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Hi all.
I have been working through a design exercise for a single transistor amplifier, operating at 8GHz. However I am having some problem calculating the values of the equivalent output components.
I have calculated the impedance of the active device to be:
\(Z = 36.153 -j27.44\) (1)
My textbook says that the output of the device looks like a 56.99 Ohm resistor in parallel with a capacitor of -75.064 Ohm reactance. I have tried calculating these values, but stumble at the last leg. My calculations are as follows.
In polar form:
\(45.39 \angle 322.80^{\circ}\) (2)
The value for the angle (equivalent to -37.2 degrees) for a parallel RC circuit is given as:
\(\varphi =\arctan (- \omega RC)\) (3)
and for the magnitude Z by:
\(\left | Z \right | = \frac{1}{\sqrt{\left ( \frac{1}{R} \right )^{2}+(\omega C)^{2}}}\) (4)
We have values for phi, |Z|, omega; only C and R to find. R would give the resistance, and C can be used to find X. I thought it'd be best to solve as simultaneous equations. So I rearranged (4) to find R:
\(R = \frac{1}{\sqrt{\left ( \frac{1}{\left | Z \right |} \right )^{2}-(\omega C)^{2}}}\) (5)
and put (5) into (3) to get:
\(\varphi =\arctan \left ( - \omega C \frac{1}{\sqrt{\left ( \frac{1}{\left | Z \right |} \right )^{2}-(\omega C)^{2}}} \right )\) (6)
However, when I stuck the numbers in, and handed it to wolfram alpha to find C, it (kind of predictably I guess) said no result. I'm guessing this is due to the approximations in the cartesian to polar conversion.
My question is, assuming the algebra was good, how would one go about figuring out C and R, given the approximate nature of values? Perhaps I have gone entirely the wrong way about it.
Many thanks for your time.
Sparky
I have been working through a design exercise for a single transistor amplifier, operating at 8GHz. However I am having some problem calculating the values of the equivalent output components.
I have calculated the impedance of the active device to be:
\(Z = 36.153 -j27.44\) (1)
My textbook says that the output of the device looks like a 56.99 Ohm resistor in parallel with a capacitor of -75.064 Ohm reactance. I have tried calculating these values, but stumble at the last leg. My calculations are as follows.
In polar form:
\(45.39 \angle 322.80^{\circ}\) (2)
The value for the angle (equivalent to -37.2 degrees) for a parallel RC circuit is given as:
\(\varphi =\arctan (- \omega RC)\) (3)
and for the magnitude Z by:
\(\left | Z \right | = \frac{1}{\sqrt{\left ( \frac{1}{R} \right )^{2}+(\omega C)^{2}}}\) (4)
We have values for phi, |Z|, omega; only C and R to find. R would give the resistance, and C can be used to find X. I thought it'd be best to solve as simultaneous equations. So I rearranged (4) to find R:
\(R = \frac{1}{\sqrt{\left ( \frac{1}{\left | Z \right |} \right )^{2}-(\omega C)^{2}}}\) (5)
and put (5) into (3) to get:
\(\varphi =\arctan \left ( - \omega C \frac{1}{\sqrt{\left ( \frac{1}{\left | Z \right |} \right )^{2}-(\omega C)^{2}}} \right )\) (6)
However, when I stuck the numbers in, and handed it to wolfram alpha to find C, it (kind of predictably I guess) said no result. I'm guessing this is due to the approximations in the cartesian to polar conversion.
My question is, assuming the algebra was good, how would one go about figuring out C and R, given the approximate nature of values? Perhaps I have gone entirely the wrong way about it.
Many thanks for your time.
Sparky
