This test was NOT correct. In the last step, he tested the battery voltage since the 9v battery was still connected. What he should have done was disconnect the jumper cables before testing the charged voltage of the capacitor.
@@111boldoo Depends on a few factors but you can calculate it if you know all the parameters. You will need to know the capacitance in farads so if you have a 1uf capacitor you would change it to its value in farads so 1uf would be 0.000001 farads. Next, you need to know the resistance of the circuit, with the setup they have, it is so small yet not 0. Use a small value if you are using wire like in this video but like I said not 0. Now the fun part! The formula for this is T = R * C * Tc T(charge time in seconds) = R(resistance in Ohms) * C(Capacitance in Farads) * Tc(Time Constant) For the Time Constant, we are only worried about the first 5 so if you want to know how long it will take to reach 63.2% charge then use 1 if you want to know how long it takes to reach 99.3% charge use 5. We are not worried about anything over 5 because a capacitor starts charging fast and then slows down and after it reaches 99.3% it charges super slow past that. So here is an example let's say we have these values Our resistance is 10 kOhms, Capacitance is 1uf, and we want to know how long it takes to reach 99.3% charge. Our formula would be T = 10,000 * 0.000001 * 5 T = 0.01 * 5 T = 0.05 So for our example, we know that after 0.05 seconds we will be at 99.3% charge close enough to know if our capacitor is good. Just so you know, if we use the same capacitor value but change the resistance from 10 kOhms to what this particular setup in the video would be less than 1 ohm, however, I'm going to use 1 just because at these values it really doesn't matter. T = 1 * 0.000001 * 5 T = 0.000001 * 5 T = 0.000005 So to finish this long post, unless you have a very high capacity capacitor and a very high resistance, you will charge the capacitor almost faster than you can clip it on and take it off. However, you may have a large value capacitor you want to test so I find it better to explain the math so you can do these calculations yourself. Hope this helps and was easy to understand, I wanted to explain it in a way that even someone not familiar with electronics could understand it.
@@111boldoo The capacitor will be fully charged almost instantaneously. When it discharges it will release 50% of its power - then 50% of the remaining power - then 50% of the remaining power and so on and so on till discharged or charged back up.
Thank you. Can I please ask, if a TV power board has 2 x 450v capacitors should they read at 450v or is it just the max tolerance? Mine both read 350v would there be a problem with a component upstream or is that normal on standby? Thank you Tim
In the voltmeter test , you got 9 volts and you said that the capacitor is good. If the capacitor was bad what value should I read in the voltmeter test ? Thank you
You don’t, and that is why you learn to read schematics. You trace the board and find the comment that doesn’t work well and perform the test. I’d you are going to remove everything on the board just to test it that means you don’t know anything about it and have zero idea what you are doing.
I just watched another video saying he replaces capacitors if they are out more than 10%. Maybe that is what technicians are told when to replace capacitors. A technician may want to swap them out at 10% just so they don't have to come back later for another service call? IDK. What do you think?
@@kevineroseThis is the case to my best knowledge, the capacitor already has demonstrated its starting to fail rather than having a callback that we have to prioritize we just replace the capacitor outright, they’re very cheap usually
@@kevinerose +- 20 % is ok, because on circuit boards manufacturers use higher rated capacitors than really at least needed. But when ever it's possible just replace every one below 10 % of its rating. You don't want to repair things more often, just because you didn't replace a cheap capacitor below 10% of its rating.
There are non-polarized capacitors with no defined + and - contact. With those it doesn't matter which side is connected to positive and which to negative. If it has no markings it's probably non-polarized sometimes also referred to as a bipolar capacitor. Most ceramic capacitors for example are non-polarized
Third test didn't make any sense to me. If you'd removed the battery wires first I'd have understood it was something to do with the cap storing a voltage, but as it was you were effectively just connected to the battery were you not? I'm just here for some help with wiring a guitar. Don't pretend to understand this stuff at all.
This test was NOT correct. In the last step, he basically tested the battery because the 9v battery was still connected. What he should have done was unplug the jumper cables first and then test the voltage of the capacitor.
When you charge a capacitor you have current going in right? The closer it gets to being full the less current flows until it stops completely. Remember the basics: high resistance means low current and vice versa. A multimeter measures resistance by applying a low testvoltage and measuring the resulting current. And with voltage and current it calculates the resistance. When measuring the resistance the capacitor gets charged with the multimeters testvoltage. While the capacitor is charging from empty to full the current decreases until it hits zero. The resistance behaves opposite to the current. So if the resistance doesn't continuously climb to infinity during the measurement it means there is a low resistance electrical connection inside the cap between the two sides that shouldn't be there and hence the test result would be: "failed"
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This test was NOT correct. In the last step, he tested the battery voltage since the 9v battery was still connected. What he should have done was disconnect the jumper cables before testing the charged voltage of the capacitor.
You are right, however, he already addresses this mistake in the video.
True. Do disconnect the battery before or else you will measure the battery voltage.
How long should we wait before disconnecting the jumper ?
@@111boldoo Depends on a few factors but you can calculate it if you know all the parameters. You will need to know the capacitance in farads so if you have a 1uf capacitor you would change it to its value in farads so 1uf would be 0.000001 farads. Next, you need to know the resistance of the circuit, with the setup they have, it is so small yet not 0. Use a small value if you are using wire like in this video but like I said not 0.
Now the fun part! The formula for this is
T = R * C * Tc
T(charge time in seconds) = R(resistance in Ohms) * C(Capacitance in Farads) * Tc(Time Constant)
For the Time Constant, we are only worried about the first 5 so if you want to know how long it will take to reach 63.2% charge then use 1 if you want to know how long it takes to reach 99.3% charge use 5. We are not worried about anything over 5 because a capacitor starts charging fast and then slows down and after it reaches 99.3% it charges super slow past that.
So here is an example let's say we have these values
Our resistance is 10 kOhms, Capacitance is 1uf, and we want to know how long it takes to reach 99.3% charge.
Our formula would be
T = 10,000 * 0.000001 * 5
T = 0.01 * 5
T = 0.05
So for our example, we know that after 0.05 seconds we will be at 99.3% charge close enough to know if our capacitor is good. Just so you know, if we use the same capacitor value but change the resistance from 10 kOhms to what this particular setup in the video would be less than 1 ohm, however, I'm going to use 1 just because at these values it really doesn't matter.
T = 1 * 0.000001 * 5
T = 0.000001 * 5
T = 0.000005
So to finish this long post, unless you have a very high capacity capacitor and a very high resistance, you will charge the capacitor almost faster than you can clip it on and take it off. However, you may have a large value capacitor you want to test so I find it better to explain the math so you can do these calculations yourself.
Hope this helps and was easy to understand, I wanted to explain it in a way that even someone not familiar with electronics could understand it.
@@111boldoo The capacitor will be fully charged almost instantaneously. When it discharges it will release 50% of its power - then 50% of the remaining power - then 50% of the remaining power and so on and so on till discharged or charged back up.
The last part i smile😅, its just like your testing the voltage of the battery
he forgot to remove the battery lol
Thank you. Can I please ask, if a TV power board has 2 x 450v capacitors should they read at 450v or is it just the max tolerance?
Mine both read 350v would there be a problem with a component upstream or is that normal on standby?
Thank you Tim
it would help if you showed a failed capacitor...what will that show?
It was plain and simple instruction thanks for your help.
The world's best thanks
Nice1..Electro & thank u
Thank you for your support!
3:02 "....make sure your buttery (power source) is off" What is buttery? If it is a typo for battery, it shows connected.!!
There are two types of people:
1) Those who can extrapolate from incomplete data
@@_LuxStein and those that are buttery
How do I test a capacitor on a PCB without removing because I want to check if the capacitor is the cause of the reason the PCB isn't working?
Nice info, thank you for sharing it :)
But how can I test a 2200uf capacitor if multimeters only go about 20uf?
In the voltmeter test , you got 9 volts and you said that the capacitor is good. If the capacitor was bad what value should I read in the voltmeter test ?
Thank you
That will read whatever the battery is, he is measuring the battery. This has nothing to do with the capacitance.
I thought the same thing
Shorted capacitor will bring 9v to much lower reading.
If done correctly, should show the same voltage as the battery if given enough time to charge the capacitor.
Has there ever been a point in history where it's worth it to remove every single part, and test every single component to fix something?
For science!
@@AeroCraftON i suppose if it helps. this made me realize i'm throwing away my old car eq though
You don’t, and that is why you learn to read schematics. You trace the board and find the comment that doesn’t work well and perform the test. I’d you are going to remove everything on the board just to test it that means you don’t know anything about it and have zero idea what you are doing.
When hvac co’s want to charge you $19k for a new system. YES! 😊
ESR meter will test capacitors in circuit multimeter is the worst thing to test a capacitor with
Appreciate.thanks
Fantastic
😊😊
What about the ESR?
can i test capacitor without removing it from the circuit board?
No
@@pontiacexchange4386 ok
@@pontiacexchange4386this is partially incorrect- you can use an ESR meter in circuit
This is a really poor video. You can’t test the ability of the capacitor to take a charge (voltage) while the battery is still connected!!
Helpful video, I love your tutorials
ITS BEST FOR THE BEGINNERS
Do i have to remove every single capacitor? Can't i do it while on the circuit board.
no
Thanx
2:00 is 8.5 a good value for a 10uf capacitor? 15% lower?
I looked up capacitor tolerance online and it says up to +/- 20% is acceptable. More expensive capacitors can have +/- 1% tolerance values.
I just watched another video saying he replaces capacitors if they are out more than 10%. Maybe that is what technicians are told when to replace capacitors. A technician may want to swap them out at 10% just so they don't have to come back later for another service call? IDK. What do you think?
@@kevinerose yeah, maybe, they probably lose value over time.
@@kevineroseThis is the case to my best knowledge, the capacitor already has demonstrated its starting to fail rather than having a callback that we have to prioritize we just replace the capacitor outright, they’re very cheap usually
@@kevinerose +- 20 % is ok, because on circuit boards manufacturers use higher rated capacitors than really at least needed. But when ever it's possible just replace every one below 10 % of its rating. You don't want to repair things more often, just because you didn't replace a cheap capacitor below 10% of its rating.
thx
but with no markings how do I test to know what side is what? + -
There are non-polarized capacitors with no defined + and - contact. With those it doesn't matter which side is connected to positive and which to negative. If it has no markings it's probably non-polarized sometimes also referred to as a bipolar capacitor.
Most ceramic capacitors for example are non-polarized
Third test didn't make any sense to me. If you'd removed the battery wires first I'd have understood it was something to do with the cap storing a voltage, but as it was you were effectively just connected to the battery were you not? I'm just here for some help with wiring a guitar. Don't pretend to understand this stuff at all.
This test was NOT correct. In the last step, he basically tested the battery because the 9v battery was still connected. What he should have done was unplug the jumper cables first and then test the voltage of the capacitor.
The voltage test is a false set up because you're just measuring the voltage of the 9V battery.
why infinity ohm on a cap? and first so low?
When you charge a capacitor you have current going in right? The closer it gets to being full the less current flows until it stops completely. Remember the basics: high resistance means low current and vice versa. A multimeter measures resistance by applying a low testvoltage and measuring the resulting current. And with voltage and current it calculates the resistance. When measuring the resistance the capacitor gets charged with the multimeters testvoltage. While the capacitor is charging from empty to full the current decreases until it hits zero.
The resistance behaves opposite to the current.
So if the resistance doesn't continuously climb to infinity during the measurement it means there is a low resistance electrical connection inside the cap between the two sides that shouldn't be there and hence the test result would be: "failed"
Checking the battery not the cap
Ha ha ha haaa! First remove the capacitor..............no, first discharge it!
The voltage test at the end was the most brain dead shit I've ever seen
🙏🙏🌹🌹
Bset
there's an "L" in soldering.
Depends where you are. The British say solder (with the L) and the Americans say sodder (without the L) 🤷🏼
@@ElectroUniversity No. Americans leave the L out. Everywhere else correctly, leave the L in.
Nice1..Electro & thank u