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Can you give me a timestamp on the video that shows the prongs and dashed lines you’re referring to?
Correctillia!
The R is only real resistance and does not account for reactance. The vectorial sum of resistance plus reactance is called impedance and it uses the symbol Z. Ohms Law still applies but you need to account for the the the total physics of the electron opposition (resistance plus reactance) for the total impedance, Z. So your ohms law equation would be E/Z or sqrtP/Z.
We just need to be aware that’s how reactive components work. We won’t be doing vector calculations in the field. Why? Because we think in terms of WATTS with AC loads, not resistance, and we use our amp clamp instead of our ohm meter.
You’ll need to keep reminding yourself of this because almost all the manufacturer literature today is dumbed down and talks in terms of ohms. They do this because most techs today do not understand even a fraction of what you’re learning in this course, eg., watts, amps, ohms law, impedance, etc. So they have to meet techs where they are. I think they also do it for liability protection because they know most techs are electrically inept and are more likely to shock themselves working on live circuits. So they give specs techs can use on dead circuits (ohms).
I explain that in the video. Start at 10 minutes into the video.
Also, here’s another webinar recording you should watch on the Ten Step Tango and the Timer Chart Cha-Cha: https://appliantology.org/topic/70571-troubleshooting-with-timing-charts-and-schematics-ten-step-tango™-and-timer-chart-ch-cha™/
on you last lesson video there are 2 loads in series one being the evaporator fan meaning that the current should and would be the same for both the def heater and the eveporator fan motor so if i put my ampmeter clamp to measure the current on different points in the circuit i would read the same amount of amps since its supposed to be equal at all points eventough one of them is non ohmic load m i right?
Yes. In a series circuit, the amps are the same at every point in the circuit regardless of the type of load.
secondly how can i calculate voltage drops across them in this case(since one of them is non homic and i cant employ equations that have R found in them)?
In that example in the webinar recording in the lesson, you should have seen that the defrost heater is essentially acting as a wire because it’s resistance relative to the evaporator fan motor is orders of magnitude less (4,000 ohms vs. 12 ohms). That’s the concept I was showing there. That means we can ignore defrost heater for troubleshooting purposes (unless it is actually open). So the evap fan motor is said to be the controlling load because its resistance is so much higher that it controls the current in the circuit. Accordingly, it is still getting virtually all of its rated power supply. Close enough that we ignore the effects of the heater.
The wattage for the evap fan motor will be stamped on the motor itself and also shown in the spec block on the tech sheet. Here again, we don’t bother calculating the resistance of the fan motor (which is also given on the tech sheet, by the way). We know the wattage, so then use the voltage supply along with the fan’s wattage rating to calculate the expected amp reading. We make our amp reading and compare.
Or, even easier, simply look/listen to see if the fan motor is actually running with voltage applied to the circuit! If the fan motor is running, the circuit is working correctly.
Don’t get hung up on non-ohmic loads. Just be aware that that’s a thing and, because of that, we don’t us ohms to make diagnostic conclusions about AC loads. Similarly, don’t fixate on resistance with AC loads. Think in terms of watts.
so if a washer is fitted with an AC asynchronous moter ,when its in the wash cycle it spins at a certian speed and after the drain it goes into spin cycle which is faster than wash in other words the motor will rotate much faster in spin cycle than wash,so how does the motor carry out faster rotation duering spin?
120 VAC asynchronous running on a 60 Hz power motors will always spin at the same speed, 3600 RPM, regardless of direction of rotation. It is a limitation of their inherent design and construction. The RPM speed reduction or increase AT THE DRUM is changed by the transmission. These are old skool machines. Newer machines will use different motors, such as BLDC motors, which can easily be controlled by an inverter to run at different RPMs, including gradual ramp up in speed, thus eliminating the need for a transmission.
another questions that came up are how does the vfd system change the direction of the rotation? does inverter board open and close the IGBT’S in reverse order than what you showed?
Yep. Control the direction of rotation of that magnetic field and you control the movement of the rotor shaft, speed and direction of rotation.
second i know what the term shorted mean when wires are involved,short to ground etc… but when you said a traic is shorted what exactly do you mean by that ? and why when a traic or diod is shorted they will constanly let current thruogh to the other side?
In the context of a specific component like a triac or a capacitor, a failure condition is “internally shorted.” For a triac, this means the PN junction between Anode 1 and Anode 2 has gone low resistance in both directions of current (electron) movement. So the triac no longer acts like a switch but more like a wire, letting electrons come and go in either direction.
Here’s a good article on triac operation for appliance techs that you should read: https://appliantology.org/blogs/entry/953-triac-operation-for-appliance-techs/
The phase shift of the temporary phase is done on-the-fly at the moment of start by the motor’s start winding. In the case of a capacitor start motor, the capacitor increases the phase angle of the temporary second phase. As soon as the motor’s rotor starts rotating, the start winding and start capacitor are taken out of the circuit by the start device or relay. In the case of a capacitor start-capacitor run motor, two capacitors are used: the start capacitor does the increased phase shift for starting– same drill; the run capacitor helps smooth out the back EMF produced by the motor’s run winding so the motor runs more smoothly and uses slightly less power.
i enjoyably watched your lesson about psc motor but one thing i didnt realy got is how the capacitor make one phaze shifted ?
Watch the first part of the video again several times. This is explained without using math.
does the capacitor fulfill the same function on an ordinary run-of-the-mill AC ASYNCHRONOUS motor?does it do the same job?
Yes.
secondly is ther any formula to work out by how many degrees the phaze is shifted by a capacitor? i know these formula 1/2×3.14xFxC has it got somthing to do with the shifted phaze?
There are several formulas. This gets into higher math involving LC circuits, vectors, and trigonometry that is beyond the scope of this course. As a tech, you will not be doing these calculations– you only need know what these capacitors do and how they affect motor operation. In other words, you will not be doing engineering calculations on sizing start and run capacitors.
In general, algorithms can be either mechanical or software. For some things, the manufacturer has to tell you about algorithms that may affect operation and troubleshooting. This can be done verbally with descriptions, or by giving pinout voltages on a board, or by giving you a timer chart in the case of a mechanical algorithm. Other times, you can determine by deduction based on your understanding of technology and circuits. This post at Appliantology has a good explanation and examples: https://appliantology.org/blogs/entry/1106-control-board-troubleshooting-inputs-outputs-and-algorithms/
A quick example would be a computer controlled refrigerator that’s not defrosting. You test amps on the defrost circuit and they’re in spec. Failed computer? Not necessarily! If the evaporator thermistor is out of spec and feeding bad information to the computer, then the computer may never initiate defrost cycle. Knowing about this algorithm, your next troubleshooting step is to text the evaporator thermistor to see if it’s in spec.
This post has another example: https://appliantology.org/blogs/entry/1209-samsung-refrigerator-ice-dispenser-opening-and-closing-repeatedly-where-do-you-begin/
This webinar recording goes into troubleshooting computer controlled appliances in more detail (you’ll need to have your free Appliantology account set up to watch it): https://appliantology.org/topic/56680-webinar-troubleshooting-strategies-for-computer-controlled-appliances/?do=findComment&comment=336620
This webinar recording explains troubleshooting mechanical algorithms using timer charts: https://appliantology.org/topic/70571-troubleshooting-with-timing-charts-and-schematics-ten-step-tango™-and-timer-chart-ch-cha™/
Reading those blog posts and watching the webinar recordings should give you a concrete understanding of how being aware of algorithms can affect troubleshooting.
That’s exactly right. You’re not going to mess around with calculating ohms. You WILL use the wattage spec of an AC load with its voltage supply spec to calculate the expected amp reading on your meter to see if the load is operating in spec electrically.
In a circuit with a single load, such as a dryer or dishwasher heating element, all of the supply voltage will be dropped across that load. In this case, voltage supply and voltage drop will be the same.
I have a good webinar recording that goes into this in more detail that you should watch: https://appliantology.org/topic/72423-voltage-voltage-drop-loads-switches-jumpers-cheaters/
Let me know if you have any other questions.
It would behave the same way. Functionally, the only difference between a 3-wire and 4-wire system is the Neutral-Ground bonding point. In the 3-wire system, Neutral and Ground are bonded at the appliance. If the 4-wire system, Neutral and Ground are bonded at the circuit breaker box. Electrically, either system would behave the same way in the failure scenarios discussed in the video.
This webinar recording at Appliantology discusses this in some detail: https://appliantology.org/topic/82369-dryer-cord-wiring-weirdness/
July 26, 2022 at 11:23 am in reply to: Circuit Components and Series Circuits (calculations series circuit) #24029You usually won’t do these types of calculations in the field but they are in the course to help you understand how electricity actually works. Since we can’t see electricity, we can only understand it by using the ohms law equations, which are essentially a few ratio relationships between volts (E), amps (I), ohms (R), and watts (P).
The other equation you mentioned is actually P=E^2/R– E X E DIVIDED BY R. You would still need to use the voltage drop across the load in question (E) and the resistance of that load (R). It’s useful when you know E and R but don’t know (or don’t need to know) amps.
It’s more correct to say that copper has very *little resistance* compared to loads. Copper actually does have some resistance. Power companies deal with this when distributing high amps through power lines. It’s call “I-squared R losses” because it’s based on the ohms law equation, P= I^2 * R. Where I is the amps moving through the lines and R is the copper resistance. This is the loss of power in heat just from moving high amps through power lines. The resistance of the copper is what it is so nothing you can do about that. But the delivery amps you can do something about. If you jack up the voltage, then you need to move fewer amps. This is from P= I*E where P is the deliverable power, I is the line amps, and E is the line voltage. That’s why power companies deliver power at high volts, say around 14,400 volts and then step it down at transformers on the pole near the house to your 120/240 supply.
With windings, you’re dealing with lots of copper coiled together. So you have some copper resistance there. But bigger than that, you have inductive reactance, which is another opposition to electron movement (amps). Your meter on ohms only measures resistance, not the reactance. Reactance only appears when electrons are moving back and forth in AC power. In motor windings, the reactance will be much higher than the copper resistance and is the main opposition to electron movement. That’s another reason you want to rely on amps (as a proxy for watts) when diagnosing AC loads like motors.
Measuring for resistance in this case is iffy because the resistance you’re looking for is so low– just a few ohms. Better is to work on the live circuit and measure voltage drop across the various connections. In a 240 VAC circuit, even a resistance of just a few ohms will show up as a big voltage drop.
Is the UI connected to the main board by a ribbon connector? Those are famous for getting some corrosion or oxidation at the ribbon connector pads. Sometimes, just cleaning the pads with a pencil eraser is all that’s needed. Also, what kind of keypad is it: membrane or capacitive touch?
These kinds of specific troubleshooting questions are best asked at Appliantology where we can link you to resources and other techs can chime in. That’s why we give free Appliantology membership to MST students. The MST forums are for questions about the course material. I’m happy to help you with this question but please keep this in mind for future questions on specific service calls.
I presume you were on the call in the first place for a warm compartment complaint. On these calls, your first troubleshooting move is to half-split between the sealed system and everything else. There are three common ways of doing this explained in this article at Appliantology: https://appliantology.org/blogs/entry/1227-3-ways-to-half-split-warm-refrigerator-problems/
From your description, it sounds like you’re dealing with a split phase compressor as opposed to a variable speed unit. If this is the case, it’s very likely that the sound you’re hearing is the famous hum-click cha-cha. This is usually caused by a failed compressor start device that is no longer powering the start winding in the compressor. Without the start winding supplying that temporary second phase, the compressor main winding is experiencing high amps and you’re hearing the 60 Hz hum because the compressor is unable to get the rotor spinning from a dead stop. Those high amps create heat. If this goes on very long, the compressor windings will get damaged from the heat and the compressor will need to be replaced.
An easy test for the start device is to remove it and shake it. If it rattles, it is fractured. I have a webinar on split phase compressors and PTC start devices at Appliantology: https://appliantology.org/topic/64098-mst-office-hours-8212017-split-phase-compressors-and-ptc-start-devices/
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