[R-390] Official specs

Sun Oct 27 11:24:18 EDT 2024

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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