Hi. I have to find Re so that Ie becomes equal to 0.5 mA. can anyone help ?

 

Im looking for next weeks lottery numbers.. Any help?



Sorry, had to.

You are looking for a resistance to achieve a particular current.. At what voltage?

 

Nice one . well we have 2 DC voltage sources + if we remove the AC voltage source can't we find that resistor value ?

 

Have you learned ohms law yet?

Resistance depends on Voltage and Current.

R=I/V

You need 2 to find the third.

A 1 ohm resistor will give you 10A at 10V But 525A and 525V

Resistors "convert" voltage into current per se.

 

How can i find this Re with only these informations ?

 

Looks like a rather odd bias circuit there.
From what I gather the TR only turns on when
the AC signal is at a voltage that is 0.6V greater
than the voltage at the emitter.

Based on TR operating rules;

   ++
   /
+-{
   \
    -

or as is here;

    +
    /
- -{
    \
   --

I'm stumped as to why there are two negative
DC supplies at -15V each.

Ie at 0.5mA ?
Ie includes Ib and Ic.
In this case Ic also goes through RL.
So Ie = Ib + Ic + Irl?
IC is ~ Ie, usually and thus Ib is neglected.
But Irl, that could be significant.

Good luck with this one.

 

Since the two supplies are of equal and opposite polarity and connected in series, they may be replaced with a short.

This circuit has no DC voltage applied to it. Thus there is no value of Re that will produce any current.

 

Perhaps we could cut Halim a break and assume their orientation of the supply connections is incorrectly drawn. Also forget about any missing coupling capacitors & so forth.

The emitter leg supply is -15V.

Presumably Halim just wants the steady state (quiescent) bias conditions - so we could assume the AC signal input is 0V for the moment.

The steady-state (quiescent) emitter current would be found from

\(15-Vbe=IeRe+Ie\frac{Rb}{(1+\beta)}\)

or

\(Ie=\frac{(15-Vbe)}{Re+\frac{Rb}{(1+\beta)}}\)

If Rb, Vbe, β are known and Ie=0.5mA one would re-arrange the equation to solve for Re. Which might be a useful exercise for Halim.

 

Thanks all but t_n_k gave the closest answer

 

I'm still keen to find out the winning lottery numbers for the coming week... anyone?