The input circuit constitutes an RC high-pass filter network, so it will cause a bit of phase shift. But it looks like it's cutoff is around 3 Hz, so I wouldn't expect mush shift at 1 kHz (and note that it is "kHz", not "Khz" -- these things can matter and you need to develop an attention to detail).

How would i go about calculating cut off frequency of the input filter?.... is there a formula for doing that?

 

How would i go about calculating cut off frequency of the input filter?.... is there a formula for doing that?

The problem with wanting to go to some formula is that it is only going to apply if the circuit you are actually working with satisfies the conditions required for that formula to be valid. Unless you understand both the formula and the circuit well enough to make that assessment, you are at high risk of using the wrong formula and getting a wildly incorrect answer without realizing it. This happens all the time when people just want to memorize a bunch of formulas to throw at problems -- they end up picking the wrong one and continue blindly pushing on.

As much as you hate doing math, you need to invest the time and effort learn how to analyze circuits. Once you have the fundamentals down, you will be able to look at many circuits and, by inspection, determine what is and isn't important and focus on just the important stuff.

In this case, you want to know what simple input impedance does the signal source see. For that, you need to have familiarity with Thevenin equivalent circuits and how to find the Thevenin impedance. With that understanding, you know that your circuit reduces to (R3) in parallel with (R4) in parallel with (the transmitter base-emitter in series with R2). Notice that R1 isn't included in that, so it's not surprising that changing R1 over a wide range of values has no effect on the phase shift.

Next, because you've spent some time understand transistor circuit characteristics, you know that, as seen looking into the base of Q1, that R2 looks like a resistor that is β times larger, and the a common rule of thumb is that the β of most small-signal transistors is in the range of 100 to 200 and that 100 is commonly used as a good engineering estimate. So that 470 Ω looks more like 47 kΩ, which is then in parallel with the 100 kΩ R4 and the 10 kΩ R3. At that point, you understanding of the behavior of series and parallel resistors tells you that the effective resistance is going to be smaller than 10 kΩ, but not by much a huge amount, so we can just call it 10 kΩ for a first pass (if you do the math, its 7.6 kΩ, so there is definitely an error, but it's within 25%, which is likely good enough for a quick estimate).

Now you have an RC time constant of (47 pF)(10 kΩ) which is 470 μs, which is 1/ω_c, where w_c is the critical frequency (in radians per second -- ω_c = 2πf_c, where f_c is the critical frequency in hertz)

At this point, you can just solve for f_c, you end up with 339 Hz. If you use the better approximation that doesn't ignore R2 and R4, you get 446 Hz, which should account for the discrepancy between my 17° and the simulations 21.5°.

Another way to come at it is to relay on your study and understanding of first-order RC circuits is that the cutoff frequency is when the capacitive reactance is equal in magnitude to the resistors.

From you study of AC circuits, you know that the capacitive reactance is

Xc = 1/(2πf_c⋅C)

So set

Xc = Rth

1/(2πf_c⋅C) = 10 kΩ (or the better approximation of 7.6 kΩ) and again solve for f_c (you'll get the same answer).

Note that the 3 Hz value in the post you quoted is wrong, as explained a few posts later. I didn't have a pen and so was doing everything in my head and so I relied on using a formula that I thought applied. But it was the wrong formula and they way I used it screwed up the units, but because I was just doing it in my head I didn't catch it.

 

The problem with wanting to go to some formula is that it is only going to apply if the circuit you are actually working with satisfies the conditions required for that formula to be valid. Unless you understand both the formula and the circuit well enough to make that assessment, you are at high risk of using the wrong formula and getting a wildly incorrect answer without realizing it. This happens all the time when people just want to memorize a bunch of formulas to throw at problems -- they end up picking the wrong one and continue blindly pushing on.

As much as you hate doing math, you need to invest the time and effort learn how to analyze circuits. Once you have the fundamentals down, you will be able to look at many circuits and, by inspection, determine what is and isn't important and focus on just the important stuff.

In this case, you want to know what simple input impedance does the signal source see. For that, you need to have familiarity with Thevenin equivalent circuits and how to find the Thevenin impedance. With that understanding, you know that your circuit reduces to (R3) in parallel with (R4) in parallel with (the transmitter base-emitter in series with R2). Notice that R1 isn't included in that, so it's not surprising that changing R1 over a wide range of values has no effect on the phase shift.

Next, because you've spent some time understand transistor circuit characteristics, you know that, as seen looking into the base of Q1, that R2 looks like a resistor that is β times larger, and the a common rule of thumb is that the β of most small-signal transistors is in the range of 100 to 200 and that 100 is commonly used as a good engineering estimate. So that 470 Ω looks more like 47 kΩ, which is then in parallel with the 100 kΩ R4 and the 10 kΩ R3. At that point, you understanding of the behavior of series and parallel resistors tells you that the effective resistance is going to be smaller than 10 kΩ, but not by much a huge amount, so we can just call it 10 kΩ for a first pass (if you do the math, its 7.6 kΩ, so there is definitely an error, but it's within 25%, which is likely good enough for a quick estimate).

Now you have an RC time constant of (47 pF)(10 kΩ) which is 470 μs, which is 1/ω_c, where w_c is the critical frequency (in radians per second -- ω_c = 2πf_c, where f_c is the critical frequency in hertz)

At this point, you can just solve for f_c, you end up with 339 Hz. If you use the better approximation that doesn't ignore R2 and R4, you get 446 Hz, which should account for the discrepancy between my 17° and the simulations 21.5°.

Another way to come at it is to relay on your study and understanding of first-order RC circuits is that the cutoff frequency is when the capacitive reactance is equal in magnitude to the resistors.

From you study of AC circuits, you know that the capacitive reactance is

Xc = 1/(2πf_c⋅C)

So set

Xc = Rth

1/(2πf_c⋅C) = 10 kΩ (or the better approximation of 7.6 kΩ) and again solve for f_c (you'll get the same answer).

Note that the 3 Hz value in the post you quoted is wrong, as explained a few posts later. I didn't have a pen and so was doing everything in my head and so I relied on using a formula that I thought applied. But it was the wrong formula and they way I used it screwed up the units, but because I was just doing it in my head I didn't catch it.

Thats alot of information some of which i have absorbed, you are right though when you say i need to brush up on my math.

I bought a scientific calculator last month thinking it would do formula for me....WRONG!!!....i couldn't figure out how to use it so i tossed it in the garbage and bought a more basic one LMAO

I found 1/2piRC formula for doing cut off frequency and came up with 338Hz so i was happy with that, i get excited when i do electronic stuff and am keen to learn more.

thank you so much for your patience and time.