Minimum photons-per-second [amplitude] required for 150 KHz?
 

On Sun, 17 Jun 2007 15:19:59 -0700, Radium
wrote:

What is the minimum amount of photons-per-second needed for a 150 KHz
AM radio carrier wave to transmit audio signals? Around 20,000-photons-
per-second?

That one is real simple:
2 photons (of appropriate amplitude, hence color) at most 3.333 (less
would be better, but not too much less) microseconds apart.

Feel free to desire more, but you asked for the minimum. If you want
more audio (sideband) content, that will certainly drive up the count
too.

Now, how's your quantum efficiency these days? (Use it to boost the
count higher.)

73's
Richard Clark, KB7QHC

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In article . com,
Radium wrote:
Hi:

What is the minimum amount of photons-per-second needed for a 150 KHz
AM radio carrier wave to transmit audio signals? Around 20,000-photons-
per-second?


1. What is the energy of a photon at 150 kHz?

2. What is the minimum discernable signal in your receiving system?
(How much power is needed at the receiver to overcome the internal
noise of the receiver system and detect the signal?)

3. What signal to noise ratio makes for a tolerable listening condition?
(How much more power than quetion #2 is needed at the receiver to decode
the modulation and yield a usable signal?)

Mark Zenier
Googleproofaddress(account:mzenier provider:eskimo domain:com)

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In article .com,
Denny wrote:

On Jun 17, 10:37 pm, Richard Clark wrote:
On Sun, 17 Jun 2007 15:19:59 -0700, Radium
wrote:

Snip

I need a beer...

Nope. You just need to be added to the kill file.

Cheers.

Plonk

--
Telamon
Ventura, California

Minimum photons-per-second [amplitude] required for 150 KHz?
 

On Jun 18, 11:05 am, (Mark Zenier) wrote:

1. What is the energy of a photon at 150 kHz?

6.2 X 10^-10 eV

Minimum photons-per-second [amplitude] required for 150 KHz?
 

John Smith I wrote:

Telamon wrote:

In article ,
"Mike Kaliski" wrote:


Ohh gawd, bizarre telemundo is back ...

PLONKERS!

JS


The irony is he's proudly plonking those who are actually more on topic
with this 'photon' thread than he often is in other threads, where the
subject is really OT.

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In article .com,
Denny wrote:

Snip

plonk

--
Telamon
Ventura, California

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In article ,
Plonk Dodger wrote:

Snip

Plonk

--
Telamon
Ventura, California

Minimum photons-per-second [amplitude] required for 150 KHz?
 

Telamon wrote:

...
plonk


My gawd man! Is your quest to plonk the world ... ;-)

JS

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In rec.radio.amateur.antenna Denny wrote:
On Jun 18, 10:53 pm, Telamon
wrote:
In article .com,



Denny wrote:
On Jun 17, 10:37 pm, Richard Clark wrote:
On Sun, 17 Jun 2007 15:19:59 -0700, Radium
wrote:

Snip

I need a beer...

Nope. You just need to be added to the kill file.

Cheers.

Plonk

--
Telamon
Ventura, California

Oh gawd, now my life is over....

denny / k8do

Maybe not.

Let's see how many more times you are added to his killfile.

--
Jim Pennino

Remove .spam.sux to reply.

Minimum photons-per-second [amplitude] required for 150 KHz?
 

In article . com,
Radium wrote:
On Jun 18, 11:05 am, (Mark Zenier) wrote:

1. What is the energy of a photon at 150 kHz?

6.2 X 10^-10 eV


You didn't answer the other questions. How much power does a good radio
need to get a listenable signal?

Here's some numbers

input impedance 50 ohms
Noise Figure 5 db
Bandwidth 6 kHz
signal to noise ratio 40 dB

(This would be what some of the newsgroups listeners here would want if
they had one of their pretty damn good radios listening to a broadcast
station. Really picky ones would probably want a 8-15 kHz bandwidth
with a 60 dB s/N ratio).

There are equations out there that will give you how much power
you need for this signal...

Mark Zenier
Googleproofaddress(account:mzenier provider:eskimo domain:com)