21. Count Yer Amps, Champs

Tech-Talk

Part 19

Welcome to part 19 of our series.

This time, we're going to take a look at how we can measure high current without having to use the heavy gauge wire necessary for so many Amps.  That's the magic of a meter shunt.  Here's how it works...
  Take a look at the diagram below.

Our power source is drawn as a battery, but it could just as easily be a typical power supply in the shack.  We'll assume a 12V supply.  It's powering a load marked as resistor R2.  This could be any device.  We'll assume it's a 25 watt transceiver drawing 10 amps at full power.  In any case, you'll see that the supply voltage and current draw won't directly affect our discussion here.   Notice that the meter is connected across resistor R1.  That's our shunt resistor, and here's how it works.

Current flowing through a resistance causes a voltage to develop across the resistance.  This is also called the voltage drop across the resistor.  Refer to parts 15 and 16 below if you want to refresh your understanding.  How do we use the resistor to calculate current?  Our old friend Mr. Ohm and his Laws ride to the rescue again!  Analog meters like the one in the picture react to changes in voltage.  (It's an oversimplification, but digital meters are similar).  So we need to know what voltage it takes to drive the meter to full scale.  Let's assume 1 volt is full scale.  And we'll further assume that 40 Amps is the biggest load we'll want to measure.  Ohm's Law tells us that resistance is voltage divided by current.  So for a full scale reading, we need a resistor that is 1 ÷ 40 (1 volt divided by 40 Amps) Ohms.  That's 0.025Ω.

There are two small problems here, though.  First of all, we've dropped 1 of our 12 volts across the shunt resistor.  That's a hefty penalty to pay, especially on battery power.  Even worse, remember that Mr. Ohm also tells us that Power in Watts is equal to Voltage times Current.  With a 40 Amp load, that resistor will need to dissipate 40 Watts of power as heat.  That calls for a mighty big resistor, eh?

Well, the math is pretty straightforward here.  If we use a meter that reads full scale at 100 millivolts -- a tenth of a volt -- we need a 0.0025Ω resistor.  That's still 4 Watts of power and a pretty big resistor to handle the heat.  But we've cut that voltage drop way down.  We could do even better with a 10 millivolt meter, but as you can imagine, that would drive the cost up. There's one final problem to solve.  Take a look at the picture below:

Those are two shunt resistors from a shipment of meters I recently received.  They are specified as having 75 mV drop at 50A current.  That's 0.0015Ω.  Now it's relatively easy to make a 1 Ohm resistor with say 1% tolerance, or 0.01Ω either way.  But think -- if the resistor above is just 0.01Ω off either way, our current reading is off by a factor of ~~7.  Not very accurate, to say the least.  Look closely at the pictures, and you'll see that the resistor on top has two small notches cut into it, and the one below has a single notch.  Each notch raises the resistance ever so slightly -- just enough to get it right.  And that's how the factory does the final calibration of the resistance.  It follows, then, that if you're using a shunt like this, you must mount it where it won't be subject to damage.  A protective cover might be a good idea.

That's it for this month.  Next time, we'll continue looking at DC Power in the shack.

73 for now
John Bee, N1GNV
Quicksilver Radio Products