Why tail transistor need to be in saturation region?

[Anachip]

Hi Guys,

I'm a new guy in the Analog Design Field. I'm in the learning curve. I have a first question here in the way of desiging. Why we need to have the tail transistor to work in the "saturation region" whenever we design a differential amplifier. Please explain to me with a practical and theoritacally as i'm quite confused on this after have read from analogs books.

Thanks,
Anachip

 

[nile_king]

Hi
the tail current must be in the sturation region as u use it as current source and when u go to the c/cs curve of the transistor u find the saturation region give u nearly flat current with voltage which the response u need for current source.

 

[eng_Semi]

U need the tail transistors to be in saturation because after fabrication of any chip there will be a process variations and also there is a temperature variations, so we want to be sure that with any variations the current will vary in a small region to have a constant performance for the amplifier.

 

[amarnath]

no the reason for the tail transistor to be in saturation is that,u are not going to allow the common mode signal to affect the output of the op amp,since when u are maintaining this transistor in saturation u are not going to allow the currents of the of the transistors in the differential arm of the differential amplifier.hope iam clear.

regards
amarnath

 

[eng_Semi]

since when u are maintaining this transistor in saturation u are not going to allow the currents of the of the transistors in the differential arm of the differential amplifier

I didn't get u ?

 

[khouly]

as the tail transistor in sat , the current passing through it idealy independent on "VDS" so so any variation in it will not change the currnt which is the bisa current for the opamp or diffpair.
also this tarsistor must be wide enough

so this point considered AC ground in differential operation

i wish this help

khouly

 

[uncle_urfi]

For Vt=v2-v1 = 400mV the differential pair current switches completely from one branch to the other as given by the tanh equation for the individual branch currents.

A transistor in saturation is a current source and that's why it has to be kept in saturation. Otherwise, you will have a voltage variable resistor at your hands and the diff.pair will not function properly.

 

[mmohsen]

no the reason for the tail transistor to be in saturation is that,u are not going to allow the common mode signal to affect the output of the op amp,since when u are maintaining this transistor in saturation u are not going to allow the currents of the of the transistors in the differential arm of the differential amplifier.hope iam clear.

regards
amarnath

i didnot understand ur reply please try to clarify !!.. thanks

my point of view is that you need it in saturation ,so the current flowing through it is constant so you have constant currents in the diff-pairs as gm=2*Id/Veff

regards,..

 

[suria3]

Hi,

When you bias your tail transistor in saturation, your total current that sinks at the tail won't depend on VDS when it change over temperature or process. Thus, your amplifier will operate linearly, meaning that your gain will be constant as the current is being constant when the tail switching takes place.

Thanks,
Suria3

 

[tok47]

In another point of view, if the tail transistor is not in saturation, it will be like a varible resistor.

So, the bias point that you set for the differential transistor will not be correct.

 

[mmohsen]

Hi,

When you bias your tail transistor in saturation, your total current that sinks at the tail won't depend on VDS when it change over temperature or process. Thus, your amplifier will operate linearly, meaning that your gain will be constant as the current is being constant when the tail switching takes place.

Suria3

i agree with this reply.

 

[elbadry]

I think the reason is:

1. You need an ideal current source; i.e. a source whose o/p impedance is (ideally) inifinite. This ensures that the the bias current and hence the gain would be independent on the voltage drop on the current source (within limits of course).
Biasing the transistor in the linear or triode region would make it have a very low output impedance-giving a highly nonlinear response. However, in saturation region the o/p impedance is large, giving acceptable performance as a currrent source. Sometimes even this performance is not satisfactory, that's why there are many modifications to current mirrors that try to improve the o/p impedance (cacscoding-Wilson mirrors..etc)


2. In a diff pair, you usually need a high CMRR. This means that you need a ver low common mode gain. In the common mode analysis, the tail current source acts as a source degeneration resistance; thus improving it's o/p impedance will reduce the common mode gain significantly..thus improving CMRR.


Hope this is useful

 

[huojinsi]

When the transistor is in the saturation region,the most key is that the transistor is operating at linear region and the transistor's transconductor is constant.

 

[muldersun]

If you want to get a right AC result,the current source must be constant.

 

[JPR]

For a pure differential pair, the device providing tail current does not need to be in saturation, and could even be a simple resistor.

However, in most applications, the differential pair is loaded by a pair of devices that convert the differential signal to single ended, which drives a second stage. For the second stage to operate properly without offset, the current densities in the load devices of the differential pair and the device they are driving in the second stage must match.

With a poor current source (resistor or device in linear region), the current will have a strong dependance on the input common mode voltage. This will make it difficult to bias the second stage to balance with the first, and without this balance, the CMRR will be terrible. It CAN be done (and HAS been done), if, for example, you need an amplifier whose GBW will vary with input common mode voltage, but in most cases, you want the amplifier to behave consistently across the operation range.

With a good current source (transistor in saturation), the tail current will be roughly independant of input common mode voltage, it will be easy to balance the first and second stages, and the amplifier performance (gain, BGW, output resistance, etc) will be consistent.

The ease of balancing the first stage with the second stage, along with the desire for consistent performance of the amplifier, are the reasons for biasing the device providing the tail current in the saturation region for most applications.

 

[surianova]

HI JPR,

How about common noise or CMRR ? Is is better for it in saturation to get
higher CMRR? Is it critcal?

Surianova

 

[layes2]

only in this region it has enough gain