# Typical capacitance of a bigger laser diode?

J

#### Joerg

Jan 1, 1970
0
Jamie said:
Sounds like you need to make the diode become part of a resonant tank
with in the power circuit. Using the diodes capacitance to contribute
to the remaining C required.

I can see that being tricky due to the C changing dynamically. At
least I would think that to be the case.

That's one of the nasties because it can make harmonics and those are
not liked in the land. But there is a whole bag of trick to overcome
this, it depends on how complex a circuit can be tolerated. The nice
thing is that RF parts are so cheap these days, thanks to an insatiable
appetite of our youngsters to always be connected.

J

#### Joerg

Jan 1, 1970
0
George said:
Oh just having my own sort of fun. I've got a few resistors and the
diode in a little pomona box.
Say, can I ask a 'silly' question. Since the probe capacitance is
limiting me, can I hang a 50 ohm resistor on the end and let it drive
a coax cable... throw away a bunch of signal.

Pulse in----+---1k---+---50R---->scope
| |
50R _
| V <-diode
| |
GND GND

I'm figuring the diode looks like a voltage source with a few ohms of
resistance.... ? Perhaps my figuring is wrong.

Then I could reduce the 1k ohm down to something that would let the
diode laser turn on.

You can do that for a big laser diode, 100mW and up. But small ones are
more in the tens of ohms.

A common trick to get a cheapo probe is to take a piece of coax with BNC
at one end, solder a 950ohms 0402 in series at the tip, and then set the
scope to 50ohm input. The coax only needs to be terminated at one end
and then it is. Gives a nice clean representation of the input signal
divided 10:1.

J

#### Jamie

Jan 1, 1970
0
Joerg said:
That's one of the nasties because it can make harmonics and those are
not liked in the land. But there is a whole bag of trick to overcome
this, it depends on how complex a circuit can be tolerated. The nice
thing is that RF parts are so cheap these days, thanks to an insatiable
appetite of our youngsters to always be connected.
Well Ok but when I decide to sit down and lets say build an amplifier,
after the cost of getting a few CX1000 series, vacuums caps, high grade
door knob caps and lots of ceramics and let us not forget the roller
inductor with it's clean copper tubing coils and all.

Of course, you could always slip some of that through your employers
accounting system.

This would be for my personal enjoyment. At work, we deal with hot
cathode 300kWatt+ custom tubes and the last one we got cost \$12k. But
that brings a smile because I don't pay for that!

Jamie

J

#### Jamie

Jan 1, 1970
0
Joerg said:
You can do that for a big laser diode, 100mW and up. But small ones are
more in the tens of ohms.

A common trick to get a cheapo probe is to take a piece of coax with BNC
at one end, solder a 950ohms 0402 in series at the tip, and then set the
scope to 50ohm input. The coax only needs to be terminated at one end
and then it is. Gives a nice clean representation of the input signal
divided 10:1.

One of our product lines is a special coax with a steel wire in the
center instead of copper. It actually is more like a alloyed used in
thermocouple wire but, it works well for lowering the Q and we sell this
to companies that make instrument cables for the most part.

Jamie

J

#### josephkk

Jan 1, 1970
0
Sounds like you need to make the diode become part of a resonant tank
with in the power circuit. Using the diodes capacitance to contribute
to the remaining C required.

I can see that being tricky due to the C changing dynamically. At
least I would think that to be the case.

Jamie
Think pulse forming networks from radar systems.

?-)

J

#### josephkk

Jan 1, 1970
0
You can do that for a big laser diode, 100mW and up. But small ones are
more in the tens of ohms.

A common trick to get a cheapo probe is to take a piece of coax with BNC
at one end, solder a 950ohms 0402 in series at the tip, and then set the
scope to 50ohm input. The coax only needs to be terminated at one end
and then it is. Gives a nice clean representation of the input signal
divided 10:1.

Hmmmm. Wouldn't that be 20:1? Or did you mean 450 Ohms?

?-)

J

#### Joerg

Jan 1, 1970
0
josephkk said:
Hmmmm. Wouldn't that be 20:1? Or did you mean 450 Ohms?

Oops, sorry, 20:1. You could use 450ohms but sometimes that's too low,
for example for tapping the collector on an RF stage.

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