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 Discussion on question by michael_kuz…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Mon 21:11
Just a thought, but you didn't connect the three groups of lights in series by any chance? Shouldn't they be wired in parallel?
Transistor
Mon 21:11
The datasheet is a little "foggy". It suggests (on top of page 4) that the 'B' model controls output current. Are you sure you haven't got an 'A' model? Just to rule stuff out, try measuring the voltage on the dimmer input as you vary the resistance across the input wires. 10k should give you 1 V. 20k should give you 2 V, etc.
Transistor
Mon 21:11
I've explained the operation of the Mean Well analog inputs in this article: lednique.com/power-supplies/dimmable-mains-psu-control. Have you got a datasheet link for the lamps? Are there bulk capacitors on the light PCBs?
Transistor
Mon 21:11
PWN? Do you mean PWM (Pulse Width Modulation)?
 

 Discussion on question by ETHAN COUSI…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Dec 17, 2024 20:29
>What do I use to cover up these wires?
You use insulated wires, same as in your house or in a car.
 

 Discussion on question by LifeIsBeati…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Apr 10, 2024 02:35
"I'm trying to measure 3dB bandwidth of LED." What do you mean by this?
 

 Discussion on answer by Andy aka: Dim…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Feb 25, 2024 15:42
@ZiphoLunika, "Zero cross detected --> turn on 555 ---> 555 turns on the photo coupler which controls the switching on and off of the triac". You can't switch off a TRIAC! (without a complex circuit). All phase-controlled dimmers delay the switch on point and let the TRIAC turn off naturally at the next zero-cross. Any basic TRIAC web tutorial will explain this.
 

 Discussion on question by AhmedH2O: 9…

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Transistor
Jan 22, 2024 13:22
'V' for volt, 'A' for ampere, 'W' for watt. I fixed your post.
 

 Discussion on question by Noobielectr…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Jan 1, 2024 03:47
You'll need to turn on the component designators (the names of the components, U1, U2, etc.) on the schematic otherwise we won't be able to discuss the circuit.
Transistor
Jan 1, 2024 03:47
Why a crooked photo instead of a screen grab? (Hit Prt-Scrn.)
Transistor
Jan 1, 2024 03:47
I fixed the capitalisation and punctuation.
Transistor
Jan 1, 2024 03:47
Please show a schematic diagram (not a breadboard wiring diagram). Otherwise we all have to guess what you have done. Thanks.
 

 Discussion on question by JRE: What i…

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Transistor
Dec 31, 2023 15:51
"In either case, full speed is reached by basically connecting the motor straight to the mains." Yes, but you're driving it from \$230\sqrt 2 = 325 \ \text {V DC}\$ (ignoring capacitor droop). I suspect that you shouldn't be running at 100% PWM. Try limiting to \$\frac 1 {\sqrt 2} = 70.1\%\$ PWM maximum.
 

 Discussion on question by Curious Dio…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Nov 27, 2023 17:10
@RohatKılıç, I know. I'm changing the world! d:^)
Transistor
Nov 27, 2023 17:10
Right, now that you've edited the question, it's clear that you are not series-connecting transformers but series connecting SMPS DC outputs. That changes everything as the outputs are smoothed and there is no need for synchronisation. Please fix the title of the question - or if you do mean to series-connect the transformer secondaries then make that clear in the post.
Transistor
Nov 27, 2023 17:10
"I want to drive the transformers individually using synced clocks (such each transformer has its own IC driver and opto-feedback) ..." For the secondaries to combine additively all the primaries would have to be driven in-phase and in-sync. Without that the secondaries will "beat" sometimes adding and sometimes cancelling each other out. Edit your question to explain further.
 

 Discussion on question by nobody_love…

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Transistor
Jun 17, 2023 14:49
"... suppliers of physics education equipment, often sell weights marked in newtons instead of kilograms [both lowercase] ..." Those markings would only be relevant in specific gravitational fields!
 

 Discussion on question by Ghost: How …

Imported from a comment discussion on electronics.stackexchang...
Transistor
May 28, 2023 16:36
What is the current through R_AB.
Transistor
May 28, 2023 16:35
That doesn't work either.
Transistor
May 28, 2023 16:35
Hmm, MathJAX doesn't work in chat. Let's try HTML markup.
What is the current through R<sub>AB</sub>?
Transistor
May 28, 2023 16:34
Right, so what is the current through \$R_{AB}\$?
Transistor
May 28, 2023 16:23
Your updated sketch with a total resistance of 4.2k and 42 V is correct. 1. It was recommended that you number all your components, R1, R2, etc., so we can discuss them. 2. You have included the \$V_o\$ resistors (even though they're irrelevant) but included them in the wrong place. They can't be connected directly to the current source terminals. Remove them. 3. Now work out the current in each branch. 4. Then work out the voltage at the junctions of the potential dividers.
Transistor
May 28, 2023 16:23
I has been pointed out already that since there is nothing connected to the \$V_o\$ terminals that there is no current flowing in the \$V_o\$ resistors. Therefore they can be ignored in your calculations.
Transistor
May 28, 2023 16:23
You could work out the total resistance between the upper and lower rails of your circuit. Then you'll have two of the variables in V = IR and be able to work out the third.
Transistor
May 28, 2023 16:23
"I wonder what type of school or college would give such a thing as homework." That's a standard electrical engineering problem so most would, I expect. Your original current has a 100 mA source. Your second circuit has a 10 mA source. Assuming that's an error and both should be 100 mA what will determine how the current splits in the two branches?
Transistor
May 28, 2023 16:23
This looks like homework so you need to show your work and where you are stuck. Otherwise the question will be closed.
 

 Discussion on question by electricala…

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Transistor
Apr 21, 2022 22:08
Few modern domestic devices are frequency sensitive and many now have universal switched mode power supplies meaning that they can be plugged directly into 230 V supply. What kind of devices do you want to power?
 

 Discussion on question by alexanderjs…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Dec 12, 2021 04:05
I’m voting to close this question because it's not really an electronics design question. It might be better suited to DIY.
 

 Discussion on question by Crafter2: M…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Oct 25, 2021 21:52
"... the fan was still working ..." What fan?
 

 Discussion on question by Deschele Sc…

Imported from a comment discussion on worldbuilding.stackexcha...
Transistor
Aug 4, 2021 15:17
@SolomonSlow, you are correct except it is capitalised - Celsius.
Transistor
Aug 4, 2021 15:17
Not quite. Since "lord" is his title it is a proper noun and gets a capital 'L'. That makes it, "Lord Kelvin's temperature ..." You could say, "Lord Kelvin's temperature was 310 kelvin" (or "310 K" or "37 degrees Celcius" or "37 °C"). Note that Celcius is not an SI unit and it is capitalised when spelt out.
Transistor
Aug 4, 2021 15:17
Tip for your writing: SI units named after a person have their symbols capitalised but are lowercase when spelt out. So, 'K' or 'kelvin' but not 'Kelvin'.
 

 Discussion on question by Sayan: freq…

Imported from a comment discussion on electronics.stackexchang...
Transistor
Jul 27, 2021 18:26
Yes, but you are expecting your readers to muddle through your writing and the lack of proper formatting is somewhat disrespectful to your audience that you are asking to give freely of their time to help you. Make it easy for your readers. See Write to the best of your ability on the site's help pages.
Transistor
Jul 27, 2021 18:26
Please capitalise your writing properly for ease of reading and accuracy. Your compiler wouldn't tolerate it ...
 

 Discussion between Transistor and gho…

Imported from a comment discussion on engineering.stackexchang...
Transistor
Jul 19, 2021 23:25
I've posted an answer. Please accept that to indicate that your question has been answered.
Transistor
Jul 19, 2021 23:18
Is past midnight in Ireland. Goodnight.
Transistor
Jul 19, 2021 23:17
Just a comment on your approach: you didn't have a systematic approach to the problem and you got fixated on 9 minutes way to early. Break it down into logical steps.
Transistor
Jul 19, 2021 23:15
Note that the question implies that you'll get a linear rate of cooling. In the real world you won't because the cooling rate will decrease as the difference between the steel and air temperature decreases.
Transistor
Jul 19, 2021 23:13
Correct.
Transistor
Jul 19, 2021 23:09
Um, you've got the volume, you've got the number of minutes. How would you calculate the volume per minute?
Transistor
Jul 19, 2021 22:59
Correct and that answers the question in the title of your post but your post has the 9 minute requirement so I suspect that you were asked for the required airflow. Calculate m^3/minute.
Transistor
Jul 19, 2021 22:52
To be clear, 41 kg is correct. Convert to m^3.
Transistor
Jul 19, 2021 22:50
Also, be careful with units. Kg is kelvin-grams. Capitals matter.
Transistor
Jul 19, 2021 22:49
Not quite. Your question asks for the volume of air. Look up the density of air ...
Transistor
Jul 19, 2021 22:41
It will be the same mass no matter how long it takes.
Transistor
Jul 19, 2021 22:28
Right, you'll need to extract 2.5 MJ of heat.
What mass of air will you need. (Same procedure means run the calculator with the previous result and the SHC of air.)
Transistor
Jul 19, 2021 22:26
Look up the SHC of steel. It's not 1012 J/(kg-K). You've chosen air.
Transistor
Jul 19, 2021 22:26
I'm stepping you through the procedure. I was bringing you onto calculating the mass of air required when I noticed that your first calculation is not correct. You have to calculate the energy to be removed from the steel first. You've used the SCH for air instead. Start again.
Transistor
Jul 19, 2021 22:26
If this is homework then you'll need to show your work. First task: calculate the energy to be removed from the steel. For this you're going to have to look up its specific heat capacity (SHC). Edit it into your question.
Transistor
Jul 19, 2021 22:26
OK, so you've got to extract 5.5 MJ. How many kg of air will that take? (Same procedure.) Actually, where did you get your SHC value from? It doesn't look right.