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Joshy

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- Homework Statement
- In addition to maximizing power transfer, impedance-transforming networks are widely used simply to enable a specific amount of power to be delivered to a load. An important example is found in the output stage of a transmitter where, owing to supply voltage limitations, a downward impedance transformation of the antenna resistance is necessary.

A common load impedance is 50 Ohms. Suppose we wish to deliver 1 W of power into such a load at 1 GHz, but the power amplifier has a maximum peak-to-peak sinusoidal voltage swing of only 6.33V because of various losses and transistor breakdown problems. Design the following matching networks to allow that 1 W to be delivered. Use low-pass versions in all cases, and assume that all reactive elements are ideal (if only that were true...).

- Relevant Equations
- (1) ##\frac{|V_R|^2}{R_L}## = ##\frac{R_L |V_S|^2}{(R_L + R_S)^2 + (X_L + X_S)^2}##, (2) ##v_{rms} = \frac{V}{\sqrt{2}}##

Hi!

I normally have a pretty good grasp of what the source and load impedance for these types of problems; however: This textbook is giving me peak to peak voltage swing 6V33 and 1 W of power. I'm a little bit confused at converting this into an input impedance and am worried I'll be off by a scalar of 2, or something like that... This problem has parts (a)-(e) dependent on having correct source impedance for my matching networks, and so it couldn't hurt to ask for a sanity check or extra set of eyes. (It's for a Tuesday/Thursday class professor shared it on Thursday and it's due on Tuesday, and so it is unlikely they'll be able to answer my question on time.)

The equation (1) given earlier in the chapter talks in terms of rms voltages ("##V_R## and ##V_S## are rms voltages across the load resistance and source, respectively") and so I'm thinking I need to convert this to rms voltage as well.

I start by taking 6V33 and divide it by 2 for V peak (not V peak to peak)

Then I need to turn this into rms voltage using equation (2)

Then power I'm calling ##P## should be (V^2)/Rs

So if the power 1 W, then Rs should be 5 Ohms.

~~Also occurred to me after using the cool latex format that seems to work in other categories... that it's not showing up in preview here? Reformatted the post to just use words :( That's strange.~~ Seemed to only be an issue in preview.

By the way if you're curious as to which book it is, then it's "The Design of CMOS Radio-Frequency Integrated Circuits" (2nd edition) by Tom Lee and the problem is Chapter 3 Problem 2.

I normally have a pretty good grasp of what the source and load impedance for these types of problems; however: This textbook is giving me peak to peak voltage swing 6V33 and 1 W of power. I'm a little bit confused at converting this into an input impedance and am worried I'll be off by a scalar of 2, or something like that... This problem has parts (a)-(e) dependent on having correct source impedance for my matching networks, and so it couldn't hurt to ask for a sanity check or extra set of eyes. (It's for a Tuesday/Thursday class professor shared it on Thursday and it's due on Tuesday, and so it is unlikely they'll be able to answer my question on time.)

The equation (1) given earlier in the chapter talks in terms of rms voltages ("##V_R## and ##V_S## are rms voltages across the load resistance and source, respectively") and so I'm thinking I need to convert this to rms voltage as well.

I start by taking 6V33 and divide it by 2 for V peak (not V peak to peak)

##Vp = \frac{6.33}{2} = 3.165##

Then I need to turn this into rms voltage using equation (2)

##V_{rms} = \frac{Vp}{\sqrt{2}} = \frac{3.165}{\sqrt{2}} \approx 2.23799##

Then power I'm calling ##P## should be (V^2)/Rs

##P = \frac{2.23799^2}{R_S} = \frac{5}{R_S}##

So if the power 1 W, then Rs should be 5 Ohms.

By the way if you're curious as to which book it is, then it's "The Design of CMOS Radio-Frequency Integrated Circuits" (2nd edition) by Tom Lee and the problem is Chapter 3 Problem 2.

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