[R-390] Official specs
Bob kb8tq
kb8tq at n1k.org
Thu Oct 24 20:41:04 EDT 2024
Hi
You can indeed find standard measures like:
db V = db above / below one volt
https://www.everythingrf.com/community/what-is-dbv#:~:text=dBV%20or%20decibel%20Volt%20is,a%20voltage%20below%201%20V.
db uV = db above / below one microvolt (and milivolt and …. )
The cable TV folks like voltage based measurements. There are other areas that seem to go that way as well.
Is this mis-use of the term? If so, it’s a *very* commonly accepted mis-use.
Bob
> On Oct 24, 2024, at 4:05 PM, Ing. Giovanni Becattini via R-390 <r-390 at mailman.qth.net> wrote:
>
> Thanks! I think I have understood now. As you understood, my mistake was thinking that dB could be used for every ratio of congruent values.
>
> Very happy.
>
>> Il giorno 24 ott 2024, alle ore 21:57, Jim Whartenby <old_radio at aol.com> ha scritto:
>>
>> Gianni
>>
>> Answer to 1 below: dB is defined as a POWER ratio. The use of voltage to calculate dB assumes that the input and output resistances of any electrical system being measured are the same. Take the formula that defined the dB: dB = 10 log (power out / power in). Substitute (voltage^2 /R) for each term "power" then simplify the equation. You should end up with 20 log (voltage out / voltage in). This simplification only works if Rout = Rin. WRT the DA-121, the out resistance does not equal the input resistance or 125 ohms does not equal 50 ohms. So the "voltage" form of the dB equation results in an error in certain cases where the power form never will.
>>
>>
>> Answer to 2 below: you make the point that Rin is 50 ohms and Rout is 125 ohms but then ignore what this means. See Answer 1 above. Also your Ohm's Law calculation for what the R-390 "sees" is in error. (50 in parallel with 68) +100 in parallel with 125 is 63.44 ohms, is it not?
>>
>>
>> Answer to 3 below: the equation you used is wrong, for voltage it is 20 log X not 10 log X so 5 dB would be correct if both input resistance and output resistance are the same but you point this out in your 2 below that they are not equal resistances. Again, see Answer 1 above.
>>
>>
>> Answer to 4 below: you are correct, the loss in the DA-121 is 9 dB when you consider that the input impedance is different from the output impedance and calculate accordingly.
>>
>>
>> Your error is in thinking that there are two different definitions for dB, power and voltage. There is only power, the voltage form of the equation has the caveat that the input resistance must be the same as the output resistance. You are not the first one to make this mistake and surely you will not be the last.
>>
>>
>> Regards, Jim
>>
>>
>> Logic: Method used to arrive at the wrong conclusion, with confidence. Murphy
>>
>>
>> On Thursday, October 24, 2024 at 12:30:03 PM CDT, Ing. Giovanni Becattini <giovanni.becattini at icloud.com> wrote:
>>
>>
>> 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> 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> 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>> 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>> 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>>
>>>>> Envoyé : 23 octobre 2024 16:49
>>>>> À : Jacques Fortin <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>>; r-390 at mailman.qth.net; Larry Haney <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>> 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> <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>> Cc : r-390 at mailman.qth.net;
>>>>>> Larry Haney <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>> 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>> 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>> 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>> 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>> 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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>>>
>>>>>>>
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>>>>>
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