# Find Re to get Ie = 0.5mA

Discussion in 'Homework Help' started by Halim, May 11, 2011.

1. ### Halim Thread Starter New Member

May 7, 2011
4
0
Hi. I have to find Re so that Ie becomes equal to 0.5 mA. can anyone help ?

2. ### retched AAC Fanatic!

Dec 5, 2009
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313
Im looking for next weeks lottery numbers.. Any help?

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

3. ### Halim Thread Starter New Member

May 7, 2011
4
0
Nice one . well we have 2 DC voltage sources + if we remove the AC voltage source can't we find that resistor value ?

4. ### retched AAC Fanatic!

Dec 5, 2009
5,201
313
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.

5. ### Halim Thread Starter New Member

May 7, 2011
4
0
How can i find this Re with only these informations ?

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6. ### dcarson7 New Member

Apr 28, 2011
7
1
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;
Code ( (Unknown Language)):
1.    ++
2.    /
3. +-{
4.    \
5.     -
or as is here;
Code ( (Unknown Language)):
1.     +
2.     /
3. - -{
4.     \
5.    --
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.

7. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,436
1,626
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.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
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.

Last edited: May 12, 2011
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9. ### Halim Thread Starter New Member

May 7, 2011
4
0
Thanks all but t_n_k gave the closest answer

10. ### nbw Member

May 8, 2011
36
10
I'm still keen to find out the winning lottery numbers for the coming week... anyone?

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