Rework of timer on coffeemaker - Page 1
Forum section: electronics / beginners

author: CatalinaWOW
author_rank: Super Contributor
timestamp: December 31, 2015, 06:56:40 pm
content: Not quite a newbie question, but close. The old coffeemaker died and a replacement was purchased. The new one is fine, EXCEPT, it incorporates a timer to shut itself off after two hours. Not consistent with my style of stewing the brew all day as I sip and work. So I dive into the innards and find a small circuit card. After reverse engineering it is a PIC processor that samples the power switch and controls the power relay. The power switch is a low power unit so I can't just bypass the board. While the PIC is unmarked it wasn't hard to find a package that can replace the current unit, and reprogramming the new processor isn't an issue. So I can fix the coffeemaker to my preference without problems. Front and back of the circuit card are shown here (it is actually a dual coffee maker so there are two relays and two switch input sections.)The problem is my curiosity. I don't fully understand the power supply section of the circuit. The attached schematic shows that segment of the circuit. Fundamentally it is a half wave rectifier driven directly from the mains with two simple zener regulated sections, one supplying 12.5 volts for the relays used and the other 5.2 volts for the PIC. R1 and the ZNR1 provide protection against line transients. C1, R1 and R2 provide the impedance required to limit the current in the regulator section.Any number of questions about this circuit come to mind.Why use a parallel RC circuit to drop the current into the zener regulators? (The value shown for C1 is currently a guess, based on providing a reasonable impedance at line frequency. There are no legible markings and I have not yet unsoldered the part to measure it. The coffeemaker is in use.) It seems that a purely capacitive approach would be better from a safety standpoint. Are the resistors merely to provide a discharge path for the capacitor if the unit is unplugged? Did this safety concern over ride concerns about isolating the mains from downstream connections?Why split the parallel resistance in the input resistance into two parts (R1 and R2)? Parts of the total value are readily available and the power dissipation is easily handled in all but the smallest SMD parts. Parts of the size used could easily handle the power.Why not use 5 volt relays and get rid of the whole 12.5 volt section?Given that there are two sections, the purpose of R9 and D5 in the circuit appears to be to supply current to the 5.2 volt section if the 12.5 volt section glitches down during relay operation. Is that really what is going on?What purpose is served by D1 and D6?Altogether this design seems far too baroque for a cost sensitive consumer item. What am I missing? I would appreciate any comments.

author: Seekonk
author_rank: Super Contributor
timestamp: December 31, 2015, 09:35:36 pm
content: We built a lot of products with this type of power supply. The first resistor limits current surges and acts as a fuseable element. These often fail from surges and show no visable sign of stress. As noted the other two resistors discharge the capacitor. This is required by certification agencies. The current through the resistors is nothing compared to that through the capacitor. Two are used to overcome flashover and use a cheaper resistor. A 5V relay takes toooooooooo much current. They would have to have a much larger capacitor.