If I were to take a standard AC to DC converter, say a laptop charger, and hook up the input side (which expects 120VAC at 60Hz) to a DC power supply of some sorts, will the electricity still be “converted,” or will it just not work at all? I am clearly very uneducated when it comes to electronics (albeit working on it) so I would very much an ELI5 answer Thanks!
Think about the circuit over time. The sine wave on the left is the input.
When the sine wave is positive, the top wire is positive and the bottom wire is negative. That means d1 and d3 are both positive and d2 and d4 are both negative.
Diodes “forward bias”, meaning they allow electricity to flow, when you have a positive on the triangle side. This also implies a negative on the straight bar side. If the positive is on the straight bar side, then it won’t conduct. You have to have a negative on the straight bar side.
Note: Keep in mind positive and negative means in relation to each value. So if one side is 5 volts, and the other is 0, then the other side is negative compared to the positive side.
So with that in mind, when the sine wave is positive, greater than 1,then d1 and d4 are conducting (there an implied loop on the left side).
When the sine wave is negative, then d2 and d3 are conducting. Because these two diodes are flipped, the high value goes to the top wire on the output which is now seen as a positive. The input is different, but because you’ve flipped the wires, the output is still the same. Now all current flows in one direction no matter the input. AC is now DC.
Shorter verison: positive side of the sine wave makes d1 and d4 conduct. Negative side makes d2 and d3 conduct. Since d2 and d3 cross over, they effectively turn the negative into a positive value.
Thank you for your response! Here’s my understanding so maybe you can point out where I’m getting confused.
When the AC is positive, there is current coming in from the top line. The current goes to D1 & D3. Since it’s positive, it gets through D1 and creates a positive on the DC side. Since it’s hitting the line part of D3, it isn’t allowed through and it stops right there.
When the AC is negative, there is current coming in from the bottom line. The current goes to D2 & D4. The D2 allows the current to go through on the positive side, and continues to provide a positive current for DC. When it gets to D4, it is hitting the line side of the diode, so it isn’t allowed through.
Both D1 & D2 allow for a positive charge, while D3 & D4 prevent a negative charge from coming. However, couldn’t that be achieve without havining D3 & D4. Since there is no negative charge on the DC side, D3 & D4 never allow current through at all. If we just cut the wires to D3 & D4, wouldn’t it be the same outcome? What are their roles?
Kind of. You’re on the right track.
What you’re missing is that there is a loop. Imagine a resistor being connected to negative and positive of the load. Since it’s about relative voltages not absolutes, d1 and d3 both conduct. A positive on the triangle, or a negative on the bar will make them conduct. For electricity to flow you must have a loop from source to load and back to source. So current would flow from the source through d1 through the load through d3 and back to the source.
Hope that makes more sense.
You could remove d1/d4 OR d2/d3 and you’ll have a half wave rectifier, so you’ll get every other peak.
Ohhh!! I think get it now. D3 & D4 are necessary to complete the loop. Otherwise, there would be alternating current on the bottom phase going into the DC negative once per cycle, causing a short. So when D1 is allowing positive power through on the top of the phase, D4 is allowing negative current out to the source. When the AC is on the bottom phase, D2 lets positive current in, and D3 let’s the negative current out to the source. Is that correct?