Mismatched load problem for EMC course. - Electronic Circuits Projects Diagrams Free

jamesmi · 14th February 2005, 06:18 AM

Hello,

I'm taking a course on electromagnetic compatability and having trouble with one of the homework problems. The instructor does not collect any of the homework so this is not a for credit assignment. The book also gives the answer to the problem, I just can't seem to get the same answer. The instructor doesn't answer questions during class and unfortunately I cant make his office hours. I can contact him by email, but I thought I might try here first.

Heres the problem from the book "Introduction to Electromagnetic Compatibility" by Clayton R. Paul.

1.17 A 50 ohm oscillator is attached to the high-impedance input of an oscilloscope (Cin = 47pF, Rin= 1Mohm). The source is tuned to 100Mhz and the level set to -30dBm. Determine the voltage level (peak) of the sinusoid seen on the oscilloscope face.

Answer given in book: 11.22 mV

My approach based an example in the book:

The source assumes a 50 ohm load, so first I determined the open circuit voltage of the source.

Vout = sqrt(50*Pout) = sqrt(50*.000001) = .007071 Vrms
Voc = 2*Vout = .01414 Vrms
Vocpeak = 1.414*Voc = .019997 Vpeak

I then determined the input impedance of the load.

Xc = 1/(2*pi*f*c) = 1/(2*pi*100E6*47E-12) = 33.86 ohms

This is in parrallel with the 1Mohm, which doesn't change value to this precision.

The voltage seen at the scope could then be calculated by using a voltage divider.

Vscope = Vocpeak * 33.86/(33.86+50) = .00807 Vpeak = 8.07 mVpeak

I also tried calculating the open circuit voltage using the power of -30dBm going into a 33.86 ohm load and a 83.86 ohm load, but that doesn't help either.

Can anyone give me an idea of where I am going wrong?

Thanks,
Jamey

FRIED · 14th February 2005, 10:09 AM

You must use vector addition to obtain the sum of the 50 ohm output impedance and the capacitive load.

Quote:

(33.86+50)

jamesmi · 14th February 2005, 10:43 AM

Ahhh...Of course.

It's been a couple semesters since I had a circuits course. I haven't had to use AC analysis techniques sense. I knew it was probably something pretty silly I was doing wrong. I wish I would of asked a little sooner for help, I've tried doing this problem a few seperate times. After redoing the problem I got the correct answer.

Vpeak = 0.019997*33.86/sqrt(50^2+33.86^2)=.011214Vpeak
or about 11.2 mV, which is close enough considering I did some rounding.
I left out the phase shift, because the problem didn't require it.

I appreciate the help FRIED.

Thanks,
James

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

14th February 2005, 06:18 AM	(permalink)
jamesmi	Mismatched load problem for EMC course. Hello, I'm taking a course on electromagnetic compatability and having trouble with one of the homework problems. The instructor does not collect any of the homework so this is not a for credit assignment. The book also gives the answer to the problem, I just can't seem to get the same answer. The instructor doesn't answer questions during class and unfortunately I cant make his office hours. I can contact him by email, but I thought I might try here first. Heres the problem from the book "Introduction to Electromagnetic Compatibility" by Clayton R. Paul. 1.17 A 50 ohm oscillator is attached to the high-impedance input of an oscilloscope (Cin = 47pF, Rin= 1Mohm). The source is tuned to 100Mhz and the level set to -30dBm. Determine the voltage level (peak) of the sinusoid seen on the oscilloscope face. Answer given in book: 11.22 mV My approach based an example in the book: The source assumes a 50 ohm load, so first I determined the open circuit voltage of the source. Vout = sqrt(50Pout) = sqrt(50.000001) = .007071 Vrms Voc = 2Vout = .01414 Vrms Vocpeak = 1.414Voc = .019997 Vpeak I then determined the input impedance of the load. Xc = 1/(2pifc) = 1/(2pi100E647E-12) = 33.86 ohms This is in parrallel with the 1Mohm, which doesn't change value to this precision. The voltage seen at the scope could then be calculated by using a voltage divider. Vscope = Vocpeak * 33.86/(33.86+50) = .00807 Vpeak = 8.07 mVpeak I also tried calculating the open circuit voltage using the power of -30dBm going into a 33.86 ohm load and a 83.86 ohm load, but that doesn't help either. Can anyone give me an idea of where I am going wrong? Thanks, Jamey

14th February 2005, 10:43 AM	(permalink)
jamesmi	Ahhh...Of course. It's been a couple semesters since I had a circuits course. I haven't had to use AC analysis techniques sense. I knew it was probably something pretty silly I was doing wrong. I wish I would of asked a little sooner for help, I've tried doing this problem a few seperate times. After redoing the problem I got the correct answer. Vpeak = 0.019997*33.86/sqrt(50^2+33.86^2)=.011214Vpeak or about 11.2 mV, which is close enough considering I did some rounding. I left out the phase shift, because the problem didn't require it. I appreciate the help FRIED. Thanks, James