**EMRFD Message Archive 1318**

MessageDateFromSubject1318 2008-01-13 12:45:24 Dimitri Aguero Still waiting for how to calculate the output impedance of the SP-1 EMRFDGood evening,

Thanks to Ed for the link for calculating Butterworth filter.

But the problem I want to solve is not calculating the Butterworth filter...

I would like to know how to figure out what value (calculated, estimated or

measured) should I use for the output impedance of this transmitter...

Then, my question is reworded as follows: "How may I calculate, estimate or

measure the output impedance of a transmitter power stage" ?

Thank you in advance for your patience...

73 Dimitri F4DYT

[Non-text portions of this message have been removed]1319 2008-01-13 12:55:51 Nick Kennedy Re: Still waiting for how to calculate the output impedance of the S I think the answer is less simple that it might seem, and even

somewhat controversial in some circles. Some people don't believe

that certain classes of transmitters can be represented as a Thevenin

equivalent (source plus resistance), especially class C but also

probably class B as well. And you don't see many class A

transmitters.

In most cases, the output network suppresses harmonics and transforms

a known load (usually 50 ohms) to the value the transmitter's output

stage requires, without regard to that output stage having any

specific source resistance.

It might be interesting to model any filter you propose to use with

SPICE. Try driving it with just a source and also with source plus

resistance and see if the harmonic rejection and filter efficiency are

adversely affected by not having a specific source resistance.

73--Nick, WA5BDU1320 2008-01-13 13:14:18 Leon Re: Still waiting for how to calculate the output impedance of the S ----- Original Message -----

1321 2008-01-13 13:59:55 Michael Neverdosk... Re: Still waiting for how to calculate the output impedance of the S From "Solid State Design", the load needed at the collector might be

R=Vcc^2/2Po

This is likely oversimplifying but should get you in the ballpark.

The voltage on the final and the power level determine the impedance. As

long as the power stays the same things are easy. As soon as you change the

power level the impedance changes.

Hope this helps.

michael N6CHV

On 1/13/08, Dimitri Aguero <daguero@free.fr> wrote:

>

> EMRFDGood evening,

>

> Thanks to Ed for the link for calculating Butterworth filter.

>

> But the problem I want to solve is not calculating the Butterworth

> filter...

>

> I would like to know how to figure out what value (calculated, estimated

> or

> measured) should I use for the output impedance of this transmitter...

>

> Then, my question is reworded as follows: "How may I calculate, estimate

> or

> measure the output impedance of a transmitter power stage" ?

>

> Thank you in advance for your patience...

>

> 73 Dimitri F4DYT

>

[Non-text portions of this message have been removed]1323 2008-01-13 15:01:50 Jim Kortge Re: Still waiting for how to calculate the output impedance of the S Dimitri Aguero wrote:

> EMRFDGood evening,Dimitri,

>

> Thanks to Ed for the link for calculating Butterworth filter.

>

> But the problem I want to solve is not calculating the Butterworth filter...

>

> I would like to know how to figure out what value (calculated, estimated or

> measured) should I use for the output impedance of this transmitter...

>

>

> Then, my question is reworded as follows: "How may I calculate, estimate or

> measure the output impedance of a transmitter power stage" ?

>

> Thank you in advance for your patience...

>

> 73 Dimitri F4DYT

If the emitter of the final transistor is grounded, then the formula that most

use is Vcc^2/2*Po, where Vcc is the supply voltage, which is squared, and Po is

the design output power. So for a 12 volt supply and 5-watt output, the needed

impedance is 144/10 or 14.4 Ohms. Step that value up to 50 Ohms using an

L-section impedance matching section or transformer and standard 50 Ohm in/out

filters after that. If the emitter of the final is not grounded, then that

voltage has to be subtracted from the supply voltage.

I think the above information is what you are seeking.

72 and kind regards,

Jim, K8IQY

>

>

>

>

> [Non-text portions of this message have been removed]

>

>