You are very organized. Makes the explanation of the circuit very clear, very visible. One of the best Electronics channels I've ever seen on RU-vid. And the best ????? Without those horrible songs that take the focus away from the explanation. Very good ! Congratulations !!! 👏👏👏 (from Brazil)
Consumers friendly technological designing are always different from business friendly. So many extra components and complicated designing are good for business.
The videos was very good and informative, you calculation and description of the components are good. If you can explain the about resistors and their values that would be great.
How did you calculate the value of the bleeder resistor and the 100ohm/2watt resistor in series with the neutral and the value of the smoothing capacitor
Thank you very much for your video, can I have a question on avoiding of shocking on touching the output: 1. Can we you a phase cut at 5 VAC wave from using Zero cross protection stuff like MOC3021 and a small micro controller, triag to handle the trigger for wave form cut so that the overall voltage will never bee too high? 2. Of course with this, the synchronous wave form will be cut and have some time no power, can we use capacitor/ inductor to supplement the loss of power during the cut of AC wave form?
In your calculation, your AC capacitor value to be used is 1uf....but practically you are using the capacitor marked 155 J the value 1.5uf. I think you should be using 105 (1uf) to be exact instead of 155 (1.5uf). I wonder if there are changes in the output voltage at point A if the capacitance of the AC capacitor are changed. Please clarify.
Sir, R1 &R2 are 270 k but in video u r telling 220 which is right? U r using one 2w resistance but in in details u have written 1 w what is correct ? Pl explain regards nice tuitor
This ckt.will certainly give a shock! 100mA output is very high. Actually such high currents should never be drawn from non-isolated sources. Fortunately Triac gating circuts that need 5 voIts on the input side will work very well even at 10 mA! So will be much safer to reduce current! Another trick to make it safer will be to have another capacitor with bleed resistor on the 2nd instead of the 100 ohm resistor! This way both wires on AC side of rectifier will become safe to touch.
For those that as of Amp rating, this is in the area of 10 to 50mA, depending of the input cap A 470nF X2 cap gives some 10mA This is not isolated power suplly, it has potential to ground or null in the range of 220V, it can shock or kill if you make a contact whit it
Hi thank you for your video and the time you took in making it. Is it possible to use this type of circuit to drop ac to a lower ac voltage. I have two 220v 30w LED flood lights. I want to reduce the voltage to about 200v for maximum led life. The LED are very bright and I think are being over driven. The units are sealed so I can not get inside them to lower the DC voltage. Is my only option to install a mains dimmer switch? Help, advise please.
Hi! the voltage at point B is much lower than the voltage at point A because the capacitor C1 has no time to completely charge?? If Im right, is not this ciruit dangerous? because if the capacitor breaks, the voltage in the output is too big... (Im not sure if the zener can keep it low)
Hi, can you please tell me what the R3 does, couldn't you connect it directly to the full bridge rectifier and ignore adding it? If you or someone else can shed some light on my question I would be forever grateful.
These are arbitrary values: The smoothing capacitor value needs to be as large as possible (but the larger the capacitance, the longer it takes to reach/charge the required voltage). The current limit resistor consumes a lot of resistive power and also not necessary during the operation, but can useful for sudden voltage spike in AC mains.
formula for that is C = I / 2 x f x Vpp, where I is the load current, f is the AC frequency (usually 50 Hz or 60 Hz, and Vpp is the allowable ripple. Choose a cap 50 volts or higher, not critical.
Not sure how you're reaching the figure of 6v across the input to the bridge rectifier. You're dealing with AC up to the point of the bridge rectifier, so your 6v would rectify to root 2 times that, minus 2x the forward voltage of the diodes you're using in the bridge. That's 8.4v minus 1.2v = 7.2v. dc output of the bridge. This would hammer your zener diode, (which should also have a series resistance to protect it?) I can't see how you can try to use DC voltages on the AC side of the rectifier. It just doesn't make sense to me.
WOULD GREATLY APPRECIATE IF YOU. COULD MAKE A VIDEO OF LED EMERGENCY LIGHT SIMILAR TO " DP LED " SERIES 'EMERGENCY UNITS THAT ARE AVAILABLE IN MARKETS.
Because high voltage across zener. Like if you apply zero increasing to 10 volts across a regular .7v diode. Out of circuit the diode measures maybe .54V then with power supply connected positive to anode and negative to cathode. Then increase the power supply from 0 to 10V and the diode goes to more than 1V unless the power supply is weak and cuts out decreasing voltage. Also diode can cook the work bench...
