Ohm? Don’t Forget Kirchhoff!

It is hard to get very far into electronics without knowing Ohm’s law. Named after [Georg Ohm] it describes current and voltage relationships in linear circuits. However, there are two laws that are even more basic that don’t get nearly the respect that Ohm’s law gets. Those are Kirchhoff’s laws.

In simple terms, Kirchhoff’s laws are really an expression of conservation of energy. Kirchhoff’s current law (KCL) says that the current going into a single point (a node) has to have exactly the same amount of current going out of it. If you are more mathematical, you can say that the sum of the current going in and the current going out will always be zero, since the current going out will have a negative sign compared to the current going in.

You know the current in a series circuit is always the same, right? For example, in a circuit with a battery, an LED, and a resistor, the LED and the resistor will have the same current in them. That’s KCL. The current going into the resistor better be the same as the current going out of it and into the LED.

This is mostly interesting when there are more than two wires going into one point. If a battery drives 3 magically-identical light bulbs, for instance, then each bulb will get one-third of the total current. The node where the battery’s wire joins with the leads to the 3 bulbs is the node. All the current coming in, has to equal all the current going out. Even if the bulbs are not identical, the totals will still be equal. So if you know any three values, you can compute the fourth.

If you want to play with it yourself, you can simulate the circuit below.

The current from the battery has to equal the current going into the battery. The two resistors at the extreme left and right have the same current through them (1.56 mA). Within rounding error of the simulator, each branch of the split has its share of the total (note the bottom leg has 3K total resistance and, thus, carries less current).

Continue reading “Ohm? Don’t Forget Kirchhoff!”

Fail Of The Week : Measuring DC Current Has To Be Easy, Right?

[DainBramage] needed a DC ammeter to check how long his amateur radio station would be able to stay powered on battery backup power. The one’s he already had on hand were a Clamp Meter, which could only measure AC, and another one that measured just a few milliamps. Since he didn’t have one which could measure up to 25A, he decided to build his own DIY DC Ammeter with parts scavenged from his parts bin. Measuring DC current is not too difficult. Pass the current to be measured through a precision resistor, and measure the voltage drop across it using a sensitive voltmeter.

I = V/R

So far, so good. If it’s late at night and you’ve had a lot of coffee, busy building your DC ammeter, things could head south soon. [DainBramage]’s first step was to build a suitable Shunt. He had a lot of old, 1Ω, 10W resistors lying around. He made a series-parallel combination using nine of them to create a hefty 1Ω, 90W shunt (well, 0.999999999 Ohms if you want to be picky). This gave him a nice 1 Volt per Amp ratio, making it easy to do his measurements.

Next step was to hook up the shunt to a suitable voltmeter. Luckily, he had a Micronta voltmeter lying around, ripped out from a Radio Shack product. Since he didn’t have the voltmeter data, he hooked up a 10k resistor across the meter inputs, and slowly increased the voltage applied to the meter. At 260mV, the needle touched full-scale and the voltage across the inputs of the voltmeter was 33mV. [DainBramage] then describes the math he used to calculate the resistors he would need to have a 10A and a 25A measurement range. He misses his chance to catch the fail. His project log then describes some of the boring details of putting all this together inside a case and wrapping it all up.

A while later, his updates crop up. First thing he probably realized was that he needed more accurate readings, so he added connectors to allow attaching a more accurate voltmeter instead of the analog Micronta. At this point, he still didn’t catch the fail although it’s staring him straight in the face.

His head scratching moment comes when he tries to connect his home made ammeter in series with the 12V DC power supply to his amateur radio station. Every time he tries to transmit (which is when the Radio is drawing some current), the Radio shuts off.  If you still haven’t spotted the fail, try figuring out how much voltage gets dropped across the 1Ω shunt resistor when the current is 1A and when it is 5A or more.