#qcx class e #pa #lpf..... or hunting for the optimum #qcx #pa


Manuel; DL2MAN
 

Hello there,

I recently built my 40m and 20m QCX, and after some issues with low output on 20m, I was good to go after exchanging C27/C28 for Glimmer Type C´s.
I optimized my Low Pass Filters, by desoldering coupling C29, and inserting Nano VNA.
This way I got my filters perfectly tuned, for lowest possible SWR (S11) and insertion loss (S21)....

At least I thought so.....

When I measured current of both QCX´s and compared it to TX Power, I realized I might be far away from class E.....
40m RX Current: 0,11A, TX Current:0,53A, Delta = 0,42A x 13,8V = 5,8W IN, but only 3,9W Out. Thats Efficiency of 67%
20m RX Current: 0,11A, TX Current: 0,72A, Delta 0,61A x 13,8V = 8,4W IN, but only 5,8W Out. That´s Efficiency of 69%

Since it´s almost impossible to measure RF-Power before LPF, I decided to look with Thermal Camera where my Power is transformed into Heat:

As you can see, L4 gets extremely warm.
R42 gets warm as well
And the LPF Coils L3-L1 get warm as well.

I don´t think, this is healthy for class E Mode, that components heat up that much....
So I had the theory, that maybe Impedance of resonant PA Circuit is not 50 Ohms, but since I optimized my LPF with VNA -which has 50 Ohms In/Out- is now out off tune.

What I tried so far and the Effects:
Internal Capacity of BS170 are around 30pF each, together with C30 = 56pF for 40m and L1 = 1µH, this LC Combination would be resonant around 13MHz.
Still assuming 1µH and 3x BS170 contribute 30pF each, I calculated an additional Capacity of 430pF was needed to be resonant @40m.
I changed C30 to 470pF. After that, I had only 3W Out but L1 did not get as warm as before, and L1-3 stayed relatively cool.
TX Current 0,53A- RX Current 0,11A = 0,42A x 13,8V = 5,8W In for 3W Out. Thats 51% Efficency.....

So I left C30 completely away (Only Capacity of Q1-Q3 left in circuit.) Result:
TX Current 0,57A - RX Current 0,11A = 0,46A x 13,8V = 6,35W IN for 4,25W OUT. Back to 67% Effieciency.....

I resoldered the original 56pF for C30, then I started to move around the windings of the coils, while measuring Current and Output Power.
After pushing together windings of L2, I was able to get 5,23W Out. When you read all the other threads regarding Low Output Power, I would be ready and happy to go at that point....
But let´s take another look at the Current and Efficiency after tweaking L2:
0,7A TX - 0,11A RX = 0,59A x 13,8V =8,14W IN for 5,23W Out. Thats Efficiency of 64%. SO even worse then before, and still far away from 80-90% Efficiency of Class E PA.

And another Obeservation: Now heating has shifted to other components:


Q6 now gets considerably warmer then before....

I do not know what do to, so I would like to ask Hans and all the others higher qualified then me, what to do next here.... I think I´m running in circles....

73 Manuel; DL2MAN


George Korper
 

Hi Manuel, 

I ran in circles for several months on the efficiency angle. It's over my pay grade,
but I think really good calibrated and top grade instrumentation is required.
I also was amazed to see how integrated the antenna load with the outcome. 
I have seen very high efficiencies under certain circumstances, but are they illusions?
This is like analyzing the motions of a dingy in high seas. 

George


Alan G4ZFQ
 

And another Obeservation: Now heating has shifted to other components:
Q6 now gets considerably warmer then before....
Manuel,

What do these pictures represent?
Just how hot do the components get?
The pictures mean nothing to those unfamiliar with IR photographs.

73 Alan G4ZFQ


Manuel; DL2MAN
 

Hello Alan,

At left side of each picture is a color scale, showing min and max temperature in picture and the temperature value.
In case of 1st picture, you see for example L4 is abt 45°C, R42 is in the same range. L2 and L3 are a little cooler then that and so on....

@George:
You are right. Accuracy of measurement equipment should be considered... I will try to duplicate results with other equipment and compare....
But as thermals show great amount of heat loss, I assume efficiency is not in the Range of what would be considered Class E. So maybe due to accuracy of measurement equipment I´m 5% Off, but in any case: I´m pretty sure L4 should not be hotter then Q1-Q3.....

Manuel DL2MAN


Colin Evans M1BUU
 

Steve Weber, KD1JV uses an asymmetrical output filter in his MTR radios. The input impedance is modelled as 10 ohms and the output as 50 ohms. This way you can get more RF output as the PA FETs can drive easier into 10ohms.

Take a look at the 4SQRP Hilltopper manual - it has a class E BS170x3 PA stage and manages healthy output from 12v. There's also a bonus in that only two toroids are needed instead of three.

I think the G-QRP filters / W3NQN are rather dated now.

73, Colin, M1BUU

On Sun, 10 May 2020, 22:24 , <DL2MAN@...> wrote:
Hello Alan,

At left side of each picture is a color scale, showing min and max temperature in picture and the temperature value.
In case of 1st picture, you see for example L4 is abt 45°C, R42 is in the same range. L2 and L3 are a little cooler then that and so on....

@George:
You are right. Accuracy of measurement equipment should be considered... I will try to duplicate results with other equipment and compare....
But as thermals show great amount of heat loss, I assume efficiency is not in the Range of what would be considered Class E. So maybe due to accuracy of measurement equipment I´m 5% Off, but in any case: I´m pretty sure L4 should not be hotter then Q1-Q3.....

Manuel DL2MAN


Shirley Dulcey KE1L
 

The figure of 80% or higher efficiency of a Class E amplifier is just the efficiency of the output stage. The actual efficiency of Class E designs varies.

You're doing your best to take some power consumption out of the calculation by subtracting out the power consumption on receive. That's still there since the receiver is not powered down during transmit; it just stops getting a signal because Q5 cuts it off. But there are additional parts of the circuit that only use power during transmit: CMOS gates IC3A and IC3B (which serve as a driver stage for the PA) and the key shaping circuit of Q4 and Q6.

What that means is that you are not measuring the efficiency of the PA alone. You're not going to see the numbers that the Class E amplifier itself can achieve.

On Sun, May 10, 2020 at 5:32 PM Colin Evans M1BUU <colin.evans2@...> wrote:
Steve Weber, KD1JV uses an asymmetrical output filter in his MTR radios. The input impedance is modelled as 10 ohms and the output as 50 ohms. This way you can get more RF output as the PA FETs can drive easier into 10ohms.

Take a look at the 4SQRP Hilltopper manual - it has a class E BS170x3 PA stage and manages healthy output from 12v. There's also a bonus in that only two toroids are needed instead of three.

I think the G-QRP filters / W3NQN are rather dated now.

73, Colin, M1BUU

On Sun, 10 May 2020, 22:24 , <DL2MAN@...> wrote:
Hello Alan,

At left side of each picture is a color scale, showing min and max temperature in picture and the temperature value.
In case of 1st picture, you see for example L4 is abt 45°C, R42 is in the same range. L2 and L3 are a little cooler then that and so on....

@George:
You are right. Accuracy of measurement equipment should be considered... I will try to duplicate results with other equipment and compare....
But as thermals show great amount of heat loss, I assume efficiency is not in the Range of what would be considered Class E. So maybe due to accuracy of measurement equipment I´m 5% Off, but in any case: I´m pretty sure L4 should not be hotter then Q1-Q3.....

Manuel DL2MAN


George Korper
 

Shirley that was well stated.

In the comment I made, I was hoping that the instrumentation would
be configured to take all the readings within the boundary of the amplifier itself.
Do you think that is possible i.e. isolating it? The signal sent to it, may be as big a factor in its efficiency,
I think Allison for instance has used a reference signal, that may give different efficiencies, 
Some may be very high actually, and maybe they are. 
Anyone concerned should bread board or ugly construct just the Amp and LPF and test  it. 
I'm certain, it will turn out to be Class E. 


ajparent1/KB1GMX
 

Shirley, is right if you subtract the standing RX power[ or current] you get
what remains and the power into the Class E finals.

FYI there is some votlage loss across Q6 (about 1V) so measuring vottage
at collector of Q6 and adjusted gross current will give a more accurate 
numbers.  So using "battery voltage" and not accounting for Q6 and the
input diode can have you easily indicating higher input power than real.

Measure at collector of Q6 for DC voltage applied, and current in series
with Q6 for highest accuracy as Q6 collector is the DC source point to the
mosfet finals.

So:
Incorrect: 40m RX Current: 0,11A, TX Current:0,53A, Delta = 0,42A x 13,8V = 5,8W IN, but only 3,9W Out. Thats Efficiency of 67%
Correct:  40m RX Current: 0,11A, TX Current:0,53A, Delta = 0,42A x 12V = 5,04W IN,  3,9W Out. Efficiency of 72%%

FYI: if you were to measure the current at the emitter there is about 9ma 
of bias current lost to R41 and R42.  If we subtract that from  DC current 
to finals its now .411a and 12V for 4.932W DC in. and for 3.9W out your
near 79%.    Methods and measurements.

Small measurement errors can be costly in evaluation as at this power level a fraction of a whatt is a real part of the power.

There is no magic in the class E system  It assumes the output devices are
being operated as switches and C30 and L4 are the energy storage.  The
key is that tuning as it has to have the RF induced going to zero as the
mosfets are full on.  If the timing [tuning] is off the efficiency decays.
Add to that the reflected impedance of the filter and its tuning.

It takes little effort and error to degrade the Class E to Class C where
the mosfets run hotter and the efficiency is lower.

Loss elements, first is the mosfets they are not perferct switches
in that they have an Rds-on of  1.8ohms typical assuming the spec
10V gate to source.  However we have maybe 5V gate to source
so that means RDS on is higher and any current represents
heating and lost power.  With 3 in parallel we dance around
enough gate drive and too much gate capacitance.
 
So if the current goes up the power will but that does not
directly mean efficiency has.  IT could even indicate in-efficiency
and oscillation.  Shooting for 3-4W out and lowest current is likely
to yield better result.

That tuning effort based on mostly hearsay and guess and by golly 
will run you in circles.  Also your getting closer to right when the
current in is lower for the power out.

Allison
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Manuel; DL2MAN
 

Alright, let's step away from the class e aspect then...
What about heating up l4 ? Much hotter then pa fets. I can't imagine that was intended....

I'm more interested, how I get rid of the heat loss. 

73 de Manuel DL2MAN


George Korper
 

If I read Allison right and she usually is, why don't you back off the voltage to where it doesn't heat up faster than the design can dissipate. Good idea to run radios at 70% of their max, even QRP radios.


On Sun, May 10, 2020, 5:58 PM <DL2MAN@...> wrote:
Alright, let's step away from the class e aspect then...
What about heating up l4 ? Much hotter then pa fets. I can't imagine that was intended....

I'm more interested, how I get rid of the heat loss. 

73 de Manuel DL2MAN


ajparent1/KB1GMX
 

OK, only partly right.

The problem is forcing more power is creating more heat and if the
efficiency is low that contributes more heat as well.

tuning the the supply votlage at low efficiency with all the values
topsy turvy will insure possible failure or at a minimum lots of heat.

Stepping away from class E means a lot of things will go wrong as
its core to the transmitter.

I did a lot of work developing and testing Class E and D power
amplifiers.

Allison
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ajparent1/KB1GMX
 

Heating L4 is likely due to the higher current your running (its only #28 wire)
and if L4/C30 are not just right the circulating currents will heat it as well.

FYI the only part that deserve concern for heating is Q6 and the three  MOSFETS.
All the others can easily stand 100C without failure.

Allison
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George Korper
 

If L4 is #28 wire what are traces equivalent to? In fact isn't over doing the 
output for long durations endangering the board itself? I haven't any problems in running WSPR for two minutes,
but prudently run at 12 volts or less. The QCX hasn't failed.

If other people had QCX overheating cause failure please describe the circumstance. 


Jim Allyn - N7JA
 

Nobody in this thread has an accurate idea what the efficiency of the QCX power amplifier is.  One, I assume you want the efficiency of ONLY the PA, not the entire radio.  The only way to do that is to accurately measure the DC power delivered to the 3 mosfet output transistors, and accurately measure the RF power at the output, then divide RF Power / DC Power X 100 = percent efficiency.  Manual assumed the voltage to the PA is 13.8, but it almost certainly isn't.  Allison assumed the  voltage is 12.0, but it almost certainly isn't.  Assuming the  power supply voltage is in fact 13.8v (which many are - more or less), then the voltage delivered to the PA is 13.8 minus the voltage drop across D3 (according to the data sheet about .38 volts), AND minus the voltage drop across Q6 (again, according to the data sheet) about .3 volts, so roughly 13.12 volts.  If the receiver draws .11 amps, then the current draw of the transmitter is .53 amps minus .11 amps for the receiver, or .42 amps at 40 meters, and .72 amps minus .11amps for the receiver, or .61 amps.  If these current draws are correct, then the power drawn at 40 meters is 13.12 volts times .42 amps, or 5.51 watts, and the power drawn at 20 meters is 13.12 volts times .61 amps, or 8.003 watts.  If these numbers are correct, then the efficiency is 75.7 percent  at 40 meters, and 72.4 percent at 20 meters.  (By the way, Allison's number of 72 percent on 40 meters is incorrect.  12 volts times .42 amps = 5.04 watts, divided by 3.9 watts is 77.3 percent efficiency.  Typo, or brain fart, I assume.)  But these numbers still aren't correct.  What's wrong now?  The PA is not the only thing that  draws current during transmit, but doesn't during receive.  There is also IC3, the 74ACT00.  If I read the schematic correctly, this chip isn't active during receive, but obviously has to be during transmit.  So to be accurate, we need to include the current draw of IC3.  The current draw of CMOS depends on how fast it is being clocked, the internal capacitance of the chip, and any external capacitance it must drive.  I won't go through the math here, but a total of 313.5 pf is being driven during transmit that isn't being driven during receive.  That includes the internal capacitance of IC3, and the loads being driven.  Going through the calculations, I find that driving that capacitance at 7 MHz draws 55mA, and at 14 MHz draws 110mA.  So, if we are interested in the efficiency of the output stage only, as is usual, we would have to take those numbers into account.  And like most things in electronics, there can be considerable variation from one chip to another.

The bottom line is that if you really want to know what the efficiency of the power amplifier is, you must accurately know what the voltage at the drains of the FETs is, and you need to know how much current they (and nothing else!) are drawing.  If I were going to measure the efficiency of a QCX, I would lift the collector of Q6 and insert an ammeter between there and the junction of C31, C32, and L4, and measure the voltage at the collector of Q6.  That eliminates everything else from the measurement.  (Well, OK, it neglects the voltage drop across L4, but that is an integral part of the Class-E amplifier, and the voltage drop across it would be very small.)

I don't have a QCX at the moment (I've got a bunch of other radios to play with), but probably will some day, and if so, I will measure the efficiency as I have described here.  And I assume one or more people will beat me to it before I get a QCX.  And likely Hans has already measured the efficiency of the QCX PA.

So, there's my two cents worth.  73


Hans Summers
 

Hi all

For efficiency calculation of the Class-E stage the correct calculation of input power must use:

Voltage: at the Q6 collector. Which is the supply voltage minus the drop in the reverse polarity protection diode, and minus the drop from the Q6 envelope shaping transistor. 

Current: the current through L4. This is NOT the same as the overall transmit current of the radio, minus the overall receive current of the radio. Other circuits also consume higher current during transmit, for example the driver stage IC3). These are not part of the calculation of efficiency of the final PA stage. 

Output power (RF) must also be measured accurately and you need to be as sure as possible, that your output low pass filter is properly done so that there is not loss at the operating frequency. 

All of these things may seem minor but they are not. They have a large effect on the calculated output efficiency. 

Regarding heat. Remember that just because an amplifier is Class E does not mean it will be cold. If your amplifier is 80% efficient and your RF output is 4W then you have 1W as heat in the amplifier components. For small, un-heatsinked components, 1W is plenty to cause a noticeable temperature rise. This is to be expected and does not necessarily mean anything is wrong. 

73 Hans G0UPL

On Mon, May 11, 2020 at 5:41 AM Jim Allyn - N7JA <jim@...> wrote:
Nobody in this thread has an accurate idea what the efficiency of the QCX power amplifier is.  One, I assume you want the efficiency of ONLY the PA, not the entire radio.  The only way to do that is to accurately measure the DC power delivered to the 3 mosfet output transistors, and accurately measure the RF power at the output, then divide RF Power / DC Power X 100 = percent efficiency.  Manual assumed the voltage to the PA is 13.8, but it almost certainly isn't.  Allison assumed the  voltage is 12.0, but it almost certainly isn't.  Assuming the  power supply voltage is in fact 13.8v (which many are - more or less), then the voltage delivered to the PA is 13.8 minus the voltage drop across D3 (according to the data sheet about .38 volts), AND minus the voltage drop across Q6 (again, according to the data sheet) about .3 volts, so roughly 13.12 volts.  If the receiver draws .11 amps, then the current draw of the transmitter is .53 amps minus .11 amps for the receiver, or .42 amps at 40 meters, and .72 amps minus .11amps for the receiver, or .61 amps.  If these current draws are correct, then the power drawn at 40 meters is 13.12 volts times .42 amps, or 5.51 watts, and the power drawn at 20 meters is 13.12 volts times .61 amps, or 8.003 watts.  If these numbers are correct, then the efficiency is 75.7 percent  at 40 meters, and 72.4 percent at 20 meters.  (By the way, Allison's number of 72 percent on 40 meters is incorrect.  12 volts times .42 amps = 5.04 watts, divided by 3.9 watts is 77.3 percent efficiency.  Typo, or brain fart, I assume.)  But these numbers still aren't correct.  What's wrong now?  The PA is not the only thing that  draws current during transmit, but doesn't during receive.  There is also IC3, the 74ACT00.  If I read the schematic correctly, this chip isn't active during receive, but obviously has to be during transmit.  So to be accurate, we need to include the current draw of IC3.  The current draw of CMOS depends on how fast it is being clocked, the internal capacitance of the chip, and any external capacitance it must drive.  I won't go through the math here, but a total of 313.5 pf is being driven during transmit that isn't being driven during receive.  That includes the internal capacitance of IC3, and the loads being driven.  Going through the calculations, I find that driving that capacitance at 7 MHz draws 55mA, and at 14 MHz draws 110mA.  So, if we are interested in the efficiency of the output stage only, as is usual, we would have to take those numbers into account.  And like most things in electronics, there can be considerable variation from one chip to another.

The bottom line is that if you really want to know what the efficiency of the power amplifier is, you must accurately know what the voltage at the drains of the FETs is, and you need to know how much current they (and nothing else!) are drawing.  If I were going to measure the efficiency of a QCX, I would lift the collector of Q6 and insert an ammeter between there and the junction of C31, C32, and L4, and measure the voltage at the collector of Q6.  That eliminates everything else from the measurement.  (Well, OK, it neglects the voltage drop across L4, but that is an integral part of the Class-E amplifier, and the voltage drop across it would be very small.)

I don't have a QCX at the moment (I've got a bunch of other radios to play with), but probably will some day, and if so, I will measure the efficiency as I have described here.  And I assume one or more people will beat me to it before I get a QCX.  And likely Hans has already measured the efficiency of the QCX PA.

So, there's my two cents worth.  73


Alan G4ZFQ
 

At left side of each picture is a color scale, showing min and max
Manuel,

Sorry, I missed that.
In the absence of any other people's thermal pictures it is difficult to make any suggestions. Your actual power output looks good. Maybe this is "normal"?

73 Alan GZFQ


Manuel; DL2MAN
 

Hello Hans, Hello All,

Thanks for summing it up in your very clear words.
What I realized on top of that:

- I did not use fixed temp scale in my thermal photos, 
- I did not use same time for keydown until taking those thermal pictures
so it is not really comparable.

What I will do as a result of all your valuable input:
I will leave L4 and C30 as they are, assuming, they are well designed, and optimize the LPF for max Output from QCX (without VNA).

TNX es 73 de Manuel; DL2MAN


ajparent1/KB1GMX
 

Jim,

Please start with a 13.8V power supply.
Then subtract out the series polarity protect diode.
Then subtract the voltage drop across Q6 during TX (if its saturated maybe a half volt).
I assure you the voltage applied to the collectors via L4 is NOT 13.8V and
likely closer to 12... ok, maybe 12.8ish

No that bias network for Q6 and Q4, amounts to about 9ma added only
during TX and contributes nothing to output power as its the keying circuit.
But if you add that to power consumed during TX that a small but measurable
error.  Take the finals out measure current, the system drain goes up alone
for the keying current for Q6.

The 74ACt00, yes it also adds current when clocked and it would require
measuring that as well as the increased current from driving the gates of the
BS170s as that's about 180pf of total gate capacitance.  For 20M thats a Cx
of 62 ohms.   So yes you found one I missed.

But I did say to measure at the collector of Q6 for accurate voltage and breaking
that path for accurate final current.

Also you do  need to know what at the drains but for all practical uses L4 is 
much less than 50 milliohms so the junction of  C34 and L4 will be extremely
close to the drain voltage applied [less than .04V error].

IN the end you have verified most of what I said and pointed outu a miss.
The voltages I used may not match yours but i did say go measure it.
In the end the trivial way is likely good enough  but far from accurate.

Allison


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Jim Allyn - N7JA
 

On Mon, May 11, 2020 at 07:12 AM, ajparent1/KB1GMX wrote:
Please start with a 13.8V power supply.
As  I stated above, I did start with 13.8 volts.

Then subtract out the series polarity protect diode.
As I stated above, I did subtract voltage drop across the series polarity protect diode, using the data from the manufacturer's data sheet.

Then subtract the voltage drop across Q6 during TX (if its saturated maybe a half volt).

As I stated above, I did subtract the voltage drop aross Q6, using the data from the manufacturer's data sheet.

I assure you the voltage applied to the collectors via L4 is NOT 13.8V and
likely closer to 12... ok, maybe 12.8ish

As I stated above, using the manufacturer's data for the drop across D3 and for the drop across Q6, I calculated 13.12 volts at the collector of Q6.  I had to interpolate these numbers off the graph, so they might be off a little.  In an earlier life, if this were my radio, and the numbers for voltage drop across D3 and Q6 were the numbers you gave instead of the numbers the manufacturer gave, I'd probably have looked for a different vendor for those parts, if this were a critical application.  (See my next paragraph.)  When I buy and build a QCX, I will probably measure those things and look for other ways to improve the performance of the QCX.  If I decide to use mine portable, I will probably replace IC11  with a well filtered and (if needed) well shielded switcher to improve battery life.

The 74ACt00, yes it also adds current when clocked and it would require
measuring that as well as the increased current from driving the gates of the
BS170s as that's about 180pf of total gate capacitance.

I didn't measure it.  Not having a QCX here, I calculated the current draw from the numbers on the manufacturer's datasheets, and the standard formula P = C X V^2 X F.  I summed up all of the capacitances that have to be driven when in transmit, including the internal capacitances of the three gates of IC3 that are being driven and neglected the microamps current draw of IC3D which is not being driven at MHz frequencies, the input capacitance of IC3B and IC3C that has to be driven by the Si5351, the input capacitance of IC3A which has to be driven by IC3B, and the input capacitances of the three BS170s which have to be driven by IC3A.  That total draws a lot more current than one might think.  (In the 1980s, I worked for a company that built alphanumeric display pagers.  Probably these days most people would have no clue what that is.  I like to say we invented texting, and really, we did.  Anyway,  in the process, I learned a HELL of a LOT about reducing power consumption.  When a new engineer at some point a few years later wrote a change order to use a bipolar transistor to turn on the receiver instead of the FET I had specified because the bipolar transistor was cheaper and he wanted to save money, I showed him my spreadsheet and told him exactly how much that would reduce the battery life of the pager.  Battery life was a big deal in those days before lithium batteries, and I had EVERYTHING in my spreadsheet.)

Also you do  need to know what at the drains but for all practical uses L4 is 
much less than 50 milliohms so the junction of  C34 and L4 will be extremely
close to the drain voltage applied [less than .04V error].

Which is why I left out the drop across L4.  I did consider calculating how much wire is in it and what the resistance of that wire would be, but decided it wouldn't be worth the trouble.


IN the end you have verified most of what I said and pointed outu a miss.
The voltages I used may not match yours but i did say go measure it.
As did I say go measure it.  My point was that though there are ways you can calculate things like that, there is no substitute for measuring it.  Or better yet, do both.  And in my earlier life, if the measurements and calculations didn't agree, I would absolutely go figure out why.

In the end the trivial way is likely good enough  but far from accurate.

And if Manuel is curious about the efficiency, he needs to learn how to measure it.  And I think he probably just learned a lot.  Manuel, I appreciate your IR photographs, and there is likely an IR camera in my near future.

Allison, I don't mean to be rude or insulting to you here, but it's just my nature to check things like this and add my two cents worth, and I know I can come across that way at times.  (Probably way too often.  I am not a "people person", and I'm old enough that that's not likely to change.)  No offense is intended, and I recognize and acknowledge that you are one of the more knowledgeable people here, and by a significant margin.  73


ajparent1/KB1GMX
 

Jim,

Seems we violently agree.  ;)
The real trick it to help others in the process.
That's hard!

Allison
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