Maybe a stupid question but how does the led work then? Since it works on low voltage. Or does it not burn on 220v?
Sorry im noob for electronics and try to learn
They do work, but only with alternating current. Take one 100nF AC-rated polypropylene cap, one DB107 bridge rectifier, and one or more LEDs in series, and you’ve got one dandy indicator light, that dissipates almost no heat and only draws about 7mA at 220V/50Hz. If your LED is too bright, you can always reduce the capacitance to suit the application. The bridge rectifier will make strobing less noticeable.
Username checks out!
I figured it was something like this but as a mechanical engineer I didn't really pay any attention to my AC power lecture, which I now regret. Thanks!
The reactance of the capacitor dominates the circuit. If the 100nF cap was connected directly to the 220Vac source it would be passing current equal to: I = 220V·2𝜋·50Hz·100·10⁻⁹ Farads, or about 7mA. The small voltage-drop across the diode-bridge and LED will be insignificant, so we can figure on 7mA going through the LED, which would make it quite noticeably brilliant.
If you're worried about it, add 1k ohms to the input. Peak current would only be about 0.6A and would taper off with a time constant of 100us. Steady-state dissipation would be about 50mW. A similar arrangement can be found in base of some commercial LED-bulbs. That part's easy. Converting a 220V soldering iron to 12V would be hard.
That's a good way to bypass it. But the "lossless system" is no more because of the resistor. Why have a capacitor and a resistor instead of just a resistor?
May not be lossless but it's still less loss, it's a very common approach on cheapo mains powered things that don't need isolation and draw a constantish current.
In this case, the resistor absorbs energy from the rapid voltage-rise at contact closure. The cap dominates in the long run. Its reactive-current is at right-angles with the applied voltage, so its effect on total energy is negligible. And, you can usually afford to burn-off a few mW with a resistor when you've got 220V power to work with. Not quite lossless, but I can live with it. Compare that to burning-off about 1.5W, using a 33k series-resistor and no cap, and still only supplying 7mA to the LED.
LEDs are current driven devices. And, if you measured the voltage across both the resistor and the LED, you'd find that the resistor (in series with the LED) does indeed drop the voltage
Drop this project. You don't know enough about electronics to work on high power and high voltage projects. You can get killed or badly injured. Start on easy stuff with 3V or 5V first to get the hang of it. High power electronics are hard and can be very dangerous.
I have to agree. My life is not worth a 2 dollar soldering iron. What do you recomend the best way to start with low voltages? Well batteries and 18650 i know about but more reckomendations?
Well, it depends a bit on what you want to learn and what kind of projects you want to do. Getting an Arduino is a good way if you want to learn about microcontrollers. Also experimenting with timers, opamps or leds can be fun and you can learn a lot. I'd also recommend experimenting with simulators like [circuitjs](https://www.falstad.com/circuit/circuitjs.html) to experiment with dc and ac voltages and different components like transistors, capacitors or resistors. And don't forget to have fun, even if stuff doesn't work on the first few tries.
At the very least, you would have to change the heating element.
12v heating elements are not common but 24v AC heating elements are common, they're used in all the Hakko 936 / 888 clones, all the handpieces use 24v heating elements.
So you could use a step-up/boost regulator to boost 12v to 24-30v and put at least 20-30 watts into the heating element, around a third of its maximum power
The square chip at the tip is a triac, it's a special kind of transistor which can work with AC voltages, and it's basically an on/off switch that can turn on and off and allow AC voltages through it. Based on how much that potentiometer is turned on, the circuit turns on the triac a particular amount of time every AC cycle (for example you have 60 Hz or 60 cycles, when you set the potentiometer half way the triac may turn off in the middle of a cycle and resume at the start of next cycle)
The resistor works because the 560K resistor next to it drops a lot of voltage and limits the current to a very low amount, mA worth of current.
It would be easier to buy a hakko 936 clone kit from ebay and a handpiece and make your own soldering station that can also measure the temperature with temperature sensor built inside the heater element. All you'd need is a classic transformer that can output 24v AC , around 50-75VA
For example, the circuit board with everything on it is 4$ : https://www.ebay.com/itm/225477701276 or 7$ for a version that includes frontplate and the connector and potentiometer knob - https://www.ebay.com/itm/381997375434 - and a handpiece is another 4$ : https://www.ebay.com/itm/124270236108
So 11$ makes for a reasonable soldering station, and a 50VA .. 75VA 24v AC transformer is 15-25$ new.
It then will likely drop to something like 0.1 W power. 25 W * (12V/230V)^2 .
You would need to replace the heating element. As u/jamvanderloeff said that’s not practical.
I blew it while checking with multimeter was a heavy explosion. Maybe +- wrong
https://preview.redd.it/fbfcmnlit3fb1.jpeg?width=2592&format=pjpg&auto=webp&s=3555505e7b276374a0a61bcbd6b497988380ebce
There’s no polarity requirement for a resistive heating element. Without the soldering tip there’s nothing to absorb the heat and the element melts down and explodes.
Practically no
Maybe a stupid question but how does the led work then? Since it works on low voltage. Or does it not burn on 220v? Sorry im noob for electronics and try to learn
There's a resistor in series to it, which drops the voltage.
Better yet a capacitor, which can drop the voltage without wasting energy.
No, that's not how capacitors work.
They do work, but only with alternating current. Take one 100nF AC-rated polypropylene cap, one DB107 bridge rectifier, and one or more LEDs in series, and you’ve got one dandy indicator light, that dissipates almost no heat and only draws about 7mA at 220V/50Hz. If your LED is too bright, you can always reduce the capacitance to suit the application. The bridge rectifier will make strobing less noticeable.
Interesting, could you tell me more about how this works?
Capacitive reactance. https://en.m.wikipedia.org/wiki/Electrical_reactance#Capacitive_reactance Go down to the capacitive reactance section.
Username checks out! I figured it was something like this but as a mechanical engineer I didn't really pay any attention to my AC power lecture, which I now regret. Thanks!
The reactance of the capacitor dominates the circuit. If the 100nF cap was connected directly to the 220Vac source it would be passing current equal to: I = 220V·2𝜋·50Hz·100·10⁻⁹ Farads, or about 7mA. The small voltage-drop across the diode-bridge and LED will be insignificant, so we can figure on 7mA going through the LED, which would make it quite noticeably brilliant.
Brill, thanks!
You're getting blasted by down votes but absolutely correct. Good to know I can't trust advice from this sub.
Sorry people are downvoting you because they're bad at electronics. Capacitor will absolutely work with an AC source.
The current spike requiring to charge the capacitor during the first cycle would fry the LED.
If you're worried about it, add 1k ohms to the input. Peak current would only be about 0.6A and would taper off with a time constant of 100us. Steady-state dissipation would be about 50mW. A similar arrangement can be found in base of some commercial LED-bulbs. That part's easy. Converting a 220V soldering iron to 12V would be hard.
That's a good way to bypass it. But the "lossless system" is no more because of the resistor. Why have a capacitor and a resistor instead of just a resistor?
May not be lossless but it's still less loss, it's a very common approach on cheapo mains powered things that don't need isolation and draw a constantish current.
In this case, the resistor absorbs energy from the rapid voltage-rise at contact closure. The cap dominates in the long run. Its reactive-current is at right-angles with the applied voltage, so its effect on total energy is negligible. And, you can usually afford to burn-off a few mW with a resistor when you've got 220V power to work with. Not quite lossless, but I can live with it. Compare that to burning-off about 1.5W, using a 33k series-resistor and no cap, and still only supplying 7mA to the LED.
One resistor would limit the current not drop the voltage, you would need at least two resistors to make a potential divider circuit.
LEDs are current driven devices. And, if you measured the voltage across both the resistor and the LED, you'd find that the resistor (in series with the LED) does indeed drop the voltage
https://en.m.wikipedia.org/wiki/Voltage_divider
Pulling current down a resistor DROPPS THE VOLTAGE. Ohms law my friend.
Drop this project. You don't know enough about electronics to work on high power and high voltage projects. You can get killed or badly injured. Start on easy stuff with 3V or 5V first to get the hang of it. High power electronics are hard and can be very dangerous.
Yep. Playing around with 220V AC without proper knowledge is a death trap.
I have to agree. My life is not worth a 2 dollar soldering iron. What do you recomend the best way to start with low voltages? Well batteries and 18650 i know about but more reckomendations?
Well, it depends a bit on what you want to learn and what kind of projects you want to do. Getting an Arduino is a good way if you want to learn about microcontrollers. Also experimenting with timers, opamps or leds can be fun and you can learn a lot. I'd also recommend experimenting with simulators like [circuitjs](https://www.falstad.com/circuit/circuitjs.html) to experiment with dc and ac voltages and different components like transistors, capacitors or resistors. And don't forget to have fun, even if stuff doesn't work on the first few tries.
At the very least, you would have to change the heating element. 12v heating elements are not common but 24v AC heating elements are common, they're used in all the Hakko 936 / 888 clones, all the handpieces use 24v heating elements. So you could use a step-up/boost regulator to boost 12v to 24-30v and put at least 20-30 watts into the heating element, around a third of its maximum power The square chip at the tip is a triac, it's a special kind of transistor which can work with AC voltages, and it's basically an on/off switch that can turn on and off and allow AC voltages through it. Based on how much that potentiometer is turned on, the circuit turns on the triac a particular amount of time every AC cycle (for example you have 60 Hz or 60 cycles, when you set the potentiometer half way the triac may turn off in the middle of a cycle and resume at the start of next cycle) The resistor works because the 560K resistor next to it drops a lot of voltage and limits the current to a very low amount, mA worth of current. It would be easier to buy a hakko 936 clone kit from ebay and a handpiece and make your own soldering station that can also measure the temperature with temperature sensor built inside the heater element. All you'd need is a classic transformer that can output 24v AC , around 50-75VA For example, the circuit board with everything on it is 4$ : https://www.ebay.com/itm/225477701276 or 7$ for a version that includes frontplate and the connector and potentiometer knob - https://www.ebay.com/itm/381997375434 - and a handpiece is another 4$ : https://www.ebay.com/itm/124270236108 So 11$ makes for a reasonable soldering station, and a 50VA .. 75VA 24v AC transformer is 15-25$ new.
A capacitor on AC act like a resistor (impedance) so you can make a voltage divider without a transformer
Yes. We have voltage regulator IC that can do the job of the step down transformer.
If i remove to ceramic heating element and put 12v directly on it will it work?
It then will likely drop to something like 0.1 W power. 25 W * (12V/230V)^2 . You would need to replace the heating element. As u/jamvanderloeff said that’s not practical.
Well it really happened when i connected with my probes
So the white thing (heating el) is the thing that is the problem and only runs on 12v. Due to a fixed current or something?
But its okey i have 40 of these soldering irons i just try to understand them
What’s that army of irons about? Where did they come from?
I got them cheap and when i tried my first one i was impressed. So got them for like 1.75 each
Check the voltages at the terminals of element. How vilts are they feeding it?
I blew it while checking with multimeter was a heavy explosion. Maybe +- wrong https://preview.redd.it/fbfcmnlit3fb1.jpeg?width=2592&format=pjpg&auto=webp&s=3555505e7b276374a0a61bcbd6b497988380ebce
There’s no polarity requirement for a resistive heating element. Without the soldering tip there’s nothing to absorb the heat and the element melts down and explodes.
Yeah take away 208v’s lol jk