Hello,
This is a summary about amplifier gain measurement.
Thanks for this to Gerry Chingas, Tom Pratum, Brian Breczinski, Jeff Walton,
Dave Cross, Bill Jarrett, Thomas Dickinson,
Kirk Marat, Rich Shoemaker and Joe Dumais.
Contents:
A) Notes on the measurement
B) Short example
C) Equations for the gain measurement and a short discussion
D) Other dB scales
E) Directions to further reading
F) Original replies
A) Notes on the measurement:
1) for voltage measurements, RF circuits must be resistively loaded on 50
Ohm output (section C discusses somewhat why this is the case)
2) if terminating resistor is used is must be rated at the AMT output power
(ours is - looks like a moped engine) and frequency - as well as all other
external elements (scope, attenuator, directional coupler, cables).
3) directional couplers are not very accurate attenuators and add additional
insertion loss. It is better to use an attenuator rated for the measured
power and arrange the circuit as
AMT -->(30-50dB precision high power attenuator, e.g. from Pasternak
Enterprises)-->oscilloscope (must be rated for the voltage after the
attenuator)
4) if power attenuator is used, select low duty cycle (<10% suggested) in
order to avoid overheating the attenuator, because it's breakdown may be
costly
5) If the scope doesn't have a 50 ohm setting you need a BNC Tee with a 50
ohm terminator.
6) Cables add their own attenuation especially at freqs > 600 MHz, so for
correct measurement use shorter and low loss cables.
7) If absolute power measurement is needed, watt-meter will be more
appropriate to use, because scope may not be accurate enough in measuring
voltage and only 10% error in Vpeak-to-peak will cause big error in the
absolute power calculation. The best will be a watt-meter that works on
pulsed inputs -
8) Scope voltage measurement may depend on the oscilloscope input gain
setting
B) Short example (section C explains the equations)
meas_gain = 20*log(Vout/Vin) = 20*log(1.264/0.30) = 12.5dB
all logs are base 10
P = Vpp^2/8*R = Vpp^2/400
meas_gain = 10 log(Pout/Pin) -> get the same number.
full_gain = meas_gain + 50 dB (due to attenuationn) = 62.5dB
C) Equations for the gain measurement and a short discussion:
Gain is measured in dB's (decibels) as defined below
dB = 10 log(Pout/Pin) (Eq. 1)
where Pout and Pin are average power on the output and input of the device.
dB's are convenient because they add up
Time-averaged power (also called rms power) can be calculated from peak to
peak voltage as:
P = Vpp^2/(8*R) (Eq. 2)
where Vpp is peak-to-peak voltage, R - load impedance.
This equation derives from Ohms law V=IR, formula for instant power P=VI,
time averaging and assumption
that there is no phase lag between current and voltage - a case of purely
resistive load.
Here is where factor of 8 is coming from:
P = <V(t)*I(t)>, where angle brackets mean time averaging and V and I are
both functions of time.
Using Ohm's law we have P= <V(t)*V(t)/R>
Here we also used an assumption that there is no phase lag of current vs
voltage,
which I guess is valid at low R.
Now we'll switch to using root-mean-square voltage.
P = <V*V/R> = Vrms*Vrms/R - by definition of root mean square value
Vrms = Vmax/sqrt(2) - a fact from mathematics assuming sine wave for V
Vmax = Vpp/2 - amplitude is half of peak to peak voltage
So Vrms*Vrms = Vpp^2/8
Subst (Eq. 2) into (Eq. 1):
dB = 20 log(Vout/Vin) (3)
both voltages here are peak-to-peak, but it does not matter, since it's a
ratio.
This will work only when R is the same in the measurement of both Vin and
Vout.
In addition, a value of 50 Ohm. is accepted for most RF circuits due to some
physics constraints.
If RF output terminates into a high-impedance load, e.g. 1 Megaohm scope or
simply cable dangling in the air part of the wave will be reflected back.
Then there will be two waves traveling in the opposite directions, together
creating a standing wave (one whose crests are not moving). In this
situation voltage measured on the line will correspond to a superposition of
the waves and will depend on the position along the cable, length of the
cable and the output impedance of the source.
The scope at 1MOhm input impedance will typically show about twice the
voltage, since it sits on the point of reflection and therefore on the crest
of the standing wave.
D) other dB scales
There's dBm - same as dB but Pin=1 mW in the formula 1.
dBW - same but relative to 1W
E) Recommended reading.
Texts that discussing RF transmission lines
Tecmag site has a chart converting voltage to power
F) Original replies
-------------------------------------------------------------------------
First of all, always measure rf power into a 50 ohm termination. The scope
will then show the voltage of the incident wave coming down the 50 ohm
cable, which is sent or generated by whatever device you're connected to.
With any other termination, some power will be reflected back from the
scope, which generally results in a standing wave in the line. The voltage
at the scope will then be a sum of incident and reflected waves,and these
will generally depend on cable length and the output impedance of the source
device! (Having said this, high impedance terminations like scope inputs
will typically show roughly 2x the incident voltage as you find here.) So
disregard the 1 Mohm results.
LONG BUT DIRECT WAY: Calculate & compare input and output powers. You need
2 formulas, the first giving RMS power into 50 ohms when the peak-to-peak
voltage is given,
p(watts) = Vpp^2/8/R =
Vpp^2/400, (Eq 1)
and the second which relates the physical power ratio (the numeric quotient
of 2 power levels) to its specification in dB:
power_dB = 10 * log10 (power_ratio) (Eq. 2)
(Note that since the power_ratio of a device is output power / input power,
amplifiers will have quotients >1 and dB >0, while attenuators & couplers
with loss will have quotients < 1 and dB<0.)
Input Power: RMS power into 50 ohms is (Epp)^2/400, so
Pin = 0.3^2/400 = 0.225 mW.
Output Power: At the coupler output you have
Pcpl = 1.264^2/400 = 3.99 mW
But this ignores the (minus) 50 dB coupler attenuation. By inverting Eq 2,
you can calculate the attenuation ratio that corresponds to -50 dB:
att_ratio := 10^(#db/10) = 10^(-50/10) =
10^(-5). (Eq 3)
So the power being supplied by the AMT is
Pamt = 3.99 mW / 10^(-5) = 399
watts.
Finally the AMT power gain in dB is
G_dB = 10* log10 (399 w / 0.225
mW) = 62.5 dB
SHORT WAY: Since the power ratio is the square of the voltage ratio when
inpu and output have the same impedances, we have from Eq. 2
power_db = 20 * log10
(voltage_ratio) (Eq. 4)
The output-to-input gain in dB of the amplifier-coupler path is
G1 = 20 * log10
(1.264/0.3) = 12.5 dB
But since the amplifier output was reduced by the 50 dB coupling factor,
the amplifier gain must be 50 dB higher than the preceding: this givest the
same 62.5 dB gain as the long way.
This illustrates why dB's are used in the first place: because they're
logarithmic factors, you can find the gain of an entire chain just by
adding up the dB's of its components. (Again, assuming that all components
are terminated with the cable impedance).
SMART WAY: Do both long & short way to check results!
----------------------------------------
I am not sure I understand the 1 M Ohm measurements, certainly it seems like
you would use 50 Ohms. Anyway, taking those measurements I would say your
gain is 50 + 20*log(1.264/0.3) = 62.5 dB. Seems high... maybe there is that
much gain in your console (Varian I am guessing); 1.264 V at 50 dB is approx
400 V.
---------------------------------------
You need to measure with 50 ohm input impedance on the 'scope as that is the
impedance at which NMR (and most RF) equipment works.
Not sure what you are doing with the directional coupler and resistor; just
(1) plug the cable going to the amp directly into the 'scope (withe 50 ohm
input impedance selected) and measure Vin, then (2) put the input cable back
into the amp, connect the 50dB attenuator (I trust it is a big power
attenuator) between the amp's output and the 'scope's input and measure
Vout.
Keep your signal's duty cycle down so you don't overheat the power
attenuator. I would make the delay about ten times as long as the pulse
width.
Both the 'scope and the power attenuator need to have sufficient bandwidth.
If you measured it correctly,
gain (dB) = 20*log(Vout/Vin) = 20*log(1.264/0.30) = 12.5dB
{factor of 20 is due to deci (10) in decibel and 2 is from measuring
voltage, power = (voltage squared divided by impedance), the two comes out
as a multiplier when you take the log}
add the 50dB attenuation for 62.5dB total gain
----------------------
I use the attached chart downloaded from Tecmag's site. The chart indicates
that from .3 volts to 1.264 volts you have 11 db. Add in the 50 db of
attenuation you put in line and you get 61 db of gain. I suspect your
amplifier is rated at 60 db of gain so I believe it is up to spec. If it is
rated at a kilowatt, you could get more out by amplifying the input.
Why the directional coupler? it probably has an insertion loss of .1 to 1 db
as well. I would go from the amp to the 50 db attenuator into the scope if
you are measuring the amplifier.
Also, to get accurate voltages, you have to match the impedance of the
source and scope - generally 50 ohms in NMR equipment. If the scope doesn't
have a 50 ohm setting you need a BNC Tee with a 50 ohm terminator. Otherwise
you get standing waves on the coax and you could be measuring at a voltage
max or min depending on the frequency and length of coax.
-------------------------
At a glance you seem to have things worked out more or less OK…….about the
only one that does not make too much sense is your input level, as this
implies about -6 dBm. Full power on most NMR amps is normally about 0 dBm on
their input, or 632mV PP. However, I would say there is probably a little
measurement error somewhere as your output power looks more or less normal
for a liquids AMT amp running 50 MHz or thereabouts for 13C, i.e. around
400W, which is your 1.26V PP plus the 50 dB pick off. (Assuming all this
stuff is also OK.)
One of the problems with RF is actually measuring it accurately. You have
probably figured this out already by having mentioned your cable was 6ft
implying you are already wary……changing this cable may well give you a
different absolute result. However, in terms of gain your amp appears to be
purring!
I am a lazy guy so I have included a scanned old RF power table for your
reference too. Why calculate when you can just read off the corresponding
powers from your voltage readings….
---------------------------
Here's what I get:
Assuming a 50 db attenuator was used for measuring drive and output:
Input power:
[square(0.3) divided by 400]*1e5 = 22.5 Wrms
Output power:
[square(1.264) divided by 400]*1e5 = 400 Wrms
Gain:
Gain = 10 log(output/input),
Gain = 10 log(400/22.5) = 12.5
This is a reasonable gain value.
---------------------------
Please find attached an RF power chart.
I typically measure RF with a 400MHz or higher O-scope on the 50 Ohm input.
I use a 20 or 30 db attenuator when measuring the output side of the AMT. No
attenuation is needed on the PTS, Transmitter or attenuator board
measurements.
For determining the AMT output:
Place a minimum 20 db attenuator on the O-scope input. Connect an RF cable
from the AMT output to the scope input attenuator.
Pulse the spectrometer and determine the peak to peak voltage output with
tpwr set to full power, 63db.
Find the p/p voltage on the chart and note the db value just left of it. Add
the 20 db attenuator to that value and follow up the chart to determine the
actual output.
Example: p/p voltage measured is 7volts = 30db on the chart. add 20 db for
the attenuator. actual db is 50 db, 70.7 vp/p = 100 watts.
At higher frequencies a fast scope must be used because of attenuation loss
and the result is faulty measurements.
---------------------------
First, only the 50 Ohm measurements have any validity. At high frequencies
(where wavelength is comparable to cable length) terminating a cable in
anything other than its characteristic impedance turns the cable into an
impedance transformer (check any text that discusses transmission lines for
a discussion).
I'm assuming that your measurements are V peak to peak. You need RMS
voltage for power measurements.
At point a, you have 0.3 V p-p. or 0.3 * 0.5 * 0.707 = 0.106 V RMS.
Since P = (E**2)/R the power is (0.106**2)/50 = 0.224 mW or -6.5 dBm.
A point b you have 1.264 * .5 * .707 = .447 V RMS = (0.447**2)/50 = 3.99 mW
= 6.0 dBm. Since you have effective a 50 dB attenuator (the directional
coupler) you have 56 dBm of output from the amplifier. This gives you a
gain of 56 - (-6.5) = 62.5 dB.
Let's calculate the output power of the amp. 56 dBm is 26 dBW, which is ca
400 W. This seems a bit high (but in the correct ballpark) for that amp.
What is the "tpwr" setting that you are using? Our AMT amplifiers typically
give 200 W or maybe 250 W if you drive them into compression. I suspect
that using the directional coupler as an attenuator may be the problem.
They are not terribly accurate in that sort of application - they could
easily be off 3 dB. A precision high power attenuator (e.g. 30 or 40 dB
with a 1 kW peak rating) is much better (They are available from Pasternak
Enterprises, which is right in your "neck of the woods"). Also, that 50 dB
figure only applies when the line is close to properly terminated.
Another potential problem with your set-up is the "50 Ohm coax resistor"
that you are using to terminate your line. Is it rated for the full power
of the AMT amplifier? The typical little ones (the size of a BNC
connector) are only rated at typically 1 W continuous, but they also have a
peak (voltage breakdown) rating. If you fry this resistor your line is no
longer terminated in the characteristic impedance (The resistor could be a
dead short or more likely completely open, depending on the breakdown mode).
rendering the "50 dB" rating of the directional coupler problematic. You
need to get a terminating resistor that can take the power. A high power
attenuator will serve as both the terminator and attenuate the signal down
to the range that your scope can handle.
------------------------------
Since you posted this on Friday, you probably already received multiple
answers, but just in case, I’ll reply to you anyway.
The AMT amplifiers are designed to give their maximum rated (linear) output
with 0 dBm input (which would be 0.632 VPP into 50 Ohms). So, in other
words, a 500 Watt amplifier should output (nominally) 500W (which is ~57
dBm) or 448 VPP into 50-Ohms, when the input is 0.632 VPP. Through a 50dB
attenuator, 448 VPP would be 1.417 VPP on your scope. (If it is calibrated
correctly, your 50dB attenuator corresponds to a voltage attenuation of
1/316.147).
In your case, your input value of 0.3V (into 50-Ohms) is about -6 dBm, and
your forward output is +56 dBm, (400 VPP after taking into account your
50dB attenuator) so your gain is closer to 62 dB. Therefore, you must be
looking at a 1KW amplifier, which should give you 60dB of gain, or 60dBm
output with 0dBm input.
Conclusion: if you made your measurements properly your amplifier seems to
be working fine, simply in terms of gain. Whether or not the Rf output is
clean is another matter altogether. Also, you should always measure Rf of
an NMR system using 50-Ohm termination. You should never use the 1M-Ohm for
Rf with a 50-Ohm impedance matched system.
-------------------------
Did you get and answer on this? I can help you if you are still in need but
I am really busy this week. One quick thing I want to mention is that you
cannot really measure power with a scope unless you know that the output
voltage is accurate at that frequency. You are more likely to be able to
ratio the input and output levels and get amplifier gain but the problem is
that attenuators are not necessarily accurate and if in the course of your
measurement you have to change the scope gain setting the exact gain for two
different settings may not be correct. The way to measure power is to use a
watt meter (and one that can handle pulsed power as opposed to the more
common CW meters) from a company like Bird but they are expensive
considering that the only thing they do is measure power.
An additional consideration is that long cables can cause significant
attenuation at higher frequencies, roughly 600 and above. Low loss cables
need to be used and a long standard cable can cut well over a dB off of the
power.
A convenient chart to have around an NMR lab is the one Chemagnetics used.
Chemagnetics shipped a 30 dB (rated at something close to 100 watts)
in-line attenuator for this purpose. This table is set up so you can read
across to see the +30 dB level and thus account for the attenuator. But you
can take your 50 ohm voltage measurement and find the dBm value, add the
appropriate dB's to that value and read across to the power. So you
measured a 1.26 volt output, that is about 6 dBm and if your attenuator is
truly 50 dB you can add 50 dB and look at 56 dBm to get 398 watts. But note
that if you are off on your voltage measurement by about 10% and that
changes your starting value by about +/- 1 db your power then ranges from
about 320 watts to 500 watts, a pretty big range
--
Evgeny Fadeev, Ph.D.
Director, BioMolecular Spectroscopy Facility
1212 Natural Sciences 1
University of California Irvine
Irvine, CA 92697
telephone: 949-824-5842
http://www.physics.uci.edu/~biomolenmr<http://www.physics.uci.edu/%7Ebiomolenmr>
http://nmrwiki.org - Share your magnetic science!
Received on Wed Jan 20 2010 - 14:54:46 MST