[R-390] Official specs

[Larry Haney]

No Jim,  We need to know what the voltage loss is across the da-121 so we
know what the voltage is that is being fed into the 390.  It can be derived
from the 'power insertion loss', but is much more complicated and time
consuming to figure it out that way.

Regards, Larry

On Sun, Oct 27, 2024 at 8:21 AM Jim Whartenby <old_radio at aol.com> wrote:

> Larry
>
> What is the insertion loss of the DA-121?  That is what we really want to
> know, isn't it?
>
> Jim
>
>
> Logic: Method used to arrive at the wrong conclusion, with confidence.
> Murphy
>
>
> On Sunday, October 27, 2024 at 08:40:26 AM CDT, 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
>
>