Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) =3D -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hern=E1n S=E1nchez

nanchez wrote:

Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

Maybe this will help explain...

http://zone.ni.com/devzone/conceptd....256811004DD454

On Fri, 03 Jun 2005 18:07:51 -0400, -ex- wrote:

nanchez wrote:

Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

Maybe this will help explain...

http://zone.ni.com/devzone/conceptd....256811004DD454

This is pretty good, but errs when it states that 0 dBm is 1 mW in a
50 ohm system. This is the usual case, but it could just as well be
70 ohm, 600 ohm or 6 3/8 ohm.

It is worse than that, Wes. I fail to understand how a device can be
specified as X dB into a capacitive load. Last time I looked, a capacitive
load couldn't dissipate ANY power.

Jim

This is pretty good, but errs when it states that 0 dBm is 1 mW in a
50 ohm system. This is the usual case, but it could just as well be
70 ohm, 600 ohm or 6 3/8 ohm.

On Sat, 4 Jun 2005 08:56:56 -0700, "RST Engineering"
wrote:

It is worse than that, Wes. I fail to understand how a device can be
specified as X dB into a capacitive load. Last time I looked, a capacitive
load couldn't dissipate ANY power.

Jim, the OP said:

"And I have a LO source that give me an output of 2.5Vpp to a
capacitive load of 5pF at 40MHz."

No dB or dBm mentioned. This sounds like a CMOS device with limited
drive capability. That's why I suggest terminating it in 50 ohm,
assuming this doesn't destroy it, and see what kind of power it can
deliver.

Most likely, a buffer will be needed, although -16 dBm isn't much and
at that level it suggests that this is an active, not passive, mixer
and that it might be driven by a higher source Z okay.

Wes


Jim

This is pretty good, but errs when it states that 0 dBm is 1 mW in a
50 ohm system. This is the usual case, but it could just as well be
70 ohm, 600 ohm or 6 3/8 ohm.

"nanchez" wrote in message
ups.com...
Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

To rejoin the real world, take the "16" figure and divide it by 20.
get "0.8"
Then find the antilog of that 0.8 [use normal 'base10' logs]
get "6.31"
This number is a multiply or divide factor that is applied to a 50 ohm 0dBm
reference voltage.
So what is this god like reference voltage?. The 50ohm 0dBm reference
voltage is in actual fact 0.223Vac.
The original number was "-"16 dBm. Just read the minus sign as meaning a
voltage less than the 0dBm reference voltage.
So that 0.223Vac reference value is divided by your 6.31 factor.
get "0.035" Vac.
So "-16dBm" is really 35mVac. This means you have more than enough drive
voltage available from your 2.5Vpp (900mVac) local oscillator signal.

Be very wary whenever you come across dBm figures. There is a minefield of
disinformation out there.
In many cases they are intended purely to obfscutate the reader and prevent
them clearly seeing that the described circuit is junk.
In many other cases they are purposely used as an extra level of abstraction
to sort out the 'RF men' from the 'boys'.
Manufacturers still use the dB concept for historical reasons. It doesn't
effect their sales as the RF people buying their kit carry in their heads
instant dB-V conversion tables.
Don't know about everyone else but all my scopes and signal generators and
sources and dc-ac-voltmeters and DVMs and signal probes etc, are marked in
Volts and Amps. So that's what I use.
(Someday I'll get round to building a real world 1:2:5:10 50ohm attenuator.
I certainly can't buy one :-)
regards
john

On Sat, 4 Jun 2005 14:53:43 +0100, "john jardine"
wrote:


"nanchez" wrote in message
oups.com...
Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

To rejoin the real world, take the "16" figure and divide it by 20.
get "0.8"
Then find the antilog of that 0.8 [use normal 'base10' logs]
get "6.31"
This number is a multiply or divide factor that is applied to a 50 ohm 0dBm
reference voltage.
So what is this god like reference voltage?. The 50ohm 0dBm reference
voltage is in actual fact 0.223Vac.
The original number was "-"16 dBm. Just read the minus sign as meaning a
voltage less than the 0dBm reference voltage.
So that 0.223Vac reference value is divided by your 6.31 factor.
get "0.035" Vac.
So "-16dBm" is really 35mVac. This means you have more than enough drive
voltage available from your 2.5Vpp (900mVac) local oscillator signal.

Except neither you or Hernan can be sure of this. His source is not
specified to work into a 50 ohm load or present a 50 ohm source
impedance to the mixer (what is really needed). Who knows what the
delivered voltage will be when driving the mixer port?

A measurement is in order. Terminate the source in 50 ohm and measure
the power and/or voltage. If it exceeds -16 dBm, attenuate
accordingly.


Be very wary whenever you come across dBm figures. There is a minefield of
disinformation out there.
In many cases they are intended purely to obfscutate the reader and prevent
them clearly seeing that the described circuit is junk.

Spoken like a real expert on bafflegab.


In many other cases they are purposely used as an extra level of abstraction
to sort out the 'RF men' from the 'boys'.
Manufacturers still use the dB concept for historical reasons. It doesn't
effect their sales as the RF people buying their kit carry in their heads
instant dB-V conversion tables.
Don't know about everyone else but all my scopes and signal generators and
sources and dc-ac-voltmeters and DVMs and signal probes etc, are marked in
Volts and Amps. So that's what I use.
(Someday I'll get round to building a real world 1:2:5:10 50ohm attenuator.
I certainly can't buy one :-)
regards
john

This may help - url:
http://www.hardware-guru.com/LabStuff%5CdBm.htm

--
CL -- I doubt, therefore I might be !


roups.com...
Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

Actually, if it is a CMOS/HCMOS output, it might be better to terminate
it into a voltage divider of, say, 270 ohms from the osc output to 56
or 62 or 68 ohms to ground. Many of the common clock oscillators are
not intended to directly drive a 50 ohm load. Then an output taken
across the resistor to ground will look like it's from a nominally 50
ohm source. You could use a larger voltage divider ratio to get the
output down further if desired. If the osc has square wave output,
that's likely OK for a mixer input, but if it's not 50% duty cycle, you
might benefit from cleaning it up a bit with a tuned circuit, for
example. And if the oscillator has a TTL output (rather than HCMOS),
you might benefit from returning the voltage divider to the osc power
supply, presumably 5V. Beware when calculating the power delivered
from the voltage divider to a (say) 50 ohm load; put that load in
parallel with the output resistor of the divider to calculate the net
division ratio. So the suggested resistors, e.g. with 68 ohm output R,
might give you about -8dBm, if the osc delivers 2.5V P-P into the
divider, yielding roughly .24V P-P output into a presumed 50 ohm load.
-- Also, the oscillator probably has DC on its output, and you might
benefit from a blocking capacitor. 1000pF would be adequate. And
beware that the osc might deliver noticably higher voltage into a
resistive load.

Perhaps the OP could provide a bit more detail...

Also, I'd ask if that -16dBm is accurate...that's pretty low even for
active mixers.

Finally, RFSim99 is a nice little free program for playing with linear
RF circuits, and includes an "RF calculator" which has a tab for signal
levels, converting among dBm, watts, volts-RMS, and volts-P-P for a
user-specified impedance level. (I wish that tab had "locks" on the
values like the resonance one does, so you could see what power level
you get when you load a fixed voltage with various resistances, but you
can always just copy-and-paste the voltage to "remember" it over a
resistance change.) I always appreciate that RFSim99 has a lot of
tools all in one place, and I don't have to remember a whole bunch of
different programs, each of some very limited scope.

Cheers,
Tom

"Wes Stewart" wrote in message
...
On Sat, 4 Jun 2005 14:53:43 +0100, "john jardine"
wrote:


"nanchez" wrote in message
oups.com...
Hi.
I'm doing some RF experimentation and I need to know the "relation"
between dBm specificatons and voltage level for a signal.

I have a RF mixer with a specification that says:

LO drive level (50 ohm) = -16 dBm

And I have a LO source that give me an output of 2.5Vpp to a capacitive
load of 5pF at 40MHz.

How can I relate both items and design a circuit to connect LO source
to RF mixer ?

If you have some web source to study about this items, I'll be glad to
hear about it.

Thanks

Hernán Sánchez

To rejoin the real world, take the "16" figure and divide it by 20.
get "0.8"
Then find the antilog of that 0.8 [use normal 'base10' logs]
get "6.31"
This number is a multiply or divide factor that is applied to a 50 ohm
0dBm
reference voltage.
So what is this god like reference voltage?. The 50ohm 0dBm reference
voltage is in actual fact 0.223Vac.
The original number was "-"16 dBm. Just read the minus sign as meaning a
voltage less than the 0dBm reference voltage.
So that 0.223Vac reference value is divided by your 6.31 factor.
get "0.035" Vac.
So "-16dBm" is really 35mVac. This means you have more than enough drive
voltage available from your 2.5Vpp (900mVac) local oscillator signal.

Except neither you or Hernan can be sure of this. His source is not
specified to work into a 50 ohm load or present a 50 ohm source
impedance to the mixer (what is really needed). Who knows what the
delivered voltage will be when driving the mixer port?

A measurement is in order. Terminate the source in 50 ohm and measure
the power and/or voltage. If it exceeds -16 dBm, attenuate
accordingly.

I imagine we would both have read Hernans post the same way. I.e that he has
some logic running at 40MHz and is seeing a rough 2.5Vpp sinewave on his
50-100MHz oscilloscope via a 10:1 probe. From this I'd also assume we both
knew that the logic drive impedance would be a couple hundred ohms at the
most and that serious mis-matching was not going to be a problem.
I'd though, suggest he just connects the parts together and see what
happens. He's experimenting. Monitoring the results (good or bad) is just
part of the due process. Doesn't look like he's got a spectrum analyser, so
how can he validate a 50ohm test measurement as exceeding -16dBm?.

His sticking point was specifically about the link between Dbs and Volts. A
common, basic electronics question yet surprisingly badly answered by the
original suggested link, which started off with the concept of a
"dimensionless gain" and went downhill from there using 3 pages of sums.
How many radio hams talk to each other about their TX powers in terms of
(ISO standard) units of dBs wrt one watt?. Why do the filter tables tell me
a Cheb' filter ripple is in dBs when all I want is percent values. Why do my
function generators handbooks tell me the sine THD is 0.2% upto 200kHz but
beyond that the distortion suddenly becomes an obscure "-30db wrt the
fundamental".
A recent EW magazine article for a AC mV meter said the response was flat
within 0.0024dB. How flat is that?.



Be very wary whenever you come across dBm figures. There is a minefield
of
disinformation out there.
In many cases they are intended purely to obfscutate the reader and
prevent
them clearly seeing that the described circuit is junk.

Spoken like a real expert on bafflegab.


Indeed, the word was intended that way (UK English).
regards
john