Question

Determine the range over which the output voltage can be varied in LM 317, if R1=2401 and R2 = 5K potentiometer.
Assume ladj = 50PA.​

Answer 1

1. $1K\Omega=1000\Omega$

2. $1pA=10^{-12} A$

3. The output voltage is calculated using the formula,

$V_{out} = 1.25 \left [ 1+\frac{R_{2}}{R_{1}} \right ]+ I_{adj}\times R_{2}$

Given data,

For $LM317$,

$R_{1} =2401\Omega$

$R_{2} =5K\Omega$

Assume, $I_{adj} =50pA$

1. when $R_{2} =0\Omega$,

$V_{out} = 1.25 \left [ 1+\frac{0}{R_{1}} \right ]+ I_{adj}\times 0=1.25+0=1.25V$

2. when $R_{2} =5K\Omega$,

$V_{out} = 1.25 \left [ 1+\frac{5\times10^{3} }{2401} \right ]+ 50\times 10^{-12} \times 5\times 10^{3}$$V_{out} =1.25\left [ 1+2.082 \right ]+250\times 10^{-9}$

$V_{out} = 3.8525+250\times 10^{-9} =3.85250025\approx3.85V$

Hence, the range is $1.25V$ to $3.85 V$.

Answer 2

1.

$1kΩ = 1000Ω$

2. $1pA = {10}^{ - 12} A$

3. The output voltage is calculated using the formula,

$Vout=1.25[1+ \frac{R2}{R1} ]+Iadj×R2</p><p>$

Given data,

For LM 317,

R1 = 2401Ω

R2 = 5kΩ

Assume, I adj = 50pA

1. when , R2 = 0Ω

$v \: out \: = 1.25(1 + \frac{0}{R1} ) + Iadj \times 0 = 1.25$

2. when , R2 = 5kΩ

$Vout = 1.25(1 + \frac{5 \times {10}^{3} }{2401}) + 50 \times {10}^{ - 12} \times 5 \times {10}^{3} = 3.85V$

Hence, the range is 1.25V to 3.85V