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Transcript
```Physics 120 - Prof. David Kleinfeld - 2017
Notes on n-channel JFETs in the Active region
1. Review of basics
1. IG = 0
2. ID = IS
In active region, device characteristics are defined by1:
3. VGS(off) ≤ VGS ≤ 0
4. VDS > VGS - VGS(off); recall that both VGS and VGS(off) are negative
5. ID is function of VGS, with ID =
IDSS
2
VGS
( )
⎡ V − VGS off ⎤
⎦
off ⎣ GS
( )
2
This implies ID = IDSS for VGS = 0.
6. ID is independent of VDS (ideal current source)
1
The turn-off gate-to-source voltage VGS(off) has a number of aliases, such as threshold voltage or pinch-off
voltage, denoted VGS(off) = VT = VTH = VP = VPO = VP0 The active region is also called the "saturation region" or
"pentode region", while the Ohmic region is also called the "linear region" or the "triode region". Sigh.
For small changes in gate voltage, we can calculate the changes in source or drain current.
The constant of proportionality is referred to as the transconductance, denoted gm, where
gm =
dIS
2I
dID
=
= 2 DSS ⎡⎣ VGS − VGS off ⎤⎦
dVGS
dVGS
VGS off
( )
( )
so that ΔIS = gm ΔVGS. We will see later that the transconductance plays a role analogous to β
with bipolar junction transistors, but is not a constant, i.e., it depends of VDS!
2. Fixed current source
Let's now consider the worlds simplest current source.
Here VGS = 0, so the current is forced to be IDS = IDSS. This is maintained so long as the load
line can maintain a value of VDS in the active region, i.e., VDS > - VGS(off).
The load line is given by writing Kirchoff's rule for voltage drops and ignoring the
transconductance 1/gm:
0 = -VDD +IDS RLoad + VDS
to yield the load line IDS =
VDD − VDS
.
This must intercept IDS = IDSS in the active region. Here, we slide along the (flat) line of ID = IDSS
so long as VDS > - VGS(off).
To find the maximum load resistance that keeps us in the active zone, we first rewrite the load
line as an expression for VDS, i.e.,
VDS = VDD - IDSS RLoad
Noting that VDS > - VGS(off), this implies
-VGS(off) > VDD - IDSS RLoad
or
( )
VDD + VGS off
IDSS
This simple device suffers only from having a value of ID that is not adjustable! Yet for
example, the (2N5485), has IDSS = 8 mA, enough to drive a typical LED with the effective
resistance of RL ~ 100 Ω.
3. Improved current source
A more sophisticated source uses a resistor between the source and ground to determine IDS.
Here we have VG = 0 but VGS < 0, since the gate is grounded. The loop equation
encompassing the gate and source satisfies (ignoring the transconductance term 1/gm):
0 = -VG + VGS + IDS RS
or, with VG = 0, IDS = −
VGS
.
RS
We need to choose a value of Rs to fix ID. The second equation that relates IDS and VGS is the
constitutive equation
IDS =
IDSS
⎡ V − VGS
2
VGS
off ⎣ GS
( )
⎛
VGS ⎞
off ⎤⎦ = IDSS ⎜ 1−
⎟
⎝ VGS off ⎠
( )
2
2
( )
We are free to pick a desired quiescent current, denoted IDSQ, with IDSQ < IDSS. Then the
required value of RS is found by substituting VGS = -IDSQRS into the constitutive equation, i.e.,
2
IDSQ = IDSS
( )
⎛
−VGS off ⎛
IDQ ⎞
IDSQ RS ⎞
1−
1+
.
Thus
R
=
S
⎜
⎟
⎜
⎟
IDQ
IDSS ⎠
VGS off ⎠
⎝
⎝
.
( )
The load line is given by writing Kirchoff's rule for voltage drops in the outer loop and ignoring
the transconductance 1/gm:
0 = -VDD + IDSQ RLoad + VDS + IDSQ RS.
We first consider the bounds of RLoad. The load-line is rearranged to
VDS = VDD - IDSQ RLoad
- IDSQ RS.
The requirement that VDS > VGS - VGS(off) in the active region leads to
VGS - VGS(off) < VDD - IDSQ RLoad
- IDS RS.
But VGS = - IDSQ RS from our analysis of the inner loop. Thus
- VGS(off) < VDD - IDSQ RLoad or
( )
VDD + VGS off
IDSQ
.
We return to the load line equation, which is reordered as
The load line for ID versus VDS is found by computing the voltage drops along the loop, i.e.,
IDS =
VDD − VDS
.
The value of VDSQ varies for IDSQ fixed as RLoad varies.
The FET current sources are independent of fluctuations in the power supply voltage and
largely independent of gm.
As an example relevant to the laboratory exercise with a 2N5485, the choice IDSQ = 0.4 mA
with IDSS = 8 mA and VGS(off) = - 3 V, we find RS = 5.8 kΩ. We use the closest value 5 %
resistor at 5.6 kΩ.
4. Source follower
The analysis is over two loops, one to define Rs and the other to define the load line.
We consider the bottom loop to relate Vin and ID:
0 = - Vin +VGS + RS IS
so ID =
Vin − VGS
RS
.
We consider the top loop to relate Vout and ID:
0 = - VDD +VDS + RS ID
so ID =
VDD − VDS
V
= Out
RS
RS
and we get
Vout = ID RS = Vin - VGS.
We see that the output follows the input with the addition of a term VGS. Recall that VGS < 0 so
the offset is positive.
The value of VGS is found from the intersection of the load line
ID =
Vin − VGS
RS
and the constitutive equation
ID = IDSS
2
⎡ VGS − VGS off ⎤
⎢
⎥
2
VGS
off
⎢⎣
⎥⎦
.
( )
( )
The critical issue is that VGS is not constant but depends on Vin! This leads to a nonlinearity,
yet is minimal for the choice Vin << VGS(off), as seen by solving for VGS, i.e.,
⎡
⎤
2
⎡ VGS off ⎤
⎢
VGS off
Vin − VGS off ⎥
VGS = VGS off ⎢1 −
+ ⎢
⎥ +
⎥
2 RSIDSS
RSIDSS
⎢⎣ 2 RSIDSS ⎥⎦
⎢
⎥
⎢⎣
⎥⎦
( )
( )
( )
( )
We'll next see how to cancel the VGS term, but before then we consider the inclusion of a
nonzero value of 1/gm as a resistance from the gate to the source. The loop equation is:
0 = - Vin +VGS + (1/gm) IS + RS IS
so ID =
Vin − VGS
RS + 1 gm
and Vout = ID RS =
gmRS
V − VGS
1+ gmRS in
(
)
.
The key is to maintain Rsgm >> 1 or Rs << 1/gm. This is critical, as gm is also a function of VGS.
4. Improved voltage follower
An improved follower may be built in which the offset voltage VGS is minimized. We use a
current source to define the current through Rs, as show below.
Here we may write an expression for the equilibrium current (ignoring gm):
0 = - Vin + VGS (Q1) + IDQ R1 + Vout
But we previously solved for the self limiting current source corresponding to the lower JFET,
V Q
or IDQ = − GS 2 .
R2
( )
Then
0 = - Vin + VGS (Q1) −
( ) R1 + Vout
VGS Q2
R2
For R1 = R2 and matched JFETs, i.e., VGS (Q1) = VGS (Q2), the output voltage is exactly the
input voltage and we have a perfect follower with a very large input impedance.
"Matched" means that IDSS and VGS(off) are the same for each of the two FETs, so that VGS
(Q1) = VGS (Q2) since the same current IDS flows through each FET. In fact, one can purchase
FETs that are manufactured as pairs, on a common substrate, for this purpose.
```
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