# Current Source

You are given a current source made of a NPN transistor, having $I_c=1 mA$ and $beta=100$. the purpose of the diodes is to create a constant voltage drop ($0.45 V$) so that $V_B=4(0.45V)=1.8 V$. design $R_E$ for $1 mA$ current through $R_L$ which is equal to $I_C$. Find the maximum value for $R_L$ with supply $V_{cc}=10V_{dc}$, $V_{c e}=0.2 V$ and $V_E=V_B-V_{BE}$.

Do the same for a PNP current source.

1.

$I_c =I_e*beta/(beta+1)$   that is $I_e =101/100*I_c=1.01 mA$

$I_e = U_e/Re =(U_b-U_{be})/Re$   that is $Re= (U_b-U_{be})/I_e =(1.8-0.65)/1.01 m =1.139 K$

$I_b =I_c/beta =1 m/100 =10 mu A$

$Ir1 =10*I_b =100 mu A =0.1 mA$

$R1 =(V_{cc}-1.8)/0.1 m=(10-1.8)/0.1 m =82 K$

$V_e =V_b-V_{be} =1.8-0.65 =1.15 V$

$R_{L max} = (V_{cc}-V_{ce} -V_{e})/I_c =(10-0.2-1.15)/1 m=8.65 K$

take $RL =0.6*RL_{max} =5.19 K$   standard value $R_L =5 K$

$V_{ce} = V_{cc}-I_c*R_L-V_e =10-1m*5 K -1.15 =3.85 V$

2.

A rule of electronic design is that a circuit with a NPN transistor is equivalent to a circuit with a (pair) PNP transistor if you reverse the power sources and reverse the polarity of diodes in circuit. This, since for this PNP circuit $beta =100$ (the same), diodes are reversed and $V_{dc} =-10 V$ (reversed), then all the Resistors values are the same as above.

$Rb=82 K$, $Re =1.139 K$ and $R_L =5.19 K$ and $V_{c e} =3.85 V$