[R-390] Official specs
Jacques Fortin
jacques.f at videotron.ca
Thu Oct 24 15:17:10 EDT 2024
Hi Gianny,
I am not sure that anybody can create the “dB they want” to express the ratio of anything as a logarithmic quantity.
Many “dB units” exists, like dBm (power related to 1 milliwatt), dBspl (sound pressure level), and many more but all they have in common is to express a power ratio.
I never thought about expressing a plank length in dB, but it made me think about how our good old slide rules worked !
For the reasoning, that’s correct.
0.5676 and -8.898 dB in fact…
Because the DA-121/U “load” to the SG is a little higher than 50 ohms.
But both figures refer to the SG “normal” output when loaded by 50 ohms
73, Jacques, VE2JFE in Montreal
De : Ing. Giovanni Becattini <giovanni.becattini at icloud.com>
Envoyé : 24 octobre 2024 14:37
À : Jacques Fortin <jacques.f at videotron.ca>
Cc : Jim Whartenby <old_radio at aol.com>; r-390 at mailman.qth.net
Objet : Re: [R-390] Official specs
Thanks Jacques, I follow your explanation.
If I correctly understand, you are telling me that dB can be used only between two power values, and not between two generic homogeneous values, like voltages (I thought that a voltage is a voltage, independently of the impedance where it is developed, and so, for example, I could use dB also for weight, lengths or whatever else, as long as homogenous).
But is the rest of the reasoning correct? I.e, 0.556 voltage ratio and -9.078 dB power loss?
Thanks again
Gianni
Il giorno 24 ott 2024, alle ore 20:25, Jacques Fortin <jacques.f at videotron.ca <mailto:jacques.f at videotron.ca> > ha scritto:
Gianni,
Sorry to mention that, but when you wrote:
"So, when two voltages has a ratio of 0.556, their ratio, expressed in dB, is 10 log (V1/V2) = -2.5 dB."
Eeeesh...
Using the dB based on Voltage Ratios is "legal" only if the two voltages occurred in the same resistance value, and the "formula" is 20 log (V1/V2).
So equal to 10 log (V1/V2)^2...
In the case of the DA-121/U, mentioning that this "matching pad" attenuation is 5 dB is an improper use of the "20 log (Vout/Vin)".
But don't worry: you are not alone (I fell into it myself too many times).
I see almost every time the gain of audio amplifiers specified in dBs, despite that the input and output impedances are totally different.
The BEL was originally a subjective sound-level measurement unit, corresponding to a 2:1 perceived sound loudness ratio.
It was later rationalised as a power ratio of 10:1 and fractioned in 10 smaller units, to give the decibel measurement unit we know now.
73, Jacques, VE2JFE in Montreal
-----Message d'origine-----
De : <mailto:r-390-bounces at mailman.qth.net> r-390-bounces at mailman.qth.net < <mailto:r-390-bounces at mailman.qth.net> r-390-bounces at mailman.qth.net> De la part de Ing. Giovanni Becattini via R-390
Envoyé : 24 octobre 2024 13:30
À : Jim Whartenby < <mailto:old_radio at aol.com> old_radio at aol.com>
Cc : <mailto:r-390 at mailman.qth.net> r-390 at mailman.qth.net
Objet : Re: [R-390] Official specs
Thanks Jim!
I try to explain my thought and I would be grateful to whom would tell me at which point I make the mistake.
1.What dB is? For what I studied, dB is just a ratio between two congruent values, so you can define a dB for voltage or for power.
2.DA-121 output voltage. Let’s stop to voltage values for now and work almost in DC, assuming that the R390 has really a 125-ohm impedance.
In this case:
the generator "sees" 50 ohms impedance (68 ohms parallel to 100+125); the receiver "sees" about 125 ohms (100 ohms + 68 ohms in parallel to 50 ohms).
The ratio between the input and the output voltage is easy to calculate just with the ohm law. The output voltage is the input voltage times 0.556. If you don’t want to make calculations see https://ohmslawcalculator.com/voltage-divider-calculator. It’s easy to practically check that this is true.
I believe that so far nobody could disagree.
3.Voltage dB value. So, when two voltages has a ratio of 0.556, their ratio, expressed in dB, is 10 log (V1/V2) = -2.5 dB. I am just applying the dB definition (forget 5 dB that was an error of mine).
4.Power dB value. If we want to calculate the power ratio we have (using 1V as input value):
P1 = input power = V^2 / R = 1V / 50 ohm =0,02
P2 = output power = 0.556^2 / 125 ohm = 0.0025
So the power ratio in db is 10 log (P1/P2) = -9.078 dB. Again, this is simply the dB definition.
Where am I wrong?
Thanks for your help.
Gianni
Il giorno 24 ott 2024, alle ore 18:08, Jim Whartenby <old_radio at aol.com <mailto:old_radio at aol.com> > ha scritto:
Gianni
The dB is defined as a measure of the ratio of powers. In electronics, it is the power output to power input.
Using the formula 10 log (Pout / Pin) will always give the correct answer. This formula considers the resistance of both the input power and output power separately since power = voltage^2 / resistance.
The 20 log Vout/Vin will only give the correct answer when both the input resistance and the output resistance are the same.
So the 5 dB calculation using input and output voltage is wrong. In this particular case the resistance transformation provided by the attenuator was not considered. If the attenuator did not cause a change in resistance (the input resistance and output resistance are the same) then using 20 log (Vout / Vin) would result in the same answer as 10 log (Pout / Pin).
Regards, Jim
Logic: Method used to arrive at the wrong conclusion, with confidence.
Murphy
On Thursday, October 24, 2024 at 02:26:08 AM CDT, Ing. Giovanni Becattini via R-390 <r-390 at mailman.qth.net <mailto:r-390 at mailman.qth.net> > wrote:
Hi Guys,
sorry I was not considered - I hate to be tiring, but an answer would be great to help me better following the discussion.
It seems to me that 5 dB in voltage and about 8.9 dB in power are very easy to demonstrate (keeping out the frequency effect), either arithmetically, either practically, either with LTSpice.
At least on this point …. do the parties agree? I could not understand this. Perhaps, if we don’t agree on this, it is hard to go further.
No problem if you don’t want to answer...I'm not touchy 😁
Gianni
Il giorno 24 ott 2024, alle ore 02:18, Barry Scott <72volkswagon at gmail.com <mailto:72volkswagon at gmail.com> <mailto:72volkswagon at gmail.com>> ha scritto:
GR-1001A. Even funner.
Thanks,
Barry - N4BUQ
On Wed, Oct 23, 2024 at 4:46 PM Jacques Fortin <jacques.f at videotron.ca <mailto:jacques.f at videotron.ca> <mailto:jacques.f at videotron.ca>> wrote:
Bob, I see your point, but we know everything about that too !
That's bring me back to the end of the '70s when my workbench signal generator was a Marconi Instruments TF955/5.
Output calibrated in Volts EMF (open circuit voltage) and 75 ohms of output impedance !
Using it, you had to compute what was the real load voltage you got all the time...
73, Jacques, VE2JFE in Montreal
-----Message d'origine-----
De : Bob kb8tq <kb8tq at n1k.org <mailto:kb8tq at n1k.org> <mailto:kb8tq at n1k.org>> Envoyé : 23
octobre 2024 16:49 À : Jacques Fortin <jacques.f at videotron.ca <mailto:jacques.f at videotron.ca>
<mailto:jacques.f at videotron.ca>> Cc : Ing. Giovanni Becattini
<giovanni.becattini at icloud.com <mailto:giovanni.becattini at icloud.com>
<mailto:giovanni.becattini at icloud.com>>; r-390 at mailman.qth.net <mailto:r-390 at mailman.qth.net> ;
Larry Haney <larry41gm2 at gmail.com <mailto:larry41gm2 at gmail.com> <mailto:larry41gm2 at gmail.com>>
Objet : Re: [R-390] Official specs
Hi
If you start with a Spice model and put in a 2V source with some series resistance. You loose 6 db when you get to 1.0 V. That’s how Spice looks at things.
If you start with the signal generator convention you have a “spice” 2V source and a 50 ohm resistor “inside the box”. Put on a 50 ohm load and you have 1V. That’s your zero db point with the signal generator.
Start one way and you are 6 db down.
Start the other way and you are at zero db.
Bob
On Oct 23, 2024, at 1:03 PM, Jacques Fortin <jacques.f at videotron.ca <mailto:jacques.f at videotron.ca> <mailto:jacques.f at videotron.ca>> wrote:
Hello Bob,
I 100% agree that the convention on a RF source is that the displayed output voltage is valid only when the SG is loaded with the correct termination impedance.
I disagree however that the outcome "depends" on what can be used to provide the answer.
If anyone is ignorant enough to use a Spice model that not include a source output impedance, it is sure that the result of such "simulation" will be different from what is obtained with properly set up test equipment.
A properly used simulation software results will not be different that what can be obtained with "real" instruments.
If ever this is the case, the inputs to the simulation program are faulty.
Bob Pease (RIP) once fell in that trap: he took revenge by throwing his computer from the top of the building into the parking lot below.
But, at the end of the day, nothing is more practical than a good theory.
73, Jacques, VE2JFE in Montreal
-----Message d'origine-----
De : r-390-bounces at mailman.qth.net <mailto:r-390-bounces at mailman.qth.net>
<mailto:r-390-bounces at mailman.qth.net>
<r-390-bounces at mailman.qth.net <mailto:r-390-bounces at mailman.qth.net>
<mailto:r-390-bounces at mailman.qth.net>> De la part de Bob Camp
Envoyé : 23 octobre 2024 12:26 À : Ing. Giovanni Becattini
<giovanni.becattini at icloud.com <mailto:giovanni.becattini at icloud.com>
<mailto:giovanni.becattini at icloud.com>> Cc :
r-390 at mailman.qth.net <mailto:r-390 at mailman.qth.net> ; Larry Haney <larry41gm2 at gmail.com <mailto:larry41gm2 at gmail.com>
<mailto:larry41gm2 at gmail.com>> Objet : Re: [R-390] Official specs
Hi
The very basic issue here turns that into a “that depends” answer:
If you are playing with a Spice model, and starting from the voltage on your ideal source, you get one answer.
If you are working at RF and using a signal generator (and doing things properly) you get a very different answer.
The convention on an RF source is that the “starting voltage” is measured with the correct termination in place.
Since we’re talking about RF …. that’s how it would be done.
Bob
On Oct 23, 2024, at 10:39 AM, Ing. Giovanni Becattini via R-390 <r-390 at mailman.qth.net <mailto:r-390 at mailman.qth.net> <mailto:r-390 at mailman.qth.net>> wrote:
Hi,
I find this topic very intriguing, so I cannot help but return to this discussion.
As I told you, I don't consider myself an RF expert or simply a 390 expert, but I have had to solve complex engineering problems many times in my life. So I think we should first create a mathematical model that is as simple as possible, i.e. without taking into account the frequency effect. Once the model works, we can try to make it more real with the right corrections.
So I would like to ask a question to see if we are on the same page:
Do you agree that if the R-390A were a perfect 125 ohm resistor and we were working at 1 kHz, the DA-121 would attenuate 5 dB in voltage and 8.98 in power?
Greetings
Gianni
Il giorno 23 ott 2024, alle ore 16:15, Larry Haney <larry41gm2 at gmail.com <mailto:larry41gm2 at gmail.com> <mailto:larry41gm2 at gmail.com>> ha scritto:
Jim, I read your referenced post a few times looking for the
answer we're all looking for, what the microvolt level is that
is going into the 390 for a given level going into the DA-121,
but I couldn't find it. All I read was a bunch of db numbers
that don't make any difference. We need to know about the uV
levels. You can talk about insertion losses all you want, but that does not tell us what the uV level is that is going into the 390.
I used my URM-25D to generate a 1 MegaHertz 50 uV signal into
the
DA-121 and got 28 uV going into the 390 (that's a 44% reduction
of signal from the 25D as measured with my HP 400FL RMS RF AC
voltmeter). Nothing else matters. The calculation is very simple:
50 - 28 = 22, 22 / 50 = .4444 or 44.44%. That means that 55.55%
of the signal from the SG is getting to the 390. The accuracy
of my 400FL is +/- 1%. All my signal measurements were in RF RMS volts measured with my HP 400FL.
The ONLY DC measurements I made were to measure the resistances
in the
DA-121 and mine are a 70 ohm shunt and a 100 ohm in series.
These are close to the documented values of 68 ohms and 100 ohms.
So, what uV level of signal do you MEASURE (not calculated or
theorized) going into the DA-121 and going into the 390. Let's
keep it simple and stick to *MICROVOLTS* because that is what
the sensitivity and signal to noise ratio measurements use, NOT db.
By the way, since this test is all about the DA-121, you should
be using the documented resistor values in it for testing (68 and 100 ohms).
And contrary to what you said, my DC circuit calculations
(resistance and estimated signal loss) do agree with my RF
measurements. The resistance calculation is: 100 ohms / (100 + 125 ohms) = .4444. That's a 44% loss.
To get the signal level at the 390, multiply the SG output by 56%.
And I did not calculate any db loss, the 5 db loss is what my 400FL says it is.
For anyone wanting to make their own DA-121, use what's
documented in it, a
68 ohm shunt and 100 ohm series resistor. Otherwise you will
get a different answer from those that use a real DA-121.
Regards, Larry
On Tue, Oct 22, 2024 at 3:36 PM Jim Whartenby <old_radio at aol.com <mailto:old_radio at aol.com> <mailto:old_radio at aol.com>> wrote:
Larry
I built a test fixture that is essentially two DA-121's
connected back to back. Photos and drawing are enclosed. This
does the conversion from 50 ohms to 125 ohms and then back to
50 ohms. I used 1% resistors to make the attenuator circuit with the values close to those found here:
https://k7mem.com/Res_Attenuator.html
The closest I could come to the 64.18 ohms result from the
attenuator calculator was 63.9 ohms. This is from the parallel
combination of 3 each
237 ohm in parallel with a 1k, in parallel with a 499 ohm resistor.
Five resistors in parallel, all 1% resistors. The result was
63.85 ohms, a 0.5% error. The sub for the 96.83 ohm resistor
is a 100 ohm 1% resistor (3%
error) and the sub for the R-390's 125 ohm impedance was a 121
ohm 1% resistor (3% error). This is still much better then the
5% resistors used in the original DA-121.
For a test oscillator I used a Helper SM-1000 signal generator
and measured the insertion loss with a Stoddart NM-25T
frequency selective voltmeter. The insertion loss was measured
at 10 MHz using two 4 foot BNC
RG-58 coax cables from Pomona Electric. 4 foot of coax from
the
SM-1000 to the test fixture and another 4 feet from the test fixture to the NM-25T.
The SG was set for a reading of 30 dB on the NM-25T signal
strength meter when measuring a BNC through connection and then
measured 11 dB when the test fixture was installed in place of
the BNC through. The insertion loss for the test fixture is 19
dB. Dividing this by two since there are essentially two
DA-121s back to back gives an insertion loss of about 9.5 dB
for a single DA-121. This closely agrees with the attenuator calculator findings.
So it seems that your DC circuit calculation do not agree with
the RF measurements. Transmission lines behave differently
then DC circuits. You calculate a 5 dB insertion loss, I measure a 9.5 dB insertion loss.
Here is an experiment that you can try. Insert a 50 ohm
resistor in parallel with the 50 ohm coax. What do you think will happen?
Perhaps nothing since the coax is 50 ohms and the resistor is
also
50 ohms? In reality, the coax has reactive elements, parallel
capacitance and series inductance that make up the coax impedance.
Neither of which will dissipate the signal carried on the coax.
The only losses are from the resistance of the conductors that
make up the coax. Adding a parallel resistor will attenuate the signal to the receiver by 3 dB.
If anyone on this list wants to make their own version of the
DA-121, I can supply the resistor values I used for a token $2
plus postage. Just DM me with your address and if you want one or two resistor sets.
Regards, Jim
Logic: Method used to arrive at the wrong conclusion, with confidence.
Murphy
On Friday, October 18, 2024 at 05:36:08 AM CDT, Larry Haney <
larry41gm2 at gmail.com <mailto:larry41gm2 at gmail.com> <mailto:larry41gm2 at gmail.com>> wrote:
Hi Jim, I just checked and I only have 1 da-121. As for
insertion loss, my coax is very short and the connections are
very good so the loss there would not be possible for me to
measure. Now for the insertion loss due to impedance mismatch
(due to resistance
variations) would also not be possible for me to measure, as I
don't have the equipment required for that. But, because the 3
resistors in the circuit are very close to the required values
for a perfect
50 ohm match to the sig gen, I am sure that the insertion loss
due to that very slight impedance mismatch is extremely small.
I have no way to measure that loss as I don't have the 3 exact
value resistors to compare it to. I could calculate it, but I believe that would be a waste of time without being able to measure it.
After all the input you have given me and the research just
done, I'm satisfied with my current measurements and calculations (IE:
the output voltage of the da-121 is 56% of the input voltage
when the load is
125 ohms).
My biggest concern about making snr measurements is for those
folks that don't have a recently calibrated sig gen or
calibrated rms AC voltmeter to verify their readings with.
Regards, Larry
On Thu, Oct 17, 2024 at 1:55 PM Jim Whartenby <old_radio at aol.com <mailto:old_radio at aol.com> <mailto:old_radio at aol.com>> wrote:
Larry
No, just one SG and one 125 ohm load. You should be able to
determine the total loss through two DA-121 attenuators
connected back to back with an o'scope and then divide the loss by two to solve for the insertion loss.
Jim
Logic: Method used to arrive at the wrong conclusion, with confidence.
Murphy
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