Or This is my design Vcc=10V, Vf(for LED)=1.5V. taking Vce=Ve=(10-1.5)/2=4.25V, Vbe=0.7V. Ic=If=Ie=50mA beta=200, Ib=Ic/beta=0.25mA. Re=Ve/Ie=85ohms. R2=Vb/Ib=19.8k . Vb=Vcc*R2/(R1+R2) voltage divider where R1=19.4k. if not correct need help from some one. Thanks pls disregard the previous post no link |
by nurafalalu
July 05, 2013 |
There are several problems. R1 & R1 are not connected to the transistor base. Hence, the base is not biased correctly. The signal input is 10V pk-pk but you only have a 10V Vcc supply. The positive part of the input voltage swing will be developed across the emitter resistor. The led will drop about 2V (Vd). The transistor will require a minimum of approx 0.25V Vce. Hence the max input swing to avoid clipping will be about Vcc-Vd-Vce. The base will have to be biased so that the whole of the input signal swing is above approx 0.5V (Vbe). |
by signality
July 06, 2013 |
Thank you signality. I corrected that and tried to rerun it is still clipping.Can you pls run it and see. |
by nurafalalu
July 06, 2013 |
Your public circuit: i.e. https://www.circuitlab.com/circuit/5j25n4/transmitter/#postsave_link_and_share is unchanged. |
by signality
July 06, 2013 |
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