# Return to Answer

 2 change the circuit for better performance edited Nov 8 '18 at 19:47 M KS 30118 If you pass 5.5mA from a 1K ohm load, then Vout is 5.5mA * 1K =5.5volt, and circuit saturate before reaching this load. your circuit works with 5 volt supply and Vcc = R1 * 5.5mA + V_ce + R2 * 5.5mA for 5.5mA output and RL=20K, you need a supply voltage 20K*5.5mA+..+..> 100V You need a supply voltage bigger than 100V, No opamp can handle this voltage and this circuit not works a for 20k load. You can use this circuit instead: refference: http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm opamp power is 5V, Resense = 100500 ohm for 5.5 mA output -> vin(VDC) 02.5575 volt VCC = 20k * 5.5ma + 100500 * 5.5ma + V_CE ~ 120 volt is ok and Q1 should handle 100 volt (minimum (120-110112.5575) volt!) If you pass 5.5mA from a 1K ohm load, then Vout is 5.5mA * 1K =5.5volt, and circuit saturate before reaching this load. your circuit works with 5 volt supply and Vcc = R1 * 5.5mA + V_ce + R2 * 5.5mA for 5.5mA output and RL=20K, you need a supply voltage 20K*5.5mA+..+..> 100V You need a supply voltage bigger than 100V, No opamp can handle this voltage and this circuit not works a for 20k load. You can use this circuit instead: refference: http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm opamp power is 5V, Resense = 100 ohm for 5.5 mA output -> vin(VDC) 0.55 volt VCC = 20k * 5.5ma + 100 * 5.5ma + V_CE ~ 120 volt is ok and Q1 should handle 100 volt (minimum (120-110.55) volt!) If you pass 5.5mA from a 1K ohm load, then Vout is 5.5mA * 1K =5.5volt, and circuit saturate before reaching this load. your circuit works with 5 volt supply and Vcc = R1 * 5.5mA + V_ce + R2 * 5.5mA for 5.5mA output and RL=20K, you need a supply voltage 20K*5.5mA+..+..> 100V You need a supply voltage bigger than 100V, No opamp can handle this voltage and this circuit not works a for 20k load. You can use this circuit instead: refference: http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm opamp power is 5V, Resense = 500 ohm for 5.5 mA output -> vin(VDC) 2.75 volt VCC = 20k * 5.5ma + 500 * 5.5ma + V_CE ~ 120 volt is ok and Q1 should handle 100 volt (minimum (120-112.75) volt!) 1 answered Nov 8 '18 at 19:26 M KS 30118 If you pass 5.5mA from a 1K ohm load, then Vout is 5.5mA * 1K =5.5volt, and circuit saturate before reaching this load. your circuit works with 5 volt supply and Vcc = R1 * 5.5mA + V_ce + R2 * 5.5mA for 5.5mA output and RL=20K, you need a supply voltage 20K*5.5mA+..+..> 100V You need a supply voltage bigger than 100V, No opamp can handle this voltage and this circuit not works a for 20k load. You can use this circuit instead: refference: http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm opamp power is 5V, Resense = 100 ohm for 5.5 mA output -> vin(VDC) 0.55 volt VCC = 20k * 5.5ma + 100 * 5.5ma + V_CE ~ 120 volt is ok and Q1 should handle 100 volt (minimum (120-110.55) volt!)