It is nice to be able to design simple circuits, without having to rely on simulators all the time.
Firstly, any bipolar junction transistor, will always need a current flow into the base input to make it work properly.
Be it electron current flow with a PNP, or conventional current flow NPN.
Looking at the circuit it would appear to be half the supply voltage, would be at the center tap of both equal value resistors, which puts it at 5v.
Remove the connection between the transistor base and the voltage divider, and that is the voltage at the tap of the divider. [VCC x R2 /(R1 + R2)] = 5v.
Now with the transistor back in the circuit it drops to 2.13v.
THIS is VERY hard to determine without actually building and taking measurements, unless you know the exact parameters of the transistor being used.
HOWEVER you can learn alot from this circuit, about how a transistor is bilateral meaning, it will effect the input signal, as well as an output.
So lets do an analysis on this circuit, EVEN if this was not a simulator, you still breadboard the circuit, and take all the imperical measurements, to come to a logical analysis of the circuit.
So now these are the measurements you have come up with, (data).
From this data you now can analyse its behavior, AND how to design it so as to eliminate these bad characteristics.
Hopefully you have a good understanding of the basic laws (ohms, kirchoffs,ect...)
Lets start out and determine the loading affect the transistor has on the input voltage.
1. without transistor VR2 = 5v. current IR2 = 500uA.
2. VR1 = 5v. @ 500uA.
3. With transistor VR2 = 2.13v. @ 213uA
4. VR1 will then be (VCC - VR2) = 7.87v. @ 787uA.
5. Delta current through R1 goes from 500uA, to 787uA. = 287uA.
6. Delta IR2 goes from 500uA down to 213uA. = 287uA.
7. If the total current is now 787uA through R1 and 213uA OF IT flows through R2
that leaves (787uA - 213uA) = 574uA. left over to flow into the base of the transistor. = (IB)
That shows how the transistor affects the input signal.
Lets carry the analysis further.
Now a measurement across the emitter resistor determines a voltage of ~ 1.43v. = VR4.
That makes the voltae Vbe = to ( 2.13v. - 1.43v.) = 0.7v.
Now with VR4 measured then the emitter current will be (1.43v. / 1K ohms) = 1.43mA.
Now to solve for collector current we take the emitter current and subtract the base current we calculated earlier.
(1.43mA - 574uA) = 856uA.
From here we can take a quick side note and determine the Beta of this particulatr transistor, at this current value. (B = IC / IB) = ( 856uA / 574uA) = ~1.5
Continuing with the analysis, looks like you took a measurement at the collector as well. = (VC)
so (VC) = 1.44v.....
Hmmmmm if the emitter voltage is at 1.43v. and the collector voltage is at 1.44v. that means a voltage drop of 10mV. is across the transistor = (VCE).
Ok, now you took your measurements and have calculated currents and Beta, ect...
And you have your conclusion, that this design is not a good amplifier.
Now it is at this point where you apply your mathematic calculations, to redesign it so that you can determine the voltages and currents on paper, and get good results in an actual prototype.
Keeping your collector resistor in your design, the same value, you need to rework the value of the base and emitter voltage to get a more workable circuit.
From basic amplifier theory, you know that VC should be around half the supply voltage.
so with VC at 5v. that leaves 5v. to be dropped across the emitter resistor and the transistor.
The collector current is now [(VCC - VC) / R3] = 500uA. = (IC)
Lets use your original 1K ohms for R4. so VR4 will be (500uA. x 1K ohms) = 0.5v.
That puts a nice 4.5v. across the transistor, so it definately is not saturated.
From experiance you know that around 0.7v = Vbe.
So now you can calculate your voltage at the base (VB) so as to turn the transistor on.
(VB) = (VR4 + Vbe) = (0.5v. + 0.7v.) = 1.2v.
NOW a good rule of thumb is, if no input impedance is specified, then it is a good practice to make R2 to be around 10 - 20 times greater than R4. So as to not allow the transistor base current upset the divider, as it did in the previous circuit design.
So lets use your original 10K ohms for R2.
Now with R2 chosen we can now calculate the current through R2, by (VB / R2) = (1.2v. / 10K) = 120uA. = (IR2)
Now to calculate the value of R1 we need this equation. [(VCC - VB) / IR2] = [(10v. - 1.2v.) / 120uA] ~= 73K, so we will use a standard 75K or a 68K ohm resistor.
Now we take the measurements. (AND ADJUST VALUES AS NEEDED)
So you use your simulator to test out your designs, but get used to calculating first then let your simulator confirm your calcuilations, or point out any errors.
Designing a circuit is as much as prototyping and taking measurements and gathering data, as well as doing the initial calculations.