[R-390] Official specs

[Bob kb8tq]

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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