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I am struggling to find out the Value of Ie and Ib here.

Ic I have taken to be Vcc/10k. If this is correct I could use Ic/alpha = Ie . I would like to know if this is the case but seems to easy to be the case?

Usually we would have Re to stabilise the heating effect and Ie would just be

Vb-0.7/Re.

I would appreciate the help.

The reason I am doing this is to try and find the Range of values for Vout based on different gain Values of beta.

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    \$\begingroup\$ Ic I have taken to be Vcc/10k Why? If that were true then the Vce of the transistor must be 0 V. Would that make sense? Instead of trying to find a solution in "one go", try to find it in steps. As the voltage at the Output supposed to vary a nice operating point for it would be Voutput = Vcc/2 so 5 V. Now use that to determine the currents and see if that can work with the given values for beta. If not, try a different Voutput and see what happens. \$\endgroup\$
    – Bimpelrekkie
    Jan 21, 2019 at 13:41
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    \$\begingroup\$ @Tyler I agree that simulators are useful but if OP starts using a simulator now he will never learn how to solve this by hand. The best way is to solve this by hand first and then confirm that solution with a simulator. \$\endgroup\$
    – Bimpelrekkie
    Jan 21, 2019 at 13:42
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    \$\begingroup\$ Why don't you write the KVL equation and solve for Ic and Vce? \$\endgroup\$
    – G36
    Jan 21, 2019 at 13:56
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    \$\begingroup\$ See transistor circuit solution. It solves this topology. \$\endgroup\$
    – jonk
    Jan 21, 2019 at 19:40
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    \$\begingroup\$ Of course, you have to assume \$\beta\$ and also \$V_\text{BE}\$. It's possible to solve the same situation, using instead the active mode Shockley equation simplification of Ebers-Moll. This assumes knowledge ot the saturation current and \$\beta\$, though. A minor improvement since the saturation current can be measured at one point and applied over a wide dynamic range. (Or go whole hog and use the full Ebers-Moll, or modified Gummel-Poon, or VBIC or MEXTRAN models, solving iteratively.) If you know the device FAB design params, you can do it with 3D integrals over the shape and design. \$\endgroup\$
    – jonk
    Jan 21, 2019 at 19:50

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Fred - you are mentioning an emitter resistor RE. You are right, such a resistor provides negative DC feedback and, thus, stabilizes the Q-point against temperature effects...AND ALSO AGAINST BETA-VARIATIONS.

And a similar effect is caused by the 470k resistor which also provides negative DC feedback (as long as the input signal to be amplified is coupled to the base through a capacitor or a suitable resistor). For your calculations you can start again with the assumption of VBE~0.7 volts.

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