Calculate resistor value of reverse engineered 120VAC to 15VDC 1A Linear PS - Page 1
Forum section: electronics / beginners

author: klen4rt
author_rank: Newbie
timestamp: May 09, 2025, 05:45:15 pm
content: I'd like to know how to calculate the resistor value of a linear power supply.I reverse engineered an Altec Lansing A1664 plug-in Class 2 Transformer power supply because it got very noisy. Turns out that the capacitor and resistor burnt out. The capacitor value is 1000uf @ 25V but the resistor is mostly unreadable but seems to be 1/2watt 4 band resistor. The first band is yellow, the last band is brown, but the middle two are burnt. I am a novice and haven't found a way to calculate the resistor value (at least that made sense to me). Attached are an image of the transformer label and another of my reverse engineered schematic. I'm just tinkering and would like to see if I can repair this wall-wart. I appreciate any help to figure our the value of the burnt resistor.

author: bdunham7
author_rank: Super Contributor
timestamp: May 09, 2025, 05:53:18 pm
content: Your schematic needs improvement, but we get the general idea. The resistor is just a bleed resistor to discharge the capacitor when the power is off so the value is not critical. 10k will work, but so will half or double that. If you want to amuse yourself, go to Microsoft Co-Pilot and ask "what value bleed resistor should I use with a 1000uF capacitor?". It actually gets it right and asks reasonable followup questions, including power dissipation.

author: KimLum
author_rank: Contributor
timestamp: May 09, 2025, 06:12:46 pm
content: You can play with values for discharging the capacitorhere. The first yellow band suggest something that starts with 4, the last brown band suggest it has a 1% tolerance. The second band is "less important" and the third band is the multiplier. My guess would be a value around 420 ohms which would bleed the capacitor from 15V to 5V in 0.5 seconds. Higher values bleeds slower and charges faster, 4200 ohms would bleed the cap to 5V in ~5 seconds.4 band resistor values.

author: mariush
author_rank: Super Contributor
timestamp: May 09, 2025, 06:41:18 pm
content: The diodes aren't 1n5932 ... 1n5932 are zener diodes.edit /// Jesus, they're 1n5392, not 5932 ... 1n5392 are regular diodes with around 1.1v drop at 1.5A of current, less at only 0.8-1A.Most likely you can use basic 1n400x diodes, we're dealing with low frequency stuff and low currents.Just for referenceA bridge rectifier converts the AC voltage to DC voltage, and you get a wavy voltage with a peak equal to :Vdc peak = sqrt(2) x Vac - 2 x (voltage drop on a single diode from bridge rectifier).So if your AC transformer outputs 15v AC, your peak voltage will be Vdc peak = 1.414 x 15 - 2 x ~ 0.7v = 20vI suspect the transformer outputs 10-12v AC and you'd get a peak voltage of around 17v DC at idle. (1.414 x 12 - 1.5 = 15.5v and most transformers output 10-20% more at idle, no load)The capacitor smooths out the dc voltage. You can approximate capacitance with formulaCapacitance (in Farads) = Desired Maximum Current / [2 x AC Frequency x (Vdc peak - Vdc minimum desired ) ]For example, assuming it's a 12v AC secondary winding giving you 15.5v peak voltage, and you want a minimum of 14.5v at the full load of 0.8A , then your capacitance would beC = 0.8 / 2 x 60 x (15.5 - 14.5) = 0.8/120 = 0.0066666 Farads or 6666 uF minimum to achieve that.without a linear regulator, you won't achieve a clean 15v output, and 1000uF is just too low to provide a clean output at 0.8A output, the maximum the adapter can supply.

author: fzabkar
author_rank: Super Contributor
timestamp: May 09, 2025, 07:02:24 pm
content: Quote from: klen4rt on May 09, 2025, 05:45:15 pmI don't understand what could have killed the bleeder resistor.

author: klen4rt
author_rank: Newbie
timestamp: May 09, 2025, 08:13:01 pm
content: Many thanks to all for the quick responses and the solutions (bleeder resistor), tools for this and future endeavors, and the detailed explanation of how the circuit works. I will order some parts.

author: klen4rt
author_rank: Newbie
timestamp: May 10, 2025, 06:46:16 am
content: @fzabkarMy only guess is that, since the capacitor leads slipped out of the solder leaving the solder ball intact with just clean holes where the capacitor leads were, maybe a loose capacitor lead shorted the bleeder resistor. Clearly a bad soldering job on the capacitor leads. Other than that, I don't have a clue.