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- Опубликовано: 28 авг 2024
- Episode 712
Back to basics: Power Supply
AC to DC conversion, -5V supply
using diodes to increase voltage in microphones: • #450 Rode NT1a and Neu...
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Great explanation! Breaking the power supply down into small sections is just the type of details I need to better understand the complete circuit. Thanks for sharing.
I wish you have been my lecturer for 'components and basic circuits of electronics' instead of the one who was ;-)
Thanks a lot. I am watching nearly every video of you.
Thank you so much for making the time to shoot this video as well as the other 1,467 ones you have done! I’ve been a big fan of the “Back To Basics” series as it has helped me learn a lot of things. This one about negative voltage was a big epiphany for me, as it never occurred to me if you turn around the diode, that will enable it to pass negative voltage rather than positive voltage. Thanks again, wishing you the best and take care!
This is super helpful. For example, the short 20 second bit where you indicate the diode "pushes up" the AC are exactly the little building blocks of understanding I need. Thanks!
Glad it helped
mankind will certainly appreciate this in the generations to come
Such a great channel man, feel like im getting more from these than any other videos. Really appreciate it! Please keep going.
Self study time, the best way to learn.
@5:20.. the circuit is called clamping circuit.. helps people research
Damn dude thxx
great
A few points, the first circuit you gave with only one diode will not work. That is because the capacitor after the 82 ohm resistor will charge up to a point where the diode will never forward conduct. Capacitors only pass ac current so you need the second diode to ground to pass the negative current through the capacitor to discharge it so that it can conduct current for the next cycle. This is a voltage doubler, so unloaded it will produce a voltage that is twice the ac peak voltage (minus a diode drops). If it is arranged in stages you will need capacitors in each stage to store charge to transfer the current in the next cycle. Also note that this kit really is under rating the 82 ohm resistor, it will dissipate alot of heat. My estimation with a 24vac transformer the power dissipated in the 82 ohm resistor is over 1W. I think the kit only has a 1/4 watt resistor.
You are right. I tried to keep the ideas very simple but I did handwave a bit too much on that one. Yes it would average to zero. I need to go look at the 82ohm you may be right. hopefully the circuit charges and the crest delta charge voltages are low enough to reduce the wattage needed for the resistor.
@@IMSAIGuy That resistor got pretty hot in my kit and I changed it out to a 2W. I think it is pushing it with the 1/4 watt resistor, but at least it should be installed with long leads above the board to increase heat dissipation. Not sure why they did not use a bit lower capacitance to and lower resistance value try to reduce the power dissipated in the resistor.
What is the purpose of capacitor before diode and after diode?
@@addaguriprashanth1127 The capacitor before the diode is basically acting like a level shifter. The capacitor after the diode is used for smoothing.
doubling and trippling voltages are usually using in CRT televisions .
fantistic video, very helpful. thank you
You are amazing, thank you for all great information. i have learned a lot
Keep the kernels coming. Biting off in small pieces.
so that was basically a clamper circuit before a clipper one if I am not wrong. But why do we need to push the voltage up or down, when we need only 5.1 V at the end?
On what basis did you calculate the values of the resistors 82 ohm and capacitor 47 micro ?
Sir please i want basic pcb layout diagram
@IMSAIGuy first of all a great video, loved and subscribed to ur channel, one thing i couldn't process was the diode which push up the ac, searched and found it was part of clamper circuit, but when I simulated the schematics of positive voltage power supply from 8:29 in Qspice(spec Sin (dc=4V, Vpp=12V 100hz) it was only showing a chopped ac without dc(-0.8V to 7.1V), not clamped as i expected, Am I doing something wrong?
did you have the capacitor?
What is that cap doing there between the resistor and the diode?
blocks DC current
@@IMSAIGuy had to look it up because I didn't think this was possible. Fascinating. Would you find this same set up in something like a solar inverter?