LED Indication

LesJones

Joined Jan 8, 2017
3,097
I can see a few problems with your circuit.
1 As your supply voltage is 48 volts and the load resistance is 48 ohms the current will be almost 1 amp so you are exceeding the current rating of the 1N914 diodes.
2 With no load on the circuit the base of the transistor is effectively open circuit so your supply voltage of 48 volts exceeds the Vceo rating of 45 volts for the BC557 This will cause base leakage current which will turn the transistor on. Adding a resistor between D1 anode and D2 cathode would help the situation. The value of the resistor would have to be calculated to have a voltage drop across it of about 1.4 volts with the minimum load current required to turn the LED on. You could also use a transistor with a higher Vceo rating. I would still prefer to have the resistor to reduce the effect of leakage current in the transistor.

Les.
 

Thread Starter

mishra87

Joined Jan 17, 2016
906
I can see a few problems with your circuit.
1 As your supply voltage is 48 volts and the load resistance is 48 ohms the current will be almost 1 amp so you are exceeding the current rating of the 1N914 diodes.
2 With no load on the circuit the base of the transistor is effectively open circuit so your supply voltage of 48 volts exceeds the Vceo rating of 45 volts for the BC557 This will cause base leakage current which will turn the transistor on. Adding a resistor between D1 anode and D2 cathode would help the situation. The value of the resistor would have to be calculated to have a voltage drop across it of about 1.4 volts with the minimum load current required to turn the LED on. You could also use a transistor with a higher Vceo rating. I would still prefer to have the resistor to reduce the effect of leakage current in the transistor.

Les.
Hi Les,

Thanks for design consideration !
Actually i have not finalized the part number. i just wanted to make it working theoretically.
 
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