Disaster with a Marconi TF893A Output Power Meter

[AlanBeckett]

Lawrence,

The Manual, and the Front Panel, says 5R. It's Transformer coupled.

Alan

[pwdrive]

Hello Alan, this sought of conundrum normally does my head in, I would guess the true test is to ensure the source and the load are correctly matched in impedance terms and then do the measurement, not to sure how accurate the Avo 8 is over a given frequency range, mines got a definate peak and roll off but can't remember the details, if the source and the load are correctly matched then maybe try measuring the peak input with a scope and work out the RMS from that and thus the power in from the resultant.

Lawrence.

[AlanBeckett]

Lawrence,

I'll have to ponder. I'll get my super bench DVM out and do some more serious measurements, starting with the Output Impedance of the J-2.

Alan

[pwdrive]

Alan, if we thought hard enough I dare say we could come up with the answer! Come in Al, he's usually good on this stuff.

Lawrence.

[Mark Hennessy]

Think I'd want to use something with an RMS converter in it - AVOs are mean-sensing. And I'd observe the waveform on a 'scope to make sure there is nothing funny going on.

How pure a resistance will the Marconi present? Observe the voltage across the test resistor (which obviously gives current), and compare it to the voltage waveform. Moving the resistor if necessary to deal with the common earth of the two 'scope channels, needless to say...

[Skywave]

While I was at it I thought I'd better make sure that the Calibration is at least reasonable. I shorted out the duff switch position, fed it with 1kHz from my J-2 and set the Output to give a reading of 100mW on the Marconi. An AVO 8 across the Input showed 0.9Vrms. Assuming that the Marconi Input Impedance really is 8R then that shows an error of 1.25% - well within what I can measure or trust. So far so good, but what is the Input Impedance of the Marconi? I put an R-Box - a very good one - between the J-2 and the Marconi and set it to show half power on the Marconi, while keeping the Output of the J-2 constant - it does vary with Load. The answer was 5R!
Anyone care to explain that?
Alan

Presumably your R-Box consisted of a single resistance inserted in series with the J2 source and the TF 893A load. In which case, my analysis proceeds as follows.

Let the source voltage within the J2 be Vs; let its internal resistance be Rs; let the value of R-Box be R, (which at the moment we will say we don't know the value of). Let the input resistance of the TF893A be Rin and equal to a constant 8 ohms.

The first measurement.
For 100 mW indicated on the meter, the voltage across Rin is 0.9 v (by measurement).
Therefore we have:
Vs / 0.9 = (Rs + 8) / 8 (ratio of potentials = ratio of resistances)
This reduces to:
Vs = 0.11Rs + 0.88 . . . . . eqn. 1

The second measurement.
For 50 mW indicated on the meter, the voltage across Rin is now 0.632 v., since
V = √(PR) = √(0.05 x 8) = 0.632 v.
We have Rs and R in series and those two in series with Rin.
Vs is developed across all of them; the 0.632 v only appears across Rin (= 8 ohms).
Therefore, as before, we have:
Vs / 0.632 = (Rs + R + 8) / 8
This reduces to:
Vs = 0.079Rs + 0.079R + 0.632 . . . . . eqn. 2

Eliminating Vs from eqn. 1 and eqn. 2, we obtain:
0.11 Rs + 0.88 = 0.079Rs + 0.079R + 0.632.
This reduces to:
0.39Rs + 3.14 = R . . . . . eqn. 3

Now we have assumed that we do not know the value of R and the value of Rs, but we have a good idea of their typical and expected values.
If R = 5, then eqn. 3 returns a value of Rs = 4.8 ohms.
However, if Rs = 5, eqn. 3 returns a value of R = 5.09 ohms.

Both of those values are well within the tolerance ranges for each of the two unknowns Rs and R. Hence your measurements are valid & consistent and, moreover, establish that the value of Rin is also 8 ohms, within an acceptable & typical tolerance band.

HTH.

Al.

[Mark Hennessy]

In other words, Alan was expecting the power to half, rather than the voltage, when Zout=Zin

I admit that when I first skimmed his post, I missed this too.

Alan, put an Avo across the input of the power amplifier, and adjust R until the voltage is half. The power reading will be a quarter of the original reading, not half.

All the best,

Mark

[Skywave]

Following on from Mark's comments above, Alan's method is akin to the standard approach to determine the input resistance of an unknown 'black box'. Let's do an analysis for the general case.

Let: the source voltage = Vs; source internal resistance = Rs; input resistance of our 'black box' = Rin.

Prior to introducing the variable resistance between Rs and Rin, let the voltage appearing across Rin be Vin.

We have:
Vs / Vin = (Rs + Rin) / Rin,
which can be re-arranged as:
Vin / Vs = Rin / (Rs + Rin),
so:
Vin = Vs[Rin / (Rs + Rin)] . . . . . eq'n. 1.

We now introduce a variable resistance R between Rs and Rin and adjust its value such that the value of Vin is reduced to Vin/2. Let the value of that variable resistance that produces that figure of Vin/2 be R.

Now we have:
Vs / (Vin/2) = (Rs + Rin + R) / Rin,
which can be re-arranged as:
Vin / (2.Vs) = Rin / (Rs + Rin + R),
so:
Vin = 2.Vs[Rin / (Rs + Rin + R)] . . . . . eq'n. 2.

Since eq'n. 1 and eq'n. 2 are both equal to Vin, we can now write:
Vs[Rin / (Rs + Rin)] = 2.Vs[Rin / (Rs + Rin + R)].
This reduces to:
Rin + Rs = R . . . . eq'n. 3

Now for Rin to be approximately equal to R (our introduced variable resistance), we require Rs << Rin.
This is the usual case where we have a 'black box' with a high input resistance and is connected to a source of voltage whose internal resistance is very much lower. However, if that condition is not satisfied, then Rin = R - Rs.

Hence, in Alan's test arrangement, what needs to be done is to choose a value of R (his R-Box) so that his initial Vin (= 0.9 v) drops to 0.45 v. And if Rs does indeed equal 5 ohms, and Rin also equals 8 ohms, eq'n. 3 predicts that the value of R will be
8 + 5 = 13 ohms.

I sincerely hope all of that (and my preceding post) was not to difficult to follow: simply some elementary electrical theory and a bit of basic algebra.

Al.

[AlanBeckett]

Sorry about vanishing, 'domestics' intervened, the curse of the amateur potterer.
Thanks for the inputs. It's a Power Meter, so I have to assume that's how it's calibrated, at least for Sin waves.
I set the Output of the J-2 to give a reading of 9dB on the Scale then adjusted the R-Box to give 6dB while keeping the Output of the J-2 constant. That's where the 5R came from.
I don't think I'm going to make any more progress until I sort that switch. As it is I'm stuck on that range unless I get the Soldering Iron out.
I think I'll suspend operations until I've sorted it.
I will return!

Alan

[Mark Hennessy]

Yes, but as I said in post #17, that's where you're going wrong.

In post #7, you don't explicitly say it, but the strong implication (backed up by the above post) is that you're doing a standard impedance test; adjust R until reading has halved, then Rin = R. But that only works for voltage, not power. You need to measure the voltage at the input of the power meter. An AVO will probably do, but use an RMS sensing DVM if you have one.

I've now explained this twice in hand-waving terms, and Al has made two posts explaining it in mathematical terms, including proof that your 5 ohms is in the right ballpark for Zin = 8 ohms. Job done

Cheers,

Mark