Widlar Current Source | what is widlar current source using mosfet , output resistance

what is widlar current source using mosfet , output resistance ? derivation , calculation .
Voltage Regulators
An unregulated power supply consists of a transformer (step down), a rectifier and a filter. these power suppliers are not good for some applications, where constant voltage is required irrespective of external disturbances. the main disturbances are

1. As the load current varies, the output voltage also varies because of its poor regulation.
2. The DC output voltage varies directly with AC input supply. the input voltage may vary over a wide range, thus DC voltage also changes.
3. The DC output voltage varies with the temperature if semiconductor devices are used.

An electronic voltage regulator is essentially a controller used along with unregulated power supply to stabillize the output DC voltage against three major disturbances
(a) Load current (I_L)
(b) supply voltage (V_I)
(c) Temperature (T)
Fig. 60. shows the basic block diagram of the voltage reguator, where
V_I = unregulated DC voltage
V_O = regulated DC voltage
Since, the output DC voltage V_LO depends on the input unregulated DC voltage V_I, load curreent I_L and the temperature T, then the change V_O in output voltage of a power supply can be expressed as follows
V_O = V_O (V_I, I_L, T)
Take partial derivative of V_O, we get
V_O = V_O / V_I V_I + V_O/I_L  I_L + V₀/T =T
V₀ = S_V  V_I + R_L  I_L + S_T T
S_V = V_O / V_I /I_L = 0 / T = 0
R₀ = V₀ / I_L / V_I = 0 / T = 0
S_T = V_O /T   / V_I = 0 / I_L = 0
S_v gives variation in output voltage only due to unregulated DC voltage. R_o gives the output voltage variation only due to load current. S_T given the variation in output voltage only due to temperture.
The smaller the value of the three coefficients, the better the regulations of power supply. the input voltage variation is either due to input supply fluctuations or presence of ripples due to inadequate filtering.
Voltage Regulator
A voltage regulator is a device desingned to maintain the output voltage of power supply nearly constant. it can be regarded as a closed loop system because it monitors the output voltage and generates the control signal to increase or decrease the supply voltage as necessary to compensate for any change in the output voltage. thus, the purpose of voltage regulator is to eliminate any output voltage variation that might occur because of changes in load, changes in supply voltage or changes in temperature.
Zener Voltage Regulator
The regulator power supply may use zener diode as the voltage controling device as shown in fig. 61. the output voltage is determined by the reverse breakdown voltage of the zener diode. this is nearly constant for a wide range of currents. the load voltage can be maintained constant by controllin the current through zener.
The zener diode regulator has limitations of range. the load current range for which regulation is maintained, is the difference between maximum allowable zener current and minimum current required for the zener to operate in breakdown region. for example, if zener diode requires a minimum current of 10 mA and is limited to a maximum of 1 A (to prevent excessive dissipation). the range is 1 – 0.01 = 0.99 A. if the load current variation exceeds 0.99 A, regulation may be lost.
Emitter Follower Regulator
To obtain better voltage regulation in shunt regulator, the zener diode can be connected to the base ciruit of a power transistor as shown in fig. 62. this amplfies the zener current range. it is also known as emitter follower regulation.
This configuration reduces the current flow in the diode. the power transisor used in this configuration is known as pass transistor. the purpose of C_L is to ensure that the variations in one of the regulated power supply loads will not be fed to the other loads. that is the capacitor effectively shorts out high frequency variations.
Because of the current amplifying property of the transistor, the current in the zener diode is small. hence, there is little voltage drop across the diode resistance and the zener approximates an ideal constant voltage source.
A Simple Current Source
The simple two transistors current source shown in fig. 63 is commonly used in ICs.
A reference current is the input to a transistor connected as a diode. the voltage across this transistor drives the second transistor, where R_E = 0, Since, the circuit has only one resistor, it can be easily fabricated on an IC chip.
The disadvantge of this circuit is that the refernce current is approximately equal to the current source. in this circuit, Q₂ is in linear mode, since the collector voltage (output) is higher than the base voltage. the transistors Q₁ and Q₂ are identical devices fabricated on the same IC chip. the emitter currents are equal, since the transistors are matched and emitters and bases are in parallel. if we add the currents of Q₂, we obtain
I_B + I_C = I_E
So,                   I_OUT = I_C1 + I_E1 B /1 + B = I_E2 B / 1 + B
Adding currents at the collector of Q₁, we obtain
I_REF = (B / 1 + B + 2 / 1 + B)I_E = B + 2 /1 + B I_E = I₀
If B is large, the current gain is approximately unity and the current mirror has reproduced the input current. one disadvantage of this current source is that its thevenin resistance (R_TH) is limited by the r₀ (1/h_oe) of the transistor. that is
R_TH = T_O = V_A / I_C~ V_A / I_ref
Widlar Current Source
Large resistors are often required to maintain small currents of the order of few micro ampere and these large resistors occupy correspondingly large areas on the IC chip. it is therefore, desirable to replace these large resistors with current sources. one such device is the widlar current source as shown in fig. 64.
The two transistors are assumed perfectly matched. for the base circuit ,
V_BE1 – V_BE2 -I_E2R₂ = 0 …………………..(1)
For a forward biased base-emitter junction diode, the emitter current is given by
I_E = I₀e^VBE/^hvt
since,               I_E ~ I_C = I_C and n = 1
I_C = I₀e^VBE/V_T
and                   V_BE = V_T In  (I_C / I₀) …………………….(2)
Substituting V_BE1 and V_BE2 from eqs. (1) to (2), we get
V_T In (I_C1 / I₀) – V_T In (I_C2 / I₀) – I_E2R₂ = 0 ………………….(3)
We have assumed that both the transistors are matched, so that I_CO, B and V_T are the same for both the transistors.
Thus,
V_T In (I_C2/I_C1) = I_E2R₂ ~ I_C2R₂
Hence,          R₂ = V_T/ I_C2 In (I_C1 / I_C2) ………………………………(4)
where,            I_C1 = V_CC – V_BE / R₁ ………………………….(5)
Example 17. Design a widlar current source to provide a constant current source of 3uA with V_CC = 12 V, R₁ = 50 K  B = 100 and V_BE = 0.7 V.
Sol. The circuit is given in figure above. applying KVL to the Q₁ transistor we get,
I_C1 = I_REF = 12 – 0.7 / 5 x 10³ = 0.226 mA
using the eq. (4) we can calculate R₂
3 x 10^-8R₂ = 0.025 in (2.26 x 10^-4 / 3 x 10^-6)
or             R₂ = 36 K
Wilson Current Source
An other current source transistor configuration that provides a very large parallel resistance is the wilson current source which uses three transistors and provides this capability and the output is almost independent of the internal transistor characteristics. the wilson current source is shown in fig. 65, uses the negative feedback provided by Q₃ to raise the output impedance.
The difference between the reference current and I_C1 is the base current of Q₂.
I_E2 = (B + 1) I_B2 = I_C3 ……………………………..(9)
Since, the base of Q₁ is connected to the base of Q₃, the currents in Q₁ are approximately independent of the voltage of the collector of Q₂. As such the collector current of Q₂ remains almost constant providing high output impedance.
Let us now see that I_C2 is approximately equal to I_REF. Applying kirchhoff’s current law at the emitter of Q₂ yields
I_E2 = I_C3 + I_B3 + I_B1 ………………………..(10)
Using the relationship between collector and base currents
I_E2 = I_C2 ( 1 + 1/B) + I_C1 /B …………………(11)
Since, all three transistors are matched, V_BE1 = V_BE2 = V_BE3 and B₁ = B₂ = B₃
With identical transistors, current in the feedback path splits equally between the bases of Q₁ and Q₃ leading so that I_B1 = I_B3 and therefore, I_C1 = I_C3
I_E2 = I_C3 (1 + 2/B) …………………….(12)
The collector current of Q₂ is
I_C2 = I_E2B / 1 + B = I_C3 (1 + 2/B)B /1 +B …………………………………(13)
Solving for I_C2 yields
I_C3 = I_C2 (1 + B) / B (1 + 2/B) = I_C2 1 + B/2 + B ………………………….(14)
Summing currents at the base of Q₂
I_C1 = I_REF – I_C2 /B ………………………(15)
I_C2 = (I_REF – I_C1) ………………………….(16)
Since, I_C1 = I_C3, we substitute I_C3 to obtain
I_C2 = BI_REF – B(1 + B) / B + 2 …………………….(17)
and solving for I_C2
I_C2 = B² + 2B / B² + 2B + 2 = I^REF
= [1 – 2 /B² + 2B + 2]I_REF  ………………………………….(18)
Eq. 10 shows that B has little upon I_C2 since, for reasonable values of B.
2 / B² + 2B + 2 <<1 ………………………….(19)
Therefore,     I_C2 = I_REF.
Multiple Current Sources using Current Mirrors
A number of current sources can be obtained from a single reference voltage. if the current is approixmately the same as the reference voltage, the simple current source can be used as shown in fig. 66 for Q₂ and Q₃.
Notice that Q₄ has an emitter resistance, which makes the current source a widlar current source. thus, the amount of current delivered by this source can be determined by the size of the emitter resistor. this type of cicuit is useful in integrated circuit ships as the one reference circuit can be used to develop current sources throughout the chip. when using the widlar circuit, the currents can be different from the reference current.
The errors in base current, however, do accumulate when multiple outputs are used and the current gain tends to deviate from unity. in these types of circuits, lateral transistors can be used, since it is not important that b be large. lateral transistor usually have a B of approximately 20. which is more than adequate for current sources.
Example 18. For the circuit shown in figure below determine the emitter current in transistor Q₃ given that B = 100, V_BE = 0.715 V.
Sol. since, all transistor are identical, therefore V_BE voltage drop will be same.
I₂ = 10 – 0.715 / 4.7 K = 1.976 mA
Let I_B be the base current of each transistor and I_C be the collector current of Q₁ and Q₂.
Therefore,    2I_C + 3I₈ = I₂
2 x I_B + 3I_B = 1.976 mA
I_B = 9.73 A
I_E = (1 + B)I_B
= 0.983 mA
Sample and Hold Circuit
A sample and hold circuit samples an input signal and holds on to its last sampled value until the input is sampled again. this type of circuit is very useful in digital interfacing and analog to digital and pulse code moduultion systems.
Log and Antilog Amplifier
Log Amplifier
A grounded base transistor is placed in the feedback path. since, the collector is held at virtual ground and the base is also grounded, the transistor’s voltage-current relationship becomes that of a diode and is given by
I_E = I_S(e^qv_e/kt – 1) ………………(1)
Since,  I_C = I_E for a ground base transistor,
I_C = I_S (e^qv_e/kt – 1)………………….(2)
I_S = emitter saturation current ~ 10^-13 A
K = Boltzmann ‘s constant
T = absolute temperature (in kelvin)
Therefore,       I_C / I_S = (e^qv_e/kt – 1) ………………………(3)
or                   e^qv_e/kt = I_C /I_S + 1 = I_C/ I_S
[as I_S ~ 10^-13 A, I_C >> I_S]
Taking natural log on both sides, we get
V_E = KT /q IN (I_C / I_S)   ………………………(4)
From figure,   I_C = V_I/R₁     OR   V_E = – V₀
So,                  V₀ = – KT /q In (V_i / R₁I_S)
= – KT /q    in  (V_I / V_REF) …………………(5)
where,                         V_REF = R₁I_S
The output voltage is thus proportional to the logarithm of input voltage.
Antilog Amplifier
The input voltage V_I for the antilog-amplifier is fed into the temperature compensating voltage divider R₂ and R_TC R_TC and then to the base of Q₂. the output V₀ of the antilog-amplifier is feedback to the inverting input of A₁ through the resistor R₁. the base to emitter voltage of transistors Q₁ and Q₂ can be written as
V_{Q1 BE} = KT /q In (V₀ / R₁I_S)    ……………………..(i)
and               V_{Q2 BE} = KT /q In (V_REF / R₁ + I_S) ………………..(ii)
since, the base of Q₁ is tied to ground, we get
V_A = – V_{Q1 BE} = – KT /q In (V_O / R₁I_S) ……………………….(iii)
The base voltage V_B of Q₂ is
V_B = [R_TC / R₂ + R_TC] V_I ………………………(1)
The voltage at the emitter of Q₂ is
V_{Q1 BE} = V_B + V_{Q2 BE}
Or               V_{Q2 BE} = [R_TC / R₂ + R_TC] V_I – KT/q in (V_REF / R₁ + I_S) ……………………(2)
But the emitter voltage of Q₂ is V_A, that is
V_A = V_{Q2 BE} ………………………..(3)
Or   – KT /q    In V₀ / R₁I_S = R_TC/R₂ + R_TC   V_I – KT/q  In V_REF /R₁I_S ……………………..(4)
or    R_TC / R₂ + R_TC V_I = – KT /q In (In  V₀ / R₁I_S – In V_REF / R₁I_S)
Or  – q / KT  R_TC / R₂ + R_TC   V_I = In (V₀ / V_REF) …………………….(5)
Changing natural log. i.e., into log₁₀ using equation log₁₀ x = 0.4343 in x, we get
– 0.4343 (q / kt) (R_TC/ R₂ + R_TC)V_I
= 0.4343 x in (V₀ / V_REF ) ………………(6)
– K’V_I = log₁₀ (V₀ / V_REF)
Or                    V₀ / V_REF = 10^-K’Vi
or                     V₀ = V_REF (10^-K’Vi)……………………….(7)
where,              K’ = 0.4343 (q/KT) (R_TC / R₂ + R_TC) ………………….(8)