Could someone point me in the right direcrion on how to work out the input resistance in the circuit

Now I know the 1st thing to work out will be VB but am strugglig as on all the circuits I have worked out there has always been an R1 And R2 coming of the base of Q1 so it would be VB = (R2/R1+R2) x VCC

 

Your signal will be going into Q1 so the input will see 220 ohms times the Beta of the transistor. Since there are no biasing resistors there is only one path for the signal and that is Q1's base into its emitter. The beta can vary per-transistor and with temperature but I would expect it to be between 100 and 200, so about 22K to 44K.

 

I see what you are saying but surely it cant be that easy. Beta is = 150 so it would be 33K. Can i just run you through my train of thought

R*** = rest of circuit which enters Q1 via the Collector

Rintot = R1//R***//Rinbase

Rinbase = Bac(re'+ R3)

re' = 25mV/Ie

Ie = VE/R3

VE = VB -VBE

VB = (???) x VCC

So I think you first work out VB, VE, IE, re', Rinbase and then Rintot. please tell me if im going about this in the totally wrong direction as at the min I cant see the wood for the trees. Im also struggling into how to work out the reistance value of R*** due to the fact of the darlington pair and the Sziklai pair

 

You are missing a Q1 base resistor, so Q1 is always off unless it is biased on by the leakage current.Q1 must be biased on by a resistor tied to a voltage. The output pairs need BE resistors too.

 

Sorry, I was being approximate. Rinbase is correct and that's your answer right there. In the circuit you presented there's nowhere else for current to flow. And re' is going to be tiny compared to R3 which is why I approximated. It's 25 ohms at 1 ma and only 0.025 ohms at 1 amp and I was figuring the variation in the beta would overshadow that.

Not sure where R1 comes in, it's not on the diagram. And as far as R***, I really have no clue but I don't normally factor that in to my own calculations. I don't *think* you need to.

If you put a resistor up to Vcc from the Q1 base, that would go in parallel with Rinbase as you seem to be indicating.

If you removed the input cap it looks like it would bias correctly around the zero volt point, but I'm not sure about the circuit as a whole, I was just really looking at the input side for the impedance calculation.

 

Another thing you can do is, if you have a simulator, look at the RMS voltage of the voltage source and divide by the RMS current that is being pulled from the voltage source into the transistor base... this will give you an impedance figure and you can compare to your own calculations.