What is the purpose of D1 Diode in the Cap Dropper Circuit here - Page 1
Forum section: electronics / beginners

author: Aakash
author_rank: Contributor
timestamp: December 14, 2021, 11:47:48 pm
content: Can anyone please let me know what is the purpose of Diode D1 in the Cap dropper circuit

author: John B
author_rank: Frequent Contributor
timestamp: December 14, 2021, 11:55:28 pm
content: It's a half wave rectifier, it's needed in order to create a net positive voltage on the output.

author: Anthocyanina
author_rank: Frequent Contributor
timestamp: December 14, 2021, 11:56:50 pm
content: D1 and D2 form a 2 diode full wave rectifieroh yeah, not a full wave rectifier, i red the schematic wrong

author: Kim Christensen
author_rank: Super Contributor
timestamp: December 15, 2021, 12:40:09 am
content: C1 needs a path for current to flow when V2 swings negative. Otherwise it would just charge up to the +peak voltage via D2 and the load/D3. Once C1 was fully charged, current would stop flowing through D2.

author: DavidAlfa
author_rank: Super Contributor
timestamp: December 15, 2021, 01:32:03 am
content: Also, it act as a voltage doubler.The negative half-wave charges the capacitor, that voltage is later in series with positive half-wave.Depending on the capacitor value, you could obtain close to 2x the peak ac voltage.This also allows to use a smaller capacitor for the dropper circuit.The bad part is that's a half wave rectifier, so the output will have plenty of ripple.

author: gcewing
author_rank: Regular Contributor
timestamp: December 15, 2021, 06:13:11 am
content: Quote from: DavidAlfa on December 15, 2021, 01:32:03 amAlthough in this particular circuit the output is limited to 12V by D3.

author: Zero999
author_rank: Super Contributor
timestamp: December 15, 2021, 10:29:39 am
content: As mentioned above, without D1, C1 will just charge up and no current will flow.There should be a1M discharge resistor connected across C1, so it isn't a shock hazard, when the mains is disconnected.A fuse, is required to protect against fire, if C1 fails.A diode can be saved, if D1 is removed and the zener, D3 is move in its place, but the output voltage will be reduced by a diode drop, about 0.7V.Consider increasing C2 to 22µF, for better smoothing and to absorb the current surge, rather than it passing through the zener diode.Warning: everything connected to this circuit should be treated with the same precautions as mains voltage. The entire circuit should be placed in an insulated enclosure, made from a flame retardant material. If the enclosure it metallic, it needs to be connected to earth/ground.

author: nfmax
author_rank: Super Contributor
timestamp: December 15, 2021, 10:50:43 am
content: In such designs, R1shouldbe a 'fusible resistor', fulfilling the functions of protection both against fast transients and short-circuit failure of C1.

author: magic
author_rank: Super Contributor
timestamp: December 16, 2021, 11:51:42 pm
content: Quote from: Zero999 on December 15, 2021, 10:29:39 amThis is the only correct answerIt also improves reliability, because if the zener gets disconnected, the PSU will simply stop working (as opposed to frying its load with overvoltage).I will add that there are 13V zeners easily available if 11~11.5V is unacceptably low.

author: Zero999
author_rank: Super Contributor
timestamp: December 17, 2021, 10:31:51 pm
content: Quote from: magic on December 16, 2021, 11:51:42 pmThat's a very good point.And as I said, C2 should be much bigger. When the power is turned on, C1 and C2 form a capacitive divider. C2 needs to be large enough so it doesn't charge to too higher voltage, when the power is first turned on.C2 > C1×(VIN√2-VOUT)/VOUTThereforeVIN= 120VACVOUT= 12VC1 = 1.5µFC2 > 1.5×(120√2-12)/12 = 1.5×(170-12)/12 = 1.5×158/12 = 19.75µFSo 22µF as I mentioned in my previous post will be ideal.