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.

"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

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.

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

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.

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

From: "nanchez" on Fri 3 Jun 2005 14:59

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 ?

A basic definition that is industry-wide, government-wide,
has "dbm" as decibels of "0 dbm" related to a power level of
1.0 milliWatts in a "50 Ohm system." That has become so
widespread that specification writers don't always include
those words. It is implicit when referring to RF components.

The RMS voltage can be quickly calculated from some identities
on the basic formula for Watts: P = E x I. Knowing R (50 Ohms)
one can substitute Ohm's Law of Resistance of I = E / R into
that to get E = SquareRoot (P x R).

For 1.0 mW in a 50 Ohm system, P x R = 0.050 and the square
root of that is 0.2236 so 0 dbm has an RMS voltage of 223.6
milliVolts.

In your mixer specification, -16 dbm is equal to 35.44 mV
RMS across a resistance of 50 Ohms.

You can't DIRECTLY use your 40 MHz source value of 2.5 V
peak-peak across a 5.0 pFd capacitance because it does not
include the characteristic RESISTIVE impedance of 50 Ohms.
Power in Watts must be related to the impedance of a load
in order to perform "work." [a basic definition of power
in Watts is "a unit of work"]

Capacitance across a load will vary its impedance depending
on the frequency. For that reason the electronics industry
has long relied on a basic resistive impedance to measure
and characterize RF components. The result is the very
common "dbm" referred to 1.0 mW across a resistive 50 Ohm
load...or the characteristic impedance of the measurement
system, both source and load impedance.

To relate your mixer specification to your RF source, you
will have to put a 50 Ohm load across the source and
measure that. If you have some stray capacitance across
that load (inevitable) and know approximately what that is,
you can calculate its effect across a resistance. At 40 MHz
a 5.0 pFd capacitance has a reactance of 796 Ohms. That is
not much but it changes the magnitude of the parallel R-C
from 50.0 Ohms resistive to 47.0 Ohms slightly capacitive.
That's a small change and can generally be neglected for
experimental bench work.

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

It's in practically every textbook on the subject of RF
electronics. What can confuse newcomers to RF is the
implicit "standard" which is not always included in
specification sheets. The definition of "dbm" is arbitrary
and probably picked (way back in time) for sake of
convenience in measurement by all concerned.

The reason for picking "50 Ohms" in a "system" is more
obscure and ties into the physics of power transfer in
coaxial cables. That's a curiosity that some can look up
if they are interested but does not apply to how to USE the
"dbm" specifications. To use "dbm" one only needs to
remember the definition and apply simple forumulas for Power
and Ohm's Law of Resistance.

I hope that was of some help to you.