Sam Brown

[Sam Brown]

For that particular test, Scott was interested in looking at the combined resistance of those two loads, which a voltage measurement wouldn’t have shown. Also, since one of his hands was stuck way back in the machine, it would have been very difficult to do that measurement without shocking himself if the machine were plugged in.

[Sam Brown]

” Cam number 1 needs to be closed in order to power the entire system. So I decide to check the cam and see it is closed, I measure O volts, this is telling me that it is closed and I am just reading one side of a closed switch to the other.”

doesn’t this question assume the rest of the circuit is in fact a circuit? wouldn’t an open neutral or bad L1 have the same result?

Excellent observation! You’re right, of course. Performing that voltage measurement would only be conclusive if you have confirmed that the circuit has a valid voltage supply and path to neutral. If either of those things are missing, you will read 0 VAC on your loading meter, just as you would if you had a closed switch with a valid power supply.

This illustrates the kind of thinking we have to do as technicians. Whenever we make a test, we have to think carefully about what that test can and cannot tell us. In the situation where you have previously confirmed that your circuit has a valid power supply, then reading 0 volts across the timer cam definitively proves that it is closed. But if you haven’t checked your power supply first, then that test doesn’t necessarily tell you anything.

Just like you have to use the right tool for the job, you have to use the correct test for your current situation. Make sense?

[Sam Brown]

Great markups! Looks like you’ve got it totally figured out — no errors that I can see. Very well done!

[Sam Brown]

Good catch! I believe that the Samurai just misspoke there — he meant to say “closed”. Well done staying on your toes! 😉

[Sam Brown]

Do you see how all those switches are drawn in bold, thick lines? That’s a common convention on schematics to show that all those switches are internal timer contacts. That’s why it doesn’t explicitly call them out as such.

We’re not interested in measuring the amperage on the black wire that brings L1 into the timer, so that’s why we don’t measure at C. That amperage would be the total current of L1, while what we’re interested in is the amperage in the heater circuit specifically.

The red wire coming off of A is an easy place to put our clamp that measures only the current passing through the heater circuit.

Make sense?

[Sam Brown]

It’s not that your meter can’t be trusted — you just have to be aware of a few things: the current setting your meter is on, the purpose of that setting, and the specifications of your meter.

When you’re doing resistance or continuity settings, your meter has a certain threshold for ohms beyond which it will simply report infinite resistance. To find out what this limit is for your meter, you have to look at its specs.

For example, if a particular meter only measures ohms up to 20,000 ohms, and you measure across a component with 20,100 ohms of resistance, then your meter will report it as open/infinite resistance.

We brought up this point in the unit for two reasons: a) you should be aware of what your meter’s specs are, and b) resistance measurements are not always conclusive. This does not mean that your meter is unreliable or that you can’t trust it — it just means that you need to be aware of your tool’s limitations and think carefully about what your measurements actually mean.

[Sam Brown]

Why exclude that 32 ohm resistor in the final equation is what I’m getting at?

Because in that particular equation, the Samurai is calculating the power output of just the loose connection, and so he only factors in the resistance of the loose connection. He did the same thing in the equation right above that one, but that time for the heating element. When calculating a particular load’s power output, you factor in the total circuit current, but only that particular load’s resistance.

If you want to calculate the power output of the entire circuit, then you would add together the resistances of all the loads present.

Make sense?

[Sam Brown]

The resistor and the controls are in parallel, but not with the heater, right?

That’s correct.

But going forward with this, is it still correct though to just add both the resistance values of the heater and resistor like they’re in series since in the earlier module/unit we learned about equivalent resistance in parallel circuits?

The heater and the resistor are in series. Remember that series-parallel circuits are a configuration that can exist. In this instance, the heater is in series with every other component in the circuit, while some of the components are in parallel with others. Your own diagram that you posted shows this quite well.

That the heater and the resistor are in series is in fact proven by the measurement that the Samurai made in the video — if they were not, how would he have measured the expected amount of ohms by reading across them?

Keep in mind too that when you make a resistance measurement, you are effectively isolating a part of the circuit. For the purposes of that test, all that matters is the section of the circuit that is between your leads.

Let me know if it’s still unclear.

[Sam Brown]

It might look that way because of the way the schematic is drawn, but don’t let that fool you. Pull up the image again and follow the lines from 1M to OR.

You see how you have to pass through both the heater and the resistor in order to get from one point to the other? That’s the path that the electrons have to follow, which means that those two components are in series — at least when the TM to OR contacts are closed on that switch by the timer motor.

Let me know if any of this still doesn’t make sense.

[Sam Brown]

Here’s the exact wording of question 4:

In a triac, the working voltage will be ____, and the control voltage will be ____.

I think you might be mixing it up with question 3, which states:

In an electromechanical relay, the working voltage will be ____, and the control voltage will be ____.

Those two questions are referring to separate pieces of technology. Triacs and relays work very differently, so it’s important to distinguish between them. Try reviewing the material in the unit to make sure that distinction is clear to you.

Once you’ve done that, write back and let me know if you’re still confused about anything.

[Sam Brown]

Hi Nate,

Remember the conditions that are set up in question 4 (question 5 is just a continuation of that same scenario). In question 4, it’s already stated that you have a good power supply. What you really need to do is make sure that the component in question is putting out the correct voltage.

Where would you put your meter’s probes to check the output of this component? (Make sure you read all the text on the schematics — text on schematics is always important.)

[Sam Brown]

Cams in timers are just the mechanical devices that make or break contacts — so yes, they function as switches. In this particular timer, each cam is capable of making two different connections. For example, cam 10 can either close the switch between contact T17 and contact T16, or between contact T17 and contact T18.

Don’t conflate the cams with the contacts (contacts are the points called out as T16, T17, T18, etc.). The contacts (also called terminals) are just points on the timer that wires from elsewhere in the machine connect to.

Are you able to see now how the Samurai used the timer chart to tell which contacts would be closed during the pump portion of the cycle?

[Sam Brown]

Hi Nate,

As we say in the text of the unit, a microprocessor board that will not enter service mode is bad by definition. This failure indicates that the logic of the board has become corrupted, and it is no longer able to execute programs correctly.

A word of caution: don’t be too quick to pronounce a board as deceased the moment you fail to go into service mode. This is only diagnostically conclusive when you have a) confirmed that the keypad is functioning properly, as the question states, and b) ensured that you are correctly executing the key presses to enter diagnostic mode. Only when you’re sure of those two things can you conclude that the board has failed.

[Sam Brown]

In a split-phase compressor, the start winding’s resistance is expected to be significantly higher than the run winding’s. What you’re seeing with your measurement looks right in line with what you should see.

As for why the first two readings don’t add up to the third: ohms readings are sloppy things, especially when dealing with low resistances like you are. A discrepancy of a few ohms like what you’re seeing is certainly within the margin of error.

The general rule is this: Common to Start will always have more resistance than Common to Run, and Start to Run will always have more resistance than either of the other two readings.

[Sam Brown]

You didn’t miss anything — this is just a place where electricity doesn’t lend itself to being understood intuitively. You have to look at the math to really get it.

Fortunately, the math involved here is just basic arithmetic — subtraction, specifically. The key concept is this: voltage is all about the difference between two points. That’s what your meter is showing when you make a measurement.

When we use a meter to make a voltage measurement between line and neutral, we place one lead on line, the other on neutral, and then our meter measures the difference in voltage between those two points. Line has 120 volts of electrical potential and neutral has 0, so our meter reads 120 volts.

Another important thing to remember: the polarity of the voltage in an AC circuit is constantly reversing from positive to negative, back and forth. So that line voltage will be at one moment positive 120 volts, and then the next it will be negative 120 volts. The difference between line and neutral remains the same, but the polarity of line is changing.

Once you get to 240 volt power supplies that have both L1 and L2, you have to take this concept of voltage difference and reversing polarity a step further. When the Samurai says that L1 and L2 are 180 degrees out of phase with each other, that basically just means that they are electrically complete opposites. Whenever L1 is at positive 120 volts, L2 is at negative 120 volts. Which line is at which polarity swaps 120 times a second, but they always remain opposite. That’s what it means to be 180 degrees out of phase.

Onto your second question: since we’ve established that voltage is all about the difference between two points, what would happen if you placed both leads on the same line? That’s what the Samurai is asking: what would you read if you put one lead on L2 and the other lead also on L2 (in other words, measuring L2 with respect to L2)?

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