[R-390] Official specs

Bob kb8tq kb8tq at n1k.org
Sun Oct 27 17:50:05 EDT 2024 

Hi

That’s sort of the point to this. Truth or not, there is a procedure:

Put in this resistor based gizmo.

Set the signal generator to 3 uV (or whatever).

Is the sensitivity in spec?

If so …. move on.

Bob

> On Oct 27, 2024, at 11:24 AM, Ing. Giovanni Becattini <giovanni.becattini at icloud.com> wrote:
> 
> Hi,
> 
> From what I read, the problem seems to have no practical solution, and the Truth is far away. 
> 
> But, more practically, why not to agree on a test set-up and procedure for R-390A sensitivity measurement? It will not tell us… the Truth, but at least we could have a common ground to check what I wanted to know when I opened this post, i.e., if I tuned up correctly my receiver or if I have to insist.
> 
> We will not talk about microvolts but at least of qthmicrovolts…😁
> 
> Gianni
> 
> 
> 
>> Il giorno 27 ott 2024, alle ore 15:09, Bob kb8tq <kb8tq at n1k.org> ha scritto:
>> 
>> Hi
>> 
>> We know that the R390() does not supply a 125 ohm load to the antenna 
>> or the test setup. Based on doc’s shown earlier, it typically is way off from
>> 125 ohms. 
>> 
>> This is not at all uncommon in the world of receivers. 
>> 
>> Thus the *assumption* that the radio supplies a 125 ohm load is suspect. 
>> 
>> Welcome to why “1 uV” out of any signal generator probably is not what the 
>> input to the radio actually has applied to it. 
>> 
>> Do people head off and work out what’s “really there”? You could work it out
>> various ways.  That’s not how the spec on the radio is written. If the signal 
>> generator says 1 uV that’s the correct number to use. 
>> 
>> How is this relevant?
>> 
>> If I hook up a 50 ohm generator directly to the input of the R390(), it is running
>> from a 50 ohm source. Based on the doc’s shown a wile back, the input to
>> the radio is *always* higher than 50 ohms (and often by quite a bit). Loading
>> will have a very different impact on that 50 ohm source than on a 125 ohm 
>> source. 
>> 
>> If you *do* want to work this out in the “real case” ( = radio hooked up) 
>> *and* you want to do it only based only on power : You have a whole lot of work 
>> to do. One (as yet unmentioned) part of that is the input to the radio has a reactive 
>> component. That messes a bit with power math. 
>> 
>> Bob
>> 
>>> On Oct 27, 2024, at 9:40 AM, Larry Haney <larry41gm2 at gmail.com> wrote:
>>> 
>>> Jim,  I agree with this posting of yours *except* for the 1st and last
>>> statements.
>>> 
>>> 1.  First you said: 'What has been overlooked is that there is an impedance
>>> transformation from 50 to 125 ohms.'  We are all very aware of this fact.
>>> 
>>> 2.  Lastly you said: 'To convert the* SG voltage output* into the voltage
>>> actually seen by the R-390, multiply the SG reading by *0.1235* or divide
>>> the SG reading by 8.097, either way works.'  That is *not right* at all.
>>> You just went through a nice step by step explanation about how to
>>> determine the power loss, then you use that power loss ratio (0.1235) to
>>> determine the voltage seen by the 390.  *Wrong, wrong, wrong.*  The last 3
>>> steps in your procedure are: 1. dB = 10 Log ^ (.00247watts / 0.02 watts),
>>> 2. dB = 10 Log ^ 0.1235, 3. dB = -9.083.  *No real disagreement there*.
>>> The input watts to the da-121 = 0.02 watts, the output watts from the
>>> da-121 = .00247 watts, that's a 12.35% loss of *power* in watts, not
>>> voltage.  You *can not* use the 0.1235 *power loss* relationship to
>>> directly calculate the *voltage loss* relationship of the da-121 as you are
>>> doing in your last statement.
>>> 
>>> One way to correctly calculate the voltage coming out of the da-121 (Vout),
>>> would be to use the formula:
>>> 
>>> Vout = Sqr rt (Pout (watts) x impedance (ohms))
>>> 
>>> Where Pout is the power coming out of the da-121 (in this case, 0.00247
>>> watts) and impedance is the da-121 load impedance provided by the 390, 125
>>> ohms.
>>> 
>>> Vout = Sqr rt (.00247 x 125) = 0.5556 Volts
>>> 
>>>   .00247 x 125 = 0.30875
>>>   Sqr rt  0.30875 = 0.5556
>>>   Vout = 0.5556 volts
>>> 
>>> Vout is what's going into the 390 (in this scenario).
>>> 
>>> Regards, Larry
>>> 
>>> 
>>> --------------------------------------------------------------------------------------------------------------------------------------------------
>>> On Wed, Oct 23, 2024 at 9:35 AM Jim Whartenby <old_radio at aol.com> wrote:
>>> 
>>>> What has been overlooked is that there is an impedance transformation from
>>>> 50 to 125 ohms.  Any calculation that ignores this transformation is in
>>>> error.  The only solution that accounts for different impedances is by
>>>> looking at the respective powers at both input and output.
>>>> 
>>>> 
>>>> 1 volt into the DA-121 gives 0.556 volts out.  Looking at the power-in
>>>> verses power-out using the respective impedances:
>>>> 
>>>> 
>>>> Power = voltage squared / resistance
>>>> 
>>>> Pin = 1 volt ^2 / 50 ohms = 0.02 watts
>>>> 
>>>> Pout = 0.556 volt ^2 / 125 ohms = .00247 watts
>>>> 
>>>> dB = 10 Log ^ (Pout / Pin)
>>>> 
>>>> dB = 10 Log ^ (.00247watts / 0.02 watts)
>>>> 
>>>> dB = 10 Log ^ 0.1235
>>>> 
>>>> dB = -9.083
>>>> 
>>>> 
>>>> To convert the SG voltage output into the voltage actually seen by the
>>>> R-390, multiply the SG reading by 0.1235 or divide the SG reading by 8.097,
>>>> either way works.
>>>> 
>>>> 
>>>> Regards, Jim
>>>> 
>>>> Logic: Method used to arrive at the wrong conclusion, with confidence.
>>>> Murphy
>>>> 
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