**EMRFD Message Archive 9149**

MessageDateFromSubject9149 2013-09-17 14:54:57 kerrypwr Calibrating RF Power Meters With DC I have a couple of home-made dummy load/power meter devices; one is a 2/20 watt unit whilst the other is a 200 watt unit.

Each is just a 50R load and a diode voltmeter with a calibration trimpot.

I calibrated them with DC when I built them; applying a DC voltage equivalent to the peak voltage of various powers and adjusting the relevant trimpot.

I don't have high power available (I have a couple of 100W transceivers but they are not accessible at present); I have a small home-made amplifier that can produce about 4 watts so I thought I'd tinker with the smaller meter.

I used 25 Mhz from a HP 8657B signal generator into the amplifier, a 30 MHz LPF, a 30 dB coupler and, finally, the power meter.

I have the choice of a DSA-815 or an AD8307 power meter to connect, via 30 dB of attenuation, to the Cpl port of the coupler; I connected my Tek TDS-320 digital 'scope via a T and a 10x probe to the power meter input and, as expected, saw a clean sine wave.

The SA confirmed that there are no measurable harmonics of the 25 MHz signal.

{I have given some detail of the test setup to indicate that I have no reason to doubt its accuracy).

The SA and the AD8307 meter agree within tenths of a dB around the -30 dBm level at which I was working.

I supplied the meter under test with DC and calibrated it; the one-watt figures are typical.

One watt at 50 ohms (the meter/load has almost-perfect 1:1 SWR at 25 MHz) is 7.1v RMS/10v peak so I injected 10v DC and adjusted it to read 1 watt.

I then injected the 25 MHz signal and adjusted it so that the meter again read one-watt. The 'scope showed the level as 8.4v RMS/12v peak, ie 12v as against 10v DC or 20%. I verified this with my 465M; not as precise as the TDS-320 of course but good enough to show that the figures were not grossly incorrect.

This difference seems a lot to me; has anyone else had experience in this area?

The only appropriate reference I can find is W7EL's article on his QRP power meter which has a side-bar description of the differences between DC and RF when applied to diodes but this seems to apply only at low levels, not to levels of a volt or more; at high levels the diode seems to behave as a resistor and passes the same current whether RF or DC is applied.

I expected a small difference caused by the diode capacitor not charging quite to the peak voltage but 20% seems a lot.

Thoughts anyone?

Kerry VK2TIL.9150 2013-09-17 22:37:22 Andy Re: Calibrating RF Power Meters With DC planningpower@iprimus.com.au wrote:

> I calibrated them with DC when I built them; applying a DC voltage equivalentThis would be OK if your power meters were peak-responding. Are they?

> to the peak voltage of various powers and adjusting the relevant trimpot.

You did mention a capacitor, but not until nearly the last sentence.

> ... at high levels the diode seems to behave as a resistor and passes theThat's not exactly true because it conducts only brief pulses at much

> same current whether RF or DC is applied.

higher peak current when RF is applied. So its voltage drop while

conducting is more.

Personally, and maybe it's just me, I don't trust oscilloscopes with

10X probes at RF unless you have verified their accuracy at frequency.

Most scopes with sufficient bandwidth (say 10 or more times the

frequency of interest) are OK without the 10X probe, so you might try

repeating the measurement without using any part of the 10X probe;

just a piece of RG-58 with a 50 ohm terminator at the scope input.

Also make sure the scope's settings do not affect its bandwidth.

I am somewhat "grasping at straws", so this might not be much help.

Andy9151 2013-09-18 00:24:29 kerrypwr Re: Calibrating RF Power Meters With DC It's a peak-reading voltmeter; here is the schematic;

http://i39.tinypic.com/s5vmkm.jpg

So the diode behaviour you describe occurs at all signal levels, not just low levels?

I read the sidebar in W7EL's article (I have it but I can't find it on the 'net; sorry); whilst the maths was way beyond me, I gathered from the accompanying graphs (I like pictures much more than I like equations :) ) that this behaviour disappeared above about 1 volt.

Your explanation, I think, explains the problem; now I'd like to attach some numbers to my results.

I tried terminating directly at the 'scope input as you suggested; this involved attaching the entire power meter to a T-piece on the 'scope input and feeding RF to the third leg of the T; fortunately the meter is small, unlike the 200W one!

First I calibrated the meter to read 1W with 7.1V DC input, ie the equivalent of 7.1V RMS/10V peak.

Then I attached the meter to the 25 MHz amp output with a T-piece; meter on one leg and a 10:1 probe on the other; it required 6.3V RMS/8.9V peak to get a 1W reading on the meter.

I changed the setup to the one suggested by you and described above; it required 5.7VRMS/8V peak to get a 1W reading.

This is not huge but it's significant. It was interesting that disconnecting the T from the 'scope caused the meter reading to drop a little, perhaps due to the 20pF shunt C

9155 2013-09-18 14:53:13 kb1gmx Re: Calibrating RF Power Meters With DC Technically the voltage read is the applied DC minus the diode drop. For the moment we will assume a 1n4148 silicon junction diode that has a forward drop of about .7V (you can measure that!).

So if you apply 7.07V DC to the load you will likely read about 6.37V on you meter for an actual power 1W applied DC.

When you apply 1W RMS power you should expect 9.3Vpeak(dc) (remember 10Vpeak minus the diode drop).

Failure to allow for the diodes forward conduction will seriously mess with your readings and you will get answer that are always a bit off. If you use Schottky diode like the 1n5711 or 1n34 germanium point contact

the offset and calibration scales will differ as they have different forward drop and also differeint series resistance.

All this gets a bit different when the applied peak AC voltage is less than the forward conductance of the diode as the rules change a bit. That's a whole topic in itself.

Allison

9156 2013-09-18 16:07:29 kerrypwr Re: Calibrating RF Power Meters With DC The diode is a 1N5711 mounted close to the load;

http://i44.tinypic.com/2qau442.jpg

I thought to avoid the question of diode-drop in my calibration; if I supply a known voltage to a known resistance I'm supplying a known power and then it's a matter of adjusting the meter cal trimpots so that that known power is indicated on the meter.

The voltages used are above the diode square-law area.

It seems that the diode drop for DC is different to the diode drop at AC/RF; it takes less RF voltage (6v RMS) than DC voltage (7V) into the 50-ohm load to produce the same meter reading (1W).

This, I think, is the reverse of the effect described by W7EL.

Incidentally, I remembered that W7EL's article to which I referred earlier is

9157 2013-09-18 18:13:29 Andy Re: Calibrating RF Power Meters With DC > So the diode behaviour you describe occurs at all signal levels, not just low levels?The diode behavior I was thinking of, is a large signal behavior, and

is like this:

To get a full scale reading from the DC source, the diode passes 1mA

continuously.

From the RF source, in the limit as it is truly peak responding, it

conducts for only a very brief portion of a cycle, when the voltage is

at its peak. To put some numbers on it, let's just say that it

conducts for 1% of each cycle. And let's call it a rectangular

current pulse just to keep it simple. To get the same 1mA average

current for the full scale meter deflection, the diode current in that

brief period needs to be 100mA, not 1mA.

(Given the actual shape of the current pulse, it is probably even greater.)

If the diode is an ideal switch and looks like a constant 0.3V or 0.7V

forward voltage drop regardless of instantaneous current, then it's no

problem. But if you include bulk resistance of the diode, plus the

diode's exponential curve, then the diode drop with 0.1A (in short

bursts) is going to be greater than the drop with 1mA (continuous).

This effect happens even at AF, if the capacitors are big enough to

keep it working as a peak responding meter at those frequencies.

A quick glance at someone's data sheet shows

Vf < 0.41V @ 1mA

Vf < 1.0V @ 15mA

Do you see what I'm getting at?

My example (1% conduction) is perhaps a bit extreme, but I think it

shows that the forward voltage drop is not the same even for the same

average current. So it's a 'fly in the ointment' and I don't know how

to compensate for it.

> I read the sidebar in W7EL's article (I have it but I can't find it on theThen I gather he was talking about small-signal effects, which I'm not.

> 'net; sorry); whilst the maths was way beyond me, I gathered from the

> accompanying graphs (I like pictures much more than I like equations :) )

> that this behaviour disappeared above about 1 volt.

> I tried terminating directly at the 'scope input as you suggested; thisYou can run some coax between the T and the terminator/meter. The

> involved attaching the entire power meter to a T-piece on the 'scope

> input and feeding RF to the third leg of the T; fortunately the meter is

> small, unlike the 200W one!

point is to have the 50 ohm termination on the far end of all the

coax, with as few and as short as possible stubs along the way. If

the scope itself doesn't have a 50 ohm input termination switch, or if

you want your power meter connected simultaneously, then plug the T

adapter right into the scope, run one coax from your source to the T,

and another coax from the T to the external terminator (your power

meter). You don't need to jam the meter in right at the T itself.

Any length of coax is fine, if its own loss doesn't impact your

measurement.

Yes the scope input capacitance will load the signal down a bit.

It's interesting that you saw it increase when connected to the scope.

(Perhaps some sort of interaction with the cable lengths?)

> First I calibrated the meter to read 1W with 7.1V DC input, ie theWait a minute ... it's a peak reading meter.

> equivalent of 7.1V RMS/10V peak.

7.1V DC is the symbolic equivalent of 7.1V peak AC = 5.0V RMS, isn't

it? To calibrate it for 1W RF, wouldn't you use 10.0V DC? (I thought

that was what you described initially.)

Regards,

Andy9158 2013-09-18 18:18:38 kerrypwr Re: Calibrating RF Power Meters With DC Apologies for the typo; I meant 10v DC which is equivalent to 7.1v RMS.

*********************************************************************

"Wait a minute ... it's a peak reading meter.

7.1V DC is the symbolic equivalent of 7.1V peak AC = 5.0V RMS, isn't

it? To calibrate it for 1W RF, wouldn't you use 10.0V DC? (I thought

that was what you described initially.)

".9160 2013-09-19 21:35:01 kerrypwr Re: Calibrating RF Power Meters With DC Andy; I'm not sure about the effect you describe.

After the first positive cycle of the sine wave, the capacitor is charged to the peak voltage less the diode drop. The meter circuit then draws current from the capacitor for a time approximately-equal (a little less actually) to the period of the sine wave; this reduces the voltage on the capacitor.

On the next positive cycle the diode doesn't conduct until the voltage equals the reduced voltage plus the diode drop. The actual point depends on the meter time constant and the frequency of the sine wave but, if the first is greater than the second by a factor of hundreds or even thousands, it is likely to be very close indeed to the peak, probably within millivolts of it.

Conduction will cease at the peak; no more can be done.

Since the amount of energy being replaced in the capacitor charge is just "topping-off" the current will be very small; it's at the point on the charging curve where the curve is almost flat, the opposite to the high inrush current that occurs at the start of capacitor charging where the curve is almost vertical.

Diode bulk resistance will be low at these voltage levels and dynamic resistance will be low since dV is of the order of millivolts.

If this effect were causing my results, it would be necessary to input more peak voltage than DC voltage to get the same meter reading; I am seeing the reverse of that.

This is an interesting exercise; it began as a simple desire to calibrate some power meters.

I think I will set up a proper test fixture and investigate diode behaviour at RF and DC further; that may take a little time as, like all of us, I do have other things to get on with.

But I think it might be worthwhile; I've done exhaustive searching for information

9161 2013-09-19 22:05:42 Andy Re: Calibrating RF Power Meters With DC Kerry, I'm not sure you understood what I was trying to describe.

For the ideal peak detector, after the first quarter-cycle where the

capacitor charges to the peak of the sine wave minus the diode forward

voltage, each cycle after that the diode conducts only at the very

peaks. I think we are in agreement on that.

In the period between those "topping off" charge bursts, the meter

conducts a nearly steady current. Current x period = charge; this is

the amount of charge removed from the capacitor per cycle.

In the brief moment where the diode conducts, it must make up all of

that charge, to maintain the same capacitor voltage from cycle to

cycle.

If the diode conducts for only 1% of the cycle when the waveform is

near its peak, then the charging current needs to be 100 times the

meter current. That is necessary in order to dump the same amount of

charge back into the capacitor.

If the ratio of charge to discharge time is 1:1000, then the diode

conducts 1000 times the meter current. For a 1mA full scale reading,

the diode would conduct 1Amp, briefly, every cycle.

This effect happens only when the waveform is AC. With a DC source,

the diode always conducts only the meter current, no more than 1mA.

The graph for the 1N5711 shows quite a bit of forward voltage change

between 1mA and 10mA or 30mA.

You are right that it should make you need to apply a greater RF peak

voltage than DC voltage, to get the same meter reading. I don't

understand why it seems to be the opposite. That has me puzzled.

Regards,

Andy9162 2013-09-20 03:54:59 Alan Melia Re: Calibrating RF Power Meters With DC I may have missed something in this discussion so excuse me is this has been

raised.

You both seem to be assuming that the "resistor" is 50+0j ......this is

very unlikely! The actual resistive element will have some inductance and

there is also an effect called "skin effect" which occurs at higher

frequencies but not at DC. Both of these will tend to give greater readings

at RF than DC, since the impedance of the load resistor will increase. Have

you put a network analyser on the load at the frequencies of interest??

Alan

G3NYK9164 2013-09-20 15:08:52 kerrypwr Re: Calibrating RF Power Meters With DC Just spent a long time thinking-out and monodigitally typing this message but Yahoo said "Couldn't send this message".

Fortunately I was able to copy it; I'l paste it below and try again.

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The load is very close to 1:1 SWR at low frequencies; it is made of two 100-ohm Diconex ceramic resistors. SWR of 2:1 is reached at about 700 MHz;

http://i43.tinypic.com/2b8gg4.png

Re the diode current; this is limited by the charging characteristic of the capacitor;

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/capchg3.gif

At the start, when the capacitor is fully-discharged, there is a large current ("inrush current" in power supplies; this circuit is really just a power supply). The dynamic resistance at any point on the curve is dV/dI; at the start, dV is small, dI is large, the curve is almost vertical and resistance is low.

As charging proceeds, the ratio changes; at some point dV = dI and, after that point, dV begins to grow much larger than dI.

The dynamic resistance is very high by the time that four or five time constants is reached; the curve is almost straight and almost level. At this point we can consider the capacitor as fully-charged.

Wiki describes this effect in delightful terms; ... "

The more a capacitor is charged, the larger its voltage drop; i.e., the more it "pushes back" against the charging current ...".

The next stage is the discharge of the capacitor through the meter; the time constant of the meter circuit (which includes the capacitor) is relevant here.

One rule-of-thumb is that this time constant should be at least 100 times the period of the RF sine wave. A 1mA meter configured for 10v FSD has a TC 1000 times greater than the period of a 10MHz sine wave.

(Hope this is correct; I did say earlier that maths is not my strong point).

The effect of this is that, in the period when the meter withdraws charge from the capacitor, it moves the capacitor voltage back along the curve until the next cycle of the RF begins its "top-up" duty as we discussed earlier.

If this point were a long way back down the curve we would be in the low dynamic resistance area and a lot of current would flow in the diode.

But, because we have done much better than the rule-of-thumb, the point at which the diode starts the new charge "top-up" is on the nearly-straight-&-level part of the charging curve where the dynamic resistance dV/dI is high and, consequently, the current through the diode is limited by this resistance.

While all this is going-

9165 2013-09-20 15:14:25 kerrypwr Re: Calibrating RF Power Meters With DC I forgot to say that I based my calculation of the meter circuit time constant on a 0.01uF capacitor.

Kerry VK2TIL.

9168 2013-09-20 20:29:41 Andy Re: Calibrating RF Power Meters With DC Kerry, I have trouble following your explanation about the charging

current. It doesn't add up. I think you are trying to reason why the

charging current at the waveform peaks would be small. But it isn't.

It is just the opposite. It HAS to be. If it isn't, then the voltage

on the capacitor ends up discharging and it does not follow the peaks

of the RF waveform.

When you write dV/dI, what V and what I are you referring to?

Also, the "dynamic resistance" of what? The capacitor or the diode?

The instantaneous current charging the capacitor is determined by the

waveform at the diode, how much above the capacitor's voltage it is at

that moment, and what the diode's i/v curve looks like.

If at any moment you try to suddenly (very quickly) increase the

capacitor voltage by even a millivolt, large current will result.

This happens even if the capacitor is nearly fully charged already.

The thing you need to keep in mind is that, when the meter is reading

full-scale, the meter draws 1mA continuously from the capacitor; so

the AVERAGE current through the diode must also be 1mA. But since the

diode current comes only in short bursts at the RF voltage peaks, with

zero current the rest of the time, its peak current HAS to be much

greater than 1mA ... so that the average is 1mA.

Andy9169 2013-09-20 22:09:48 kerrypwr Re: Calibrating RF Power Meters With DC dV & dI are the incremental change in V & I; this concept is used to examine the dynamic resistance of a diode but it seems equally-applicable to the curve for capacitor charging.

The charging curve at close to the end of charging is almost flat; the meter draw brings the charge back a little but it is still on this nearly-flat part of the curve.

This is the randomly-chosen graph to which I linked earlier;

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/capchg3.gif

Observe the dotted curve; the charging current. It begins at the top LH and is very high initially but decreases as the capacitor gains charge (the red curve).

After about 5 RC periods the capacitor is near-enough to charged; the red line has levelled-off and has reached the top RH corner (full charge).

Simultaneously the dotted line has also levelled-off and has reached the bottom RH corner (zero current).

The charging process then stops and the meter is supplied with its 1mA from the capacitor.

Suppose that this applies for a very short time (that is why the meter/capacitor TC should be much greater than the RF period) and the capacitor storage falls to the point on the red curve at 4RC.

The dotted current curve is at almost zero at this 4RC point and is likewise almost level so, if dV is 1 mV, the current cannot "rush-in"; the capacitor is presenting a high resistance (dV/dI).

The diode current will be the 1 mA drawn by the meter plus a small current that recharges the capacitor; the amount of charge to be replaced is not large in proportion to the total charge because the capacitor value is such (ie large enough) that it retains almost all of the complete charge that it acquired on the very first cycle.

If the capacitor were small, the meter discharge might reduce its charge to a point at, say, RC on the graph; in this event the diode would pass a high current as the opposition of the capacitor is much lower at this point.

I think that the diode current during this period would be the 1 mA required by the meter plus a small current to re-charge the capacitor from 4RC to 5RC as shown

9170 2013-09-21 19:47:52 Andy Re: Calibrating RF Power Meters With DC > dV & dI are the incremental change in V & I; this concept is used to examinedv/di makes sense for capacitors, properly applied. But to describe

> the dynamic resistance of a diode but it seems equally-applicable to the curve

> for capacitor charging.

it as a "dynamic resistance" is probably not a good way to put it. It

doesn't have anything with resistance.

> The charging curve at close to the end of charging is almost flat ...That curve in the figure does not describe how the voltage on the

> The dotted current curve is at almost zero at this 4RC point and is likewise

> almost level so, if dV is 1 mV, the current cannot "rush-in"; the capacitor is

> presenting a high resistance (dV/dI).

> The diode current will be the 1 mA drawn by the meter plus a small current

> that recharges the capacitor ...

capacitor gets "topped off" at the waveform peaks in your peak

detector.

You seem to not understand how the meter current must exactly equal

the average charging current through the diode, and that this requires

the peak charging current to be greater than 1mA.

You seem to not understand how to apply the figure of capacitor

voltage and current. The figure is for an input voltage step function

(which thereafter doesn't change), through a fixed, linear resistor.

None of those things are happening here.

> If the capacitor were small, the meter discharge might reduce its chargeNo. The average diode current must equal the average meter current.

> to a point at, say, RC on the graph; in this event the diode would pass a

> high current as the opposition of the capacitor is much lower at this point.

There is no other place for the current to go. If the capacitor were

small so that it discharges significantly each cycle, then the meter

current will be less ... which means that the average diode current

will also be less.

Please, I urge you to stop thinking of capacitors as having some sort

of "opposition."

Andy9328 2013-11-07 08:20:29 tttpppggg Re: Calibrating RF Power Meters With DC The difference happens to be .7 Volts which happens to be the drop of a silicon diode. You haven't described the circuit so it's hard to say, but diode offset matters. A diode detector is nonlinear so weak signals are lower due to the exponential curve response. For instance