The AM Forum
May 19, 2024, 07:41:04 AM *
Welcome, Guest. Please login or register.

Login with username, password and session length
 
   Home   Help Calendar Links Staff List Gallery Login Register  
Pages: [1]   Go Down
  Print  
Author Topic: Receiver Antenna in SPiCE  (Read 4635 times)
0 Members and 1 Guest are viewing this topic.
jdirga
Guest
« on: April 15, 2009, 10:17:39 PM »

I need to include a receiver antenna in SPICE simulation. Should I model the antenna as a current source or voltage source? Does this depend on the type of the antenna(whether it's a dipole, loop, etc)? Please help. Thx.
Logged
Rob K2CU
Contributing
Member
*
Offline Offline

Posts: 346


« Reply #1 on: April 15, 2009, 11:15:22 PM »

The simplest, and probably the best approach would be to use a voltage source with the antenna resistance (50 Ohms) in series with it. I assume you are modeling the circuit that the antenna connect to, such as a receiver simulation, or a filter. Modeling an antenna to reflect the type of antenna, gain, etc is much more complex. For that type of modeling, I would suggest MMANA-GAL, an easy to use freebee.

I have been working on models of various vacuum tubes for evaluating circuits such as the Pullen mixer versus a 6BE6 type mixer. I have just today been working on dual plate deflection tubes such as the 7360 and have it just about done. Most Spice simulators include just a few basic tubes, if any.

If anyone else is interested, I would be happy to share the models I have collected as well as schematic symbols that I have created for LTSpice. LTSpice is a free simulator available from Linear Technologies. IT works very well and is not too difficult to learn to use if you have ever used Spice simulators. I have used it to simulate several Class E transmitters, and a 6J6 pullen mixer as well as many other circuits. It can be daunting if you have never used Spice. But it is well worth it to learn to use.

Logged
jdirga
Guest
« Reply #2 on: April 15, 2009, 11:51:17 PM »

Thanks for the reply. Could you give a reason why it should be modeled as a voltage source instead of a current source?

My original thought was if it were a loop antenna, it should be modeled as a voltage source because the signal is generated on the received antenna thru Faraday's law of induction, i.e. emf=change of flux. However if it were a dipole, it should be modeled as a current source because signal is generated due to lenz's law?(but I'm not confident with this).
Logged
Rob K2CU
Contributing
Member
*
Offline Offline

Posts: 346


« Reply #3 on: April 16, 2009, 12:10:18 PM »

Classic receiving antenna analysis bases the induced terminal voltage upon the strength of the arriving time variant electric field. A good explanation can be found at:

http://en.wikipedia.org/wiki/Antenna_measurement

Look at the section "Calculation of Antenna parameters in reception".


It provides a Thevinen equivalent circuit for an antenna based upon a voltage source and series impedance. This is why I suggested a voltage source and series impedance (or resistance) for your spice simulation. A current souce could be used by converting the voltage source and series impendance to an equivalent current source with parallel conductance.  A lot of what you do has to do with what you are trying to study with the Spice simulation.

From the referencd website:
The induced voltage is proportional to: 1. the Square root of the real part of antenna impednace times the antenna gain; 2. the wavelength; 3. the cosine of the angle between the antenna elements and the incoming wave; 4. the strenght of the incoming waves electric field; and inversly proportional to the square root of Pi times the impedance of of free space (approx 377 Ohms). This equivalent circuit is valid for all types of antennas.

I can understand the thought that a dipole would be a current source as the radiation from a dipole is derived by the integration over a length of minute current dipole elements. 

Lenz's law has to do with deriving the polarity of the induced current in a conductor. the induced current is of a polarity such that if it were appplied to the conductor, it would create a magnetic field opposite to the applied magnetic field. It sort of like the idea that if you push on a wall with a one pound force, it pushes in the opposite direction with a one pound force. it was to explain why the current polarity is what it is.

Logged
Pages: [1]   Go Up
  Print  
 
Jump to:  

AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
 AMfone © 2001-2015
Powered by SMF 1.1.21 | SMF © 2015, Simple Machines
Page created in 0.061 seconds with 18 queries.