[HomeBrew] linear scale analog watt meter

[Steven Weber]

John,

>First, you should be able to improve the detector compensation accuracy 
>at low power levels by optimizing the values of the detector load 
>resistors (R5, R10) and the compensating diode load resistors (R6, 
>R11).  

Actually, I've found a 1:2 ratio works best in the 0-10 watt range. Its a
bit of a compermise between having too little compensation at the low end
and too much at the high end. But when using an analog meter, its hard to
notice much difference bewteen the optimized values and using same values.
 
>"Calibrating Diode Detectors" in QEX for August, 1990.  The short 
>version is that the DC current in the compensating diode should equal 
>the peak current in the detector diode.  Since the detector diode only 
>conducts for a small part of the cycle, its peak current will be much 
>higher than if it were driven by the same DC voltage, and thus its 
>forward drop will also be higher.  R6 and R11 want to be around 10 - 20 
>percent of R5 and R10 to improve compensation.

The way I see it is the detector diode has a much smaller current than the
compensation diode. The peak current in the detector diode will initally be
high, but once the filter cap on its output charges, which only takes a few
cycles, it will be very small. OTOH, there is a constant current flowing
through the compensation diode, and that gets higher as the detector
voltage gets larger. Therefore, the compensation resistor needs to have a
larger value than the detector load resistor.  


>Second, matched transistors.  It might be cheaper to use a single 
>matched quad than two matched pairs.  For example, Mouser part number 
>526-NTE2321 is a replacement for MPQ3904, a matched quad of 2N3904's in 
>a 14 pin DIP.  Price is $3.80 each, not cheap but better than two duals 
>at $5 each.

True, I didn't look very long or hard for other alturnatives. If your not
afraid of using SMT parts, their about 42 cents for a pair. Cheaper still
if you can use a 100 of them :-)

>Third, for those who might want to power this device from a 12 volt 
>supply at home (or 12 volt battery in the field), a 7660 or similar 
>charge pump circuit might be used to get the negative supply.

I found that the circuit works with just a -5 volt supply, so can do with
out the third 9V battery. Did require making the input resistor larger
though, as the neg supply affects the full scale. The - supply needs to be
well regulated, but there isn't much current involved, so a charge pump
might work okay. 

>Finally, I'm wondering if there might be a way to do the squaring 
>circuit with rail to rail op amps and avoid the negative supply 
>altogether.  Hmm, it's past beer time now so I'd better not think about 
>that until tomorrow.

Now that you mension it, one should be able to tie all the points marked
"ground" to the +5 regulator and run the op amp on 12 volts. The
compensation amps need to be rail to rail or a type which have input and
outputs which work to ground. 

BTW, I use a digital 0 to 10 watt meter with 1 mW resolution, based on an
Analog Devices True Power rms detector chip for my standard. The test RF
source is a 20 watt linear amp with AGC stablized output so the output
level stays constant as parts heat up and is adjustable from 0 to 10 watts
out.
 
72,
Steve, KD1JV
"Melt Solder"
White Mountains of New Hampshire
http://www.qsl.net/kd1jv/