Measure Your Crypto Rig's Power Usage

As most of you know, cryptocurrency mining has been quite popular over the past several years. And following that trend, my brother put together a mining computer. After ethereum’s switch to proof-of-stake, the older mining methods no longer work, which is what my brother was doing. So he had to switch to other coins like ergo and flux. The problem now is that these coins earn less than Ethereum used to. And with a desire to add more graphics cards, power usage is an issue too. So, the question is whether his crypto rig is both within a reasonable power limit and if the rig is currently profitable? Let’s find out.

The first thing we need to do is determine what our power limits are. To help us out, my grandfather came over. He is an electrician, so he knows a good deal about this. And he brought an ammeter. Anyways, by going to the breaker box, we can find that the limit is 15 amps. We can also measure that the voltage coming in is 120 volts AC. The first thing to do is determine whether the mining rig is drawing less than 15 amps of current. By putting the current clamp around the wire, we read 8 amps of current flowing through it. Using the power equation, we get a power draw of 960 watts. So it is completely within the limits, so there aren’t any worries there. Funnily enough, his rig is basically the only thing in the house drawing any current. The main breaker is not much higher than 8 amps. Although, the dryer does use a few amps when it is turned on too.

Anyways, the next step is to determine whether his mining rig is currently generating a profit. The way we can do that is determining electricity costs vs the amount of crypto his mining brings in. Looking at his mining control panel, we can see that he earns 76 cents per day. But that’s just earnings, let’s look at costs too. Where we live, the cost of electricity is 7 cents per kWh. This means that if he uses one thousand watts for an hour straight, he will be charged 7 cents. So, let’s see how much power he is actually using. The easiest way to find out his power usage is to just go to the mining control panel. Here it shows that he is using 530 watts. That means that it costs 3.7 cents per hour to run. Converting his daily profits to hourly profits, we get 3.2 cents per hour. Which is less than the electricity that it generates. So it runs at a deficit of half a cent per hour, or 12 cents per day. There’s just one problem though, the power calculated from the breaker is different. So are the calculations correct? Let me explain.

When it comes to DC voltages, it is simple to calculate the power usage. Just multiply current by voltage and you will have your answer. This is likely how the control panel is measuring the power useage of the system. And this measurement is correct, but it isn’t what the AC measurement is showing. It comes down to the behavior of AC currents and voltages. And the difference between true and apparent power. In general, DC circuits have similar true and apparent powers. But AC is different: you may have heard of the term ‘phase shift’. Let’s take a look at a sine wave, which is what AC typically uses. We can, for example, represent the voltage waveform in green using the equation: sin(t). We will also represent the current flowing with red in the equation sin(t) as well. So far, the true and apparent powers are the same, since the current and voltage align, just like they would in a true DC circuit. But the problem comes with a phase shift. Let’s modify the current waveform and place it out of phase by 90 degrees. Now we can see that the current and voltages both have the same RMS value, but they do not align. Now, the phase shift itself does not cause the power difference, rather it is a good way of visually showing what is happening. These phase shifts are caused by the reactive components, which are inductors and capacitors. Resistors do not cause phase shifts, since they resistive, not reactive.

We can simulate a circuit like this on the breadboard. Let’s grab two oscilloscope probes to visualize it. The yellow will be the voltage and the purple is the current. With only a resistor in the way, the current and voltage curves are completely aligned. With either a capacitor or an inductor, the current will be 90 degrees out of phase, but in different directions. A combination of capacitors and inductors gives us varying results. A perfect combination will give us resonance, and align the voltage and current.

So, in order to determine the true power from the apparent power, we can use a power factor. The power factor is simply the ratio between the true power and the apparent power: TP/AP. The power factor is also equal to the cosine of the phase angle. Which is also equal to the resistance divided by the impedance. Since the apparent power isn’t consumed like the true power is, it is denoted with the units VA as opposed to watts.

But where is all of this impedance coming from? Well, remember that the real world is not perfect. So, inductance comes from the wire itself. Capacitance comes from the close proximity of the wires. Not to mention the actual power supply itself which contains multiple capacitors and inductors inorder to create a switch mode power supply. This is a big consideration, since the apparent power is the thing that all of the components have to be rated for. The breakers will break at their specified currents. The same can be said with the wires. But, despite all of this, the electric bill is still the same. The reactive power is not actually used since inductors and capacitors will return their power. So while that explains the difference between the powers calculated, the cost doesn’t change.

So, we have arrived at a verdict: his rig is unfortunately running at a deficit. Although I know that he is waiting for the price to go up, then it will be a net profit. On the other hand, his rig is completely within the operating limits of the circuity in the house. So really, it’s up to him whether to keep it running.

Anyways, I hope you learned something from this video and understand how much your current electricty usage costs. If you enjoyed, please consider subscribing so that you can see my other videos. Have a good one!