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Design a $0.5$ A current source using IC 7805 , for $\mathrm{RL}=10 \mathrm{ohms}$.?
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Solution:

enter image description here

3 terminal fixed voltage regulators can be used as a current source.

The above figure shows the circuit where 7805 has been wired to supply a current of $0.5 \mathrm{~A}$ to $10 \Omega$ load, kilowatt load.

$ I_0=I R_1+I_Q $

where, $I Q$ is the Quiseent current and is about $4.2 \mathrm{~mA}$ for 7805 .

$ I_0=\frac{V_R}{R,}+I_Q $

$ \text { since } \begin{aligned} I_0 &=0.5 \mathrm{~A} \\ \frac{V_R}{R_1} & \simeq 0.5 \mathrm{~A} \quad\left[I_Q \ll I_0\right] \end{aligned}\\ $

Also, $V_R=5 v$ [Voltage between terminal Land] so, value of $R_{\text {, required is }}$

$ \begin{aligned} &R_1=\frac{5 \mathrm{~V}}{0.5 \mathrm{~A}} \\ &R_1=10 \Omega \end{aligned} $

Thus choose $R_1=10 \Omega$ to deliver $0.5 \mathrm{~A}$ current to a load of $10 \Omega$

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