That is, it happens when the voltage at the collector is very close the the voltage at the base. Specifically, when Vce = Vbe - 0.4 (same as Vce = 0.2).
All "saturation" means in a transistor,is that the device is turned "hard on".
All "saturation" means in a transistor,is that the device is turned "hard on".
Terminology becomes critical here. When talking about BJT regions of operation, the saturation region is asymptotic to a MOSFET's triode region. A BJT that's "hard on" is operating in the foward-active region aka saturation region for a MOSFET, which is the typical region of operation for most small-signal applications.
Sorry,you've lost me on that one!
Sorry,you've lost me on that one!
"Hard on" is a colloquial expression for where any device is biased "on"to the point that the device current cannot be increased by any further change in the bias.(This sounds like a pretty good definition of saturation to me!)
Typical cases are in multivibrator circuits,where the devices go from saturation to cutoff.
That is, it happens when the voltage at the collector is very close the the voltage at the base. Specifically, when VCE = VBE - 0.4 (same as Vce = 0.2).
You are confusing cause and effect. A transistor with no base current is not saturated regardless of what the collector to emitter voltage is. The collector to emitter voltage of a saturated transistor depends on the base current, collector current and the transistor. Datasheets usually specify it as Vce(sat) at specified base and collector currents.
Sorry,you've lost me on that one!
"Hard on" is a colloquial expression for where any device is biased "on"to the point that the device current cannot be increased by any further change in the bias.(This sounds like a pretty good definition of saturation to me!)
Typical cases are in multivibrator circuits,where the devices go from saturation to cutoff.
Ditto.
Am I correct in understanding that what you refer to informally as "hard on" is a biasing condition which results in max DC beta, viz. mathematically, max(IC/IB)?
To be sure, "saturation" used within the context of the OP's description is not "saturation" in the normal sense of its definition.
That is, it happens when the voltage at the collector is very close the the voltage at the base. Specifically, when VCE = VBE - 0.4 (same as Vce = 0.2).
You are confusing cause and effect. A transistor with no base current is not saturated regardless of what the collector to emitter voltage is. The collector to emitter voltage of a saturated transistor depends on the base current, collector current and the transistor. Datasheets usually specify it as Vce(sat) at specified base and collector currents.
Nope, this is straight out of the book. But IB>0 is also a condition.
There may be some confusion between theoretical transistors and real world ones.
could i ask this way every one?
GIVEN:
BJT small analog signal amplification application, setup as per picture.
DEFINITION:
let's just say the definition of saturation is when Ic no longer changes with Vb.
SYMPTOMS:
hard on, small steady collector current, not enough Vce, collector voltage close to base voltage, changed beta, and what not...
QUESTIONS:
1) what mechanism triggers saturation, Vcb, Vce, Ib, Ic, and how?
2) what happens to Ic when a BJT goes into saturation mode from forward active mode? for example, if a BJT is biased correctly with its Q point at 1/2 the supply voltage, it is likely to have a relatively large Ic, because that's what happens when it is amplifyitng. now, say, if the input signal is getting too large, sending the BJT into saturation, (don't know how yet at this point, the actual mechanism is still unclear to me, hope answers to question 1) will clear this one up), will Ic suddenly drop to a rather small constant value, because that's what a BJT does when saturated?
thank you for all the replies guys!
(1)Saturation is not "triggered".
If you increase the value of Ib progressively,the curve of Ib versus Ic will slowly begin to level off,until the same percentage change in Ib which earlier caused a large change in Ic produces a very much smaller change in Ic.
Eventually a point is reached where the increase is so small,you may regard the device as saturated.
Your "symptoms" are in error-the value of Ic is at its maximum at saturation.
Look at your circuit,& imagine the transistor as a variable resistor,which I will call Rq1
Initially,the variable makes up a largish proportion of the series resistance RL+Rq1, & Ie = V/(RL+Rq1).
As you reduce the resistance of Rq1,this proportion becomes progressively less,until,its contribution is so minimal, that we can
say,RL+VRq1 is approximately = RLso Ie approx =V/RL.
Further reduction of Rq1 won't make much difference to Ie.
Saturation in this sense,is affected by the circuit constants,as well as device characteristics.
(2) If the transistor is biased as per your example& is driven by an AC signal,
If the drive signal's amplitude is great enough,it will drive the transistor into saturation on the positive 1/2 cycle,and to cut off on the negative 1/2 cycle,(assuming an NPN transistor),so the result is a distorted ("clipped") signal at the output .
The classic method of design of amplifiers is to draw a "loadline" on the characteristic curves,with one extremity being the current which would flow through the load with a short circuit in place of the device,& the other extremity being zero current (which would flow with an open circuit in place of the device).
If you have modelling software,you should be able to model the various conditions,or get some real transistors & play around with them!
(1)Saturation is not "triggered".
If you increase the value of Ib progressively,the curve of Ib versus Ic will slowly begin to level off,until the same percentage change in Ib which earlier caused a large change in Ic produces a very much smaller change in Ic.
Eventually a point is reached where the increase is so small,you may regard the device as saturated.
Your "symptoms" are in error-the value of Ic is at its maximum at saturation.
Look at your circuit,& imagine the transistor as a variable resistor,which I will call Rq1
Initially,the variable makes up a largish proportion of the series resistance RL+Rq1, & Ie = V/(RL+Rq1).
As you reduce the resistance of Rq1,this proportion becomes progressively less,until,its contribution is so minimal, that we can
say,RL+VRq1 is approximately = RLso Ie approx =V/RL.
Further reduction of Rq1 won't make much difference to Ie.
Saturation in this sense,is affected by the circuit constants,as well as device characteristics.
(2) If the transistor is biased as per your example& is driven by an AC signal,
If the drive signal's amplitude is great enough,it will drive the transistor into saturation on the positive 1/2 cycle,and to cut off on the negative 1/2 cycle,(assuming an NPN transistor),so the result is a distorted ("clipped") signal at the output .
The classic method of design of amplifiers is to draw a "loadline" on the characteristic curves,with one extremity being the current which would flow through the load with a short circuit in place of the device,& the other extremity being zero current (which would flow with an open circuit in place of the device).
If you have modelling software,you should be able to model the various conditions,or get some real transistors & play around with them!
i did, i'm trying to understand it now.
to the left of the red line, in the picture, is the saturation region, correct? so saturation is sliding down the shoulder of those curves, and so Vce and Ic will both get smaller, no matter what Ib is, yeah?
Sorry,you've lost me on that one!
Unfortunately in FET theory, the terminology is confusing and likely backwards from what you are used to. At low drain-source voltages, a MOSFET acts as a voltage variable resistor controlled by the gate. For a fixed gate voltage, the current is proportional to the drain voltage. This is the ohmic region. If you increase the source voltage, beyond a certain point the current does not increase, and this is called the saturation region. However, saturation here means that the current cannot be increased by further increasing the *source*, not the gate voltage.
Therefore, when a mosfet is used as a "fully on" switch, it is used in the ohmic region -- and data sheets specify their switching behavior in terms of an on resistance (Rds on). Confusingly, linear amplifiers are operated with their FETs in the saturation region.
Which only goes to show that you have to be careful and clear when using the word saturation, since there is potential for confusion.