# Digital Circuits Analysis

Does anyone understand where 2.5v , 10 v, and 5V ( highlighted in yellow) are coming from? I know that Vcc=5V, and the two resistors R1 and R2 are in parallel so Req=0.5Kohm, and the voltage divided by 2 which makes it 2.5V . How did they find IE=2.59 mA ? Also, if you can explain the obtained KVL too, it would be great. I can see it but still cannot wrap my head around it.Thanks

• The problem has a capacitor drawn on the output and mysterious red-ink person (your TA?) has drawn a voltage cell (battery). This makes me suspicious that mysterious red-ink person may not know what's going on. Commented Nov 29, 2015 at 22:53
• This is a mess! Commented Nov 29, 2015 at 22:54
• @ andy, I know , sorry about that. it is coming from lecture notes. Commented Nov 29, 2015 at 22:58
• @user65652 if you think my answer is complete and satisfying then please click on the tick mark ,so that we can close this. Commented Nov 30, 2015 at 11:39

First of all , when Q4 is ON and Q3 is OFF (so it is a open circuit ideally) , we can use thevenin equivalent at the emitter node of Q1 ,which gives you VTh= 2.5V and Rth =0.5k. under the assumption that R1 is connected to 5V

Now applying KVL at the input loop from 5V input to the ground node,we have

5 - (10K) Ib - Vbe - Ie (0.5K) -2.5 = 0

Rearranging,

1.8V = (10K) Ib + (0.5k) Ie

but Ie =(1 + beta) Ib;

So,

1.8V = [ (10k /51) + 0.5k ] Ie

which gives Ie = 1.8/0.696k =2.585 amps.

edit: and regarding 10v 2.5v and 5v confusion, think of it like this,

we have got resistive divider with supply voltage 5V at R1 which is in series with R2 giving us 2.5 volts at emitter node.

The 10v used in the circuit seems vague as the supply itself is 5V,so thats wrong i guess.