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

By   August 8, 2020

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 (IL)

(b) supply voltage (VI)

(c) Temperature (T)

Fig. 60. shows the basic block diagram of the voltage reguator, where

VI = unregulated DC voltage

VO = regulated DC voltage

Since, the output DC voltage VLO depends on the input unregulated DC voltage VI, load curreent IL and the temperature T, then the change VO in output voltage of a power supply can be expressed as follows

VO = VO (VI, IL, T)

Take partial derivative of VO, we get

VO = VO / VI VI + VO/IL  IL + V0/T =T

V0 = SV  VI + RL  IL + ST T

SV = VO / VI /IL = 0 / T = 0

R0 = V0 / IL / VI = 0 / T = 0

ST = VO /T   / VI = 0 / IL = 0

Sv gives variation in output voltage only due to unregulated DC voltage. Ro gives the output voltage variation only due to load current. ST 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 CL 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 RE = 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

IB + IC = IE

So,                   IOUT = IC1 + IE1 B /1 + B = IE2 B / 1 + B

Adding currents at the collector of Q1, we obtain

IREF = (B / 1 + B + 2 / 1 + B)IE = B + 2 /1 + B IE = 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 (RTH) is limited by the r0 (1/hoe) of the transistor. that is

RTH = TO = VA / IC~ VA / Iref

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 ,

VBE1 – VBE2 -IE2R2 = 0 …………………..(1)

For a forward biased base-emitter junction diode, the emitter current is given by

IE = I0eVBE/hvt

since,               IE ~ IC = IC and n = 1

IC = I0eVBE/VT

and                   VBE = VT In  (IC / I0) …………………….(2)

Substituting VBE1 and VBE2 from eqs. (1) to (2), we get

VT In (IC1 / I0) – VT In (IC2 / I0) – IE2R2 = 0 ………………….(3)

We have assumed that both the transistors are matched, so that ICO, B and VT are the same for both the transistors.

Thus,

VT In (IC2/IC1) = IE2R2 ~ IC2R2

Hence,          R2 = VT/ IC2 In (IC1 / IC2) ………………………………(4)

where,            IC1 = VCC – VBE / R1 ………………………….(5)

Example 17. Design a widlar current source to provide a constant current source of 3uA with VCC = 12 V, R1 = 50 K  B = 100 and VBE = 0.7 V.

Sol. The circuit is given in figure above. applying KVL to the Q1 transistor we get,

IC1 = IREF = 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 IC1 is the base current of Q2.

IE2 = (B + 1) IB2 = IC3 ……………………………..(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 IC2 is approximately equal to IREF. Applying kirchhoff’s current law at the emitter of Q2 yields

IE2 = IC3 + IB3 + IB1 ………………………..(10)

Using the relationship between collector and base currents

IE2 = IC2 ( 1 + 1/B) + IC1 /B …………………(11)

Since, all three transistors are matched, VBE1 = VBE2 = VBE3 and B1 = B2 = B3

With identical transistors, current in the feedback path splits equally between the bases of Q1 and Q3 leading so that IB1 = IB3 and therefore, IC1 = IC3

IE2 = IC3 (1 + 2/B) …………………….(12)

The collector current of Q2 is

IC2 = IE2B / 1 + B = IC3 (1 + 2/B)B /1 +B …………………………………(13)

Solving for IC2 yields

IC3 = IC2 (1 + B) / B (1 + 2/B) = IC2 1 + B/2 + B ………………………….(14)

Summing currents at the base of Q2

IC1 = IREF – IC2 /B ………………………(15)

IC2 = (IREF – IC1) ………………………….(16)

Since, IC1 = IC3, we substitute IC3 to obtain

IC2 = BIREF – B(1 + B) / B + 2 …………………….(17)

and solving for IC2

IC2 = B2 + 2B / B2 + 2B + 2 = IREF

= [1 – 2 /B2 + 2B + 2]IREF  ………………………………….(18)

Eq. 10 shows that B has little upon IC2 since, for reasonable values of B.

2 / B2 + 2B + 2 <<1 ………………………….(19)

Therefore,     IC2 = IREF.

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, VBE = 0.715 V.

Sol. since, all transistor are identical, therefore VBE voltage drop will be same.

I2 = 10 – 0.715 / 4.7 K = 1.976 mA

Let IB be the base current of each transistor and IC be the collector current of Q1 and Q2.

Therefore,    2IC + 3I8 = I2

2 x IB + 3IB = 1.976 mA

IB = 9.73 A

IE = (1 + B)IB

= 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

IE = IS(eqve/kt – 1) ………………(1)

Since,  IC = IE for a ground base transistor,

IC = IS (eqve/kt – 1)………………….(2)

IS = emitter saturation current ~ 10-13 A

K = Boltzmann ‘s constant

T = absolute temperature (in kelvin)

Therefore,       IC / IS = (eqve/kt – 1) ………………………(3)

or                   eqve/kt = IC /IS + 1 = IC/ IS

[as IS ~ 10-13 A, IC >> IS]

Taking natural log on both sides, we get

VE = KT /q IN (IC / IS)   ………………………(4)

From figure,   IC = VI/R1     OR   VE = – V0

So,                  V0 = – KT /q In (Vi / R1IS)

= – KT /q    in  (VI / VREF) …………………(5)

where,                         VREF = R1IS

The output voltage is thus proportional to the logarithm of input voltage.

Antilog Amplifier

The input voltage VI for the antilog-amplifier is fed into the temperature compensating voltage divider R2 and RTC RTC 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

VQ1 BE = KT /q In (V0 / R1IS)    ……………………..(i)

and               VQ2 BE = KT /q In (VREF / R1 + IS) ………………..(ii)

since, the base of Q1 is tied to ground, we get

VA = – VQ1 BE = – KT /q In (VO / R1IS) ……………………….(iii)

The base voltage VB of Q2 is

VB = [RTC / R2 + RTC] VI ………………………(1)

The voltage at the emitter of Q2 is

VQ1 BE = VB + VQ2 BE

Or               VQ2 BE = [RTC / R2 + RTC] VI – KT/q in (VREF / R1 + IS) ……………………(2)

But the emitter voltage of Q2 is VA, that is

VA = VQ2 BE ………………………..(3)

Or   – KT /q    In V0 / R1IS = RTC/R2 + RTC   VI – KT/q  In VREF /R1IS ……………………..(4)

or    RTC / R2 + RTC VI = – KT /q In (In  V0 / R1IS – In VREF / R1IS)

Or  – q / KT  RTC / R2 + RTC   VI = In (V0 / VREF) …………………….(5)

Changing natural log. i.e., into log10 using equation log10 x = 0.4343 in x, we get

– 0.4343 (q / kt) (RTC/ R2 + RTC)VI

= 0.4343 x in (V0 / VREF ) ………………(6)

– K’VI = log10 (V0 / VREF)

Or                    V0 / VREF = 10-K’Vi

or                     V0 = VREF (10-K’Vi)……………………….(7)

where,              K’ = 0.4343 (q/KT) (RTC / R2 + RTC) ………………….(8)

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