Sam Brown

[Sam Brown]

Hi Steve,

Each of those different prefixes (kilo, Mega, milli, micro, and so on) is just a shorthand name for a certain amount of something.

For example, the “Mega” prefix means one million (as shown on the chart), so 1 Megavolt is equal to 1,000,000 volts. Another example would be a kilometer — since “kilo” means one thousand, 1 kilometer just means 1,000 meters.

And it works the same way with prefixes indicating small amounts, like milli or micro. The only difference is that these are names for fractional amounts. Milli means 0.001 (one one-thousandth) of something. So a millivolt is 0.001 volts, or one one-thousandth of a volt.

The math to convert between these units is (fortunately) very simple. If you want to move down a step in the table, just multiply your amount by 1,000. To move up a step, just divide by 1,000.

Here’s an example: I have 15 Megawatts of power, and I want to know how many watts that is. A “normal” unit (with no prefix) is two steps below Mega on the table. So that means that I have to multiply 15 by 1,000 twice. The first time gives me 15,000, which is the number of kilowatts that I have, and then the second time gives me 15,000,000, which is the number of plain old watts that is equal to 15 Megawatts.

Let me know if anything still isn’t making sense.

[Sam Brown]

Hi Joseph,

Here’s the equation in question: P = I^2 * R

A couple of things I can point out. First, that’s actually the letter “I”, not the number “1”. “I” of course stands for “current”, as it does in every Ohm’s law equation.

Second, the “^” symbol is used to denote exponents. So I^2 means “I to the second power”. In case you’re rusty on your exponents, that just means I * I. If it were I^3, that would mean I * I * I, I^4 would be I * I * I * I, and so on.

Let me know if there’s still anything about that equation that’s confusing you.

[Sam Brown]

Hi Phillip,

This question is about how to calculate equivalent resistance. That was covered in detail in the Equivalent Resistance video from the lesson — the same one I mentioned in my last post. Review that video if you haven’t already.

I’ll give you hint — you don’t even need to do any math to answer question 18. You just need to understand equivalent resistance and the general rule about how large the equivalent resistance is compared to the resistance of the loads in parallel. That rule was also mentioned in the video on equivalent resistance.

If you’re still unsure about the answer to this question after reviewing that video, please write me back telling me exactly what’s still confusing you.

[Sam Brown]

Hi Phillip,

Question 17 is asking why a shunt or a short in parallel with a load is not an example of a parallel circuit. This was explained in one of the videos in that unit — the one titled Equivalent Resistances in Parallel Circuits. I recommend you review that video if you haven’t already.

If you’re still unsure about the answer to this question after reviewing that video, please write me back here in the forums telling me exactly what about the question is confusing you.

[Sam Brown]

Hi Lhodo,

There are a couple of problems with using the temperature display as part of your testing.

First, you would have to determine whether the display is showing the actual current compartment temperature, of if it’s just showing the set temperature. More often than not, it will only display the set temperature.

Second, even if you are able to determine that the display is showing the actual temperature, then you still have to contend with the fact that the display is showing what the control board interprets from the input it gets from the thermistor. Sometimes the sensing circuit on the board can fail in such a way that the thermistor is in spec, but the board doesn’t interpret the data it receives from it correctly.

Long story short, the only way to do a surefire test on the thermistor is to take a measurement on it directly.

[Sam Brown]

Hi Phillip,

Could you please tell me what about question #1 is causing you trouble? I can help you better if I know where your confusion lies.

[Sam Brown]

Hi Apprentice,

Scott defined all the letters used in that equation in his post:

PI = the constant, PI (about 3.14, rounded)
F = frequency, Hz
L = inductance of the inductor, Henrys

As he showed there, L stands for the inductance of an inductor (such as a motor winding), which is measured in units called Henrys. The more inductive a component is, the more energy it stores as a magnetic field, and the more it resists changes in current.

All of this is explained in those two videos Scott linked to — if you haven’t already, you should definitely watch those, as they explain induction better than I could through text.

If you’re still unclear on this, or if you have more questions, please let me know.

[Sam Brown]

Hi Lhodo,

When it comes to thermistor specifications, you’re pretty reliant on manufacturers for specifications. Usually the only place you can find them is in the service manual for a particular model.

The best you could do with a part number is to find a model that uses that part, and then see if the manual for that model provides specifications for the thermistor.

[Sam Brown]

Hi Lhodo,

You’re right! It looks like Danfoss made some updates to their data to be more accurate. The pressure values they give now are about 0.2 psi lower than they were when we made that video. It’s a small difference — and wouldn’t really be consequential in troubleshooting — but still good to know.

Thanks for pointing it out to us!

[Sam Brown]

Hi Lhodo,

Each manufacturer distributes their training materials differently, but as for Whirlpool, you can apply for an account at https://servicematters.com/ to get access to training materials for them and all the brands they own.

Sam

[Sam Brown]

There may be no return path, but the spark can still discharge into the burner head — just like it would if we were dealing with a dumb ignition module.

However, the lack of a valid return path does mean that the reignition module can’t sense whether there is a flame, since it does that by monitoring the return current.

[Sam Brown]

Hi Abe,

Why is it that you assume we’re dealing with a simple ignition system instead of a reignition system?

Consider the problem we’re facing — continual sparking after the burner is lit. A simple ignition system only sparks when the burner knob is in the “ignite” position. A reignition system, however, will spark when it thinks that the burner isn’t lit. Given this, we must be dealing with a reignition system. A simple ignition system wouldn’t be capable of manifesting this failure.

Knowing now that you’re dealing with a reignition system — and knowing what you do about reignition technology — what component do you think could cause it to spark continuously like it is? That’s your LOI.

[Sam Brown]

Hi Abe,

I’ll tackle your questions one at a time.

1)So how does it send out this pulses with no return path. ?

A “dumb” spark module doesn’t need a return current, since it doesn’t sense whether the flame is lit. The spark sent out by the module simply grounds out in the burner head once it’s jumped over from the electrode. This is why there’s no need for a simple spark module to have a ground connection.

2 ) but on this same unit quiz Question #1: “In a gas burner spark ignition system, how does the spark current return to the spark module after it leaves the electrode?” I guess you DO need a return

Good point! If we were talking about a simple spark ignition system then, as I said before, it wouldn’t need a return path. A return path is only necessary specifically for a reignition system.

We reviewed the quiz, and we see how some of the phrasing could have been confusing. We’ve restructured it a bit, so please feel free to go back and re=take it.

3) question #8 what would be the LOI. I am not finding it in the options

Trust me, one of the answers to that question is the correct one! A couple things to think about:

– What kind of ignition system are we dealing with in this range that we’re troubleshooting?
– Considering the problem that’s manifesting in the range (continuous sparking while the burner is lit), what could possibly cause this problem?

See if you can puzzle it out. If you’re still stumped, let me know, and I’ll guide you further.

[Sam Brown]

Hi Abe,

When we talk about Electrically Equivalent Points, we’re not talking about taking a measurement where our leads are at two points that have the same voltage. I believe that’s where your confusion lies.

Let’s say you want to test the voltage of that broil element. To make that measurement, you have to put one of your meter leads on the L1 side of the element, and the other lead on the L2 side. One way to to this would be to disassemble the range and take our measurement directly on either side of the broil element, but that’s inconvenient and inefficient.

Instead, we’ll look for Electrically Equivalent Points for both the L1 and L2 legs of the broil element. The easiest place to get to these would be at the relay board. You’d simply find the points where the board supplies L1 and L2 to the broil element and do your test there.

Those convenient test points at the relay board are Electrically Equivalent to the test points you could have used directly on either side of the broil element. Does that make sense?

[Sam Brown]

Good question! There are a few telling factors.

First, this kind of failure would never occur in a non-reignition system, since the only time those systems spark is when they’re specifically being told to spark by the user via the burner switch.

Second, this is a single-point ignition system (you can see that on the schematic). You’ll probably never find a non-reignition system that’s single-point (we certainly have never run into one).

And finally, if you were doing a real service call on one of these models, you would be able to tell by looking at the controls. The burner switches for reignition systems don’t have an “ignite” position.

[Author]

Posts