC chip (transistor require less space on the chip).

I've used this circuit in chips. It is very useful. Since the advent of silicide block high-density resistors have become more common on chips. You can get better than 1000 Ohms/sq now, easy. Resistors are still bigger than transistors but for analog circuits it's not that big a deal anymore.

 

[brainbaby]

Hey guys i am not getting over this phrase...please shed light on it...
Widlar current source is used in analog integrated circuit design to produce a small driving current without the use of large resistors. This is particularly important because it is difficult to produced large resistors on chip that have good tolerances..thus enabling the current source to generate low currents using only moderate resistor values.

now since if emitter resistor is high then the output resistance is high which can then obviously decrease the output current...but since incorporating large resistor is difficult..then how widlar current source compensate this by using moderate emitter resistance values...??

 

[analogdesign]

now since if emitter resistor is high then the output resistance is high which can then obviously decrease the output current...but since incorporating large resistor is difficult..then how widlar current source compensate this by using moderate emitter resistance values...??

Be careful you don't confuse small-signal and large-signal quantities. A large emitter resistor does increase the output resistance (because of negative feedback) but that in-and-of itself has NOTHING to do with the output current. This is because if you want to use an emitter resistor to increase output resistance you put it in both emitters. This doesn't change the output current if you have enough headroom to support the emitter resistors, you still have a standard current mirror, just with more output resistance.

Now, a Widlar Current Source is another beast entirely. If you try to solve for the output current you wind up with a transcendental equation due to the exponential relation between VBE and IC. Luckily you can iterate this with a hand calculator because it converges quickly, but removing that one emitter resistor makes it a TOTALLY different circuit. This is second only to the bandgap reference when it come to Wildlar's genius.

Edit: try it out for a few currents. A Widlar Current Source gives you lower current for a given resistor and power supply voltage than you could do with an opamp current source that enforces I= V/R. It is a bit finicky but look up "peaking current source" if you're interested in how you can reduce R even more.

 

[brainbaby]

R2 of course. And this circuit has been used only in IC. But today no one use this circuit anymore. Because value of the resistor is proportional to its length on the chip. So resistors with large values will take up an inconveniently large area on the chip. And this is why transistors replace the resistors in todays IC chip (transistor require less space on the chip).

I think so they are also talking about R1..because in absence of R2(emitter resistor) and in order to generate low output current the R1 value should be very high...since now high value of any resistor is difficult to integrate on chip..thus in order to compensate its value ..an extra emitter resistor R2 is added...that further decrease the output current...hence a very high value of R1 is divided in two moderate values of R1 and R2...which are comparitively much easier to fabricate on chip...

 

[analogdesign]

I think so they are also talking about R1..because in absence of R2(emitter resistor) and in order to generate low output current the R1 value should be very high...since now high value of any resistor is difficult to integrate on chip..thus in order to compensate its value ..an extra emitter resistor R2 is added...that further decrease the output current...hence a very high value of R1 is divided in two moderate values of R1 and R2...which are comparitively much easier to fabricate on chip...

They are not talking about R1. That is put there just to simply the circuit for an explanation. No one ever generates the input current in that way. You use an actual opamp to generate a stable current. Generating the input current using R1 would be terrible in practice because all the noise on the power supply would appear across R1 so it would also appear in your output current.

 

[brainbaby]

They are not talking about R1. That is put there just to simply the circuit for an explanation. No one ever generates the input current in that way. You use an actual opamp to generate a stable current. Generating the input current using R1 would be terrible in practice because all the noise on the power supply would appear across R1 so it would also appear in your output current.

Isn't R1 the programmable resistor...pls refer following url
'limitation of current mirror'
https://books.google.co.in/books?id...epage&q=uses of widlar current source&f=false

 

[analogdesign]

Isn't R1 the programmable resistor...pls refer following url
'limitation of current mirror'
https://books.google.co.in/books?id=EDB9BAAAQBAJ&pg=PA85&lpg=PA85&dq=uses+of+widlar+current+source&source=bl&ots=gxPoNBHQrm&sig=8wv81Lsjy3xotujCt34GkWjsoE4&hl=en&sa=X&ei=dJhMVf7XH8XluQSpq4HQDQ&ved=0CDsQ6AEwBjge#v=onepage&q=uses of widlar current source&f=false

Like I said, R1 is put there to make the current source understandable. The key point of the Widlar current source is that there is a resistor in only one emitter. That's it. Don't confuse yourself with R1. Typically a chip has a lot of bias currents all over the place and when you need a very small current, one option would be to use one of these bias currents (which is large to improve noise and matching) with a Widlar current source to locally generate your low current.

 

[brainbaby]

Like I said, R1 is put there to make the current source understandable. The key point of the Widlar current source is that there is a resistor in only one emitter. That's it. Don't confuse yourself with R1. Typically a chip has a lot of bias currents all over the place and when you need a very small current, one option would be to use one of these bias currents (which is large to improve noise and matching) with a Widlar current source to locally generate your low current.

If we talk about discrete current mirror circuit(not widlar for a moment) then what provision should we make in the circuit to get a very low output current..??

 

[analogdesign]

If we talk about discrete current mirror circuit(not widlar for a moment) then what provision should we make in the circuit to get a very low output current..??

For a discrete current mirror you either make the ratio of input device to output device size very large, or you make the input current very small. Both have their issues.

 

[brainbaby]

For a discrete current mirror you either make the ratio of input device to output device size very large, or you make the input current very small. Both have their issues.

refer following...url
https://books.google.co.in/books?id...DgK#v=onepage&q=widlar current source&f=false
In fig 5.24 does Iref considered as input current...?? If yes then how can we make it small?