Electronics > Repair
BERNINA power supply board (Switch Mode) issue diagnosis and beyond (fixed)
max.wwwang:
--- Quote from: inse on December 25, 2024, 12:41:50 pm ---If we break it down to the output stage as in your equivalent circuit, it’s definitely NOR.
As long as we agree that 0=0V and 1=Vcc (or at least >0V).
The truth table is shown for example here:
https://www.build-electronic-circuits.com/nor-gate/
The Transistor cannot be blown as the base current is limited by the emitter resistor (1k) and whatever circuit is behind it.
--- End quote ---
I realise we are actually agreeing with the same end result only relative to different inputs. What I’m saying is, if without the knowledge of the comparator’s internal circuit, the outputs of two tied together can be treated as these two outputs passing through an invisible (and non-existing) logic AND gate.
Your view point is from the signals on the bases of the two transistors (each through a resistor). Relative to these two ‘inputs’ (let’s call them B1 and B2), then the combined outputs of the two transistors is certainly NOR(B1,B2). Note that B1/B2 are only internal business of the comparators.
We are saying the same thing, because —
O1=NOT(B1), O2=NOT(B2)
AND(O1,O2)=NOT(OR(NOT(O1),NOT(O2)))
=NOT(OR(B1,B2))
=NOR(B1,B2).
On the transistor after the op amp, all after P190-2 is shown in the shaded box nearby, an LED going to GND, which is the only channel sending signal back to the main business centre of the machine. On what condition will this transistor turn on and off then?
My analysis is here (based on the behaviour of ideal op-amps):
1. When T125 (on the PSU board) is off, i.e. the motor is off (T126 is either on or off), P190-10 (connected to U1A-3) is 'subordinate' and will mirror whatever voltage on U1A-2, which will be 'driving' and is determined by the R ladder (along the dash line), and is 0.1V.
2. When T125 is on, i.e. the motor is on (T126 must be off), P190-10 (U1A-3) will be forced to what is determined by the HVDC voltage (assuming 320V) and the R ladder (R16/R17/R18), which gives 2.1V (marked as "3.8V" in the schematic). U1A-2 is also flexible and is able to mirror this voltage (from 0.1V to 2.1V) by increase of current through the last 1k R on the R ladder. This transit will present a momentarily high level on U1A-1, which turns on the transistor and in turn the LED D109 (signalling to the business centre that the motor is being turned on). As soon as the levels on U1A-2/3 are (quickly )balanced, LED D109 is off.
3. When motor continues to be on, the levels on the steps of the R ladder (along the dash line) are at different corresponding values, which will have an effect (as inputs) on the corresponding comparators. (The logic is complex, and more is to be figured out.)
Similar to the case of two comparators' outputs tied together, but slightly different, is where U3-1 and U3-2 are connected through a diode. In this case, the latter can pull the former down, but not vice versa. In other words, whenever U3-1 is LO, U3-2 will be LO regardless of the level of U3-2 when not tied together. But the level of U3-2 will not affect U3-1 at all. (It's puzzling what the function the relative independent part, top right around U3B, is.)
My feeling about the overall working of this daughterboard circuit is like this (starting from the idle status – motor is not running). The motor on instruction comes in through phototransistor T128, on the PSU board (copied to the bottom left corner), turning on the 3F transistor, starting the charging of the C1 (which is part of the RC oscillator). Through complex log, ultimately, square wave is output through P190-6 (bottom right). This will turn on (and off) transistor T125 through transformer U146. The motor speed is able to be sensed outside of the PSU board, which can be used to adjust the intensity of light opposite of T128 (D108).
MathWizard:
--- Quote from: inse on December 21, 2024, 12:29:42 pm ---There is an auxiliary winding between U146-5 and -6 that is generating the supply voltage on the primary side, just like the secondary windings except that it‘s tied to primary GND.
There is an example circuit of a flyback converter in the Onsemi datasheet of UC3844 which shows a lot of the features of your patient.
Except its using the aux winding for both supply and feedback, not an optocoupler.
Or look at this app note if you really want to dive deeper: https://ww1.microchip.com/downloads/en/Appnotes/AN18-APID.pdf
--- End quote ---
Ok they're using an AUX winding for feedback too. Yeah I forget SMPS can use windings for control, and not just opto's, or even high DC resistance networks between sides too I guess.
A lot of the smaller switching transformers I've seen over the years, in things like DVD players and TV's, are just general parts, with extra pins for windings that are missing. Sometimes they would still have a winding, and just short both ends to GND. In some other model, they probably use that winding. But for some reason in this model, it's still economical just to use the same trans. with extra copper/etc, but short it.
max.wwwang:
--- Quote from: MathWizard on December 26, 2024, 07:10:29 am ---
--- Quote from: inse on December 21, 2024, 12:29:42 pm ---There is an auxiliary winding between U146-5 and -6 that is generating the supply voltage on the primary side, just like the secondary windings except that it‘s tied to primary GND.
There is an example circuit of a flyback converter in the Onsemi datasheet of UC3844 which shows a lot of the features of your patient.
Except its using the aux winding for both supply and feedback, not an optocoupler.
Or look at this app note if you really want to dive deeper: https://ww1.microchip.com/downloads/en/Appnotes/AN18-APID.pdf
--- End quote ---
Ok they're using an AUX winding for feedback too. Yeah I forget SMPS can use windings for control, and not just opto's, or even high DC resistance networks between sides too I guess.
--- End quote ---
Feedback is still through the secondary side, with octo-coupler. There is auxiliary winding on the primary, which is more commonly on the secondary side and isolated, so is 'hot' (but is not a problem in this application, because it ultimately serves the motor, which is expected to be 'hot').
This circuit is very interesting, and I'm deeply intrigued.
--- Quote from: MathWizard on December 26, 2024, 07:10:29 am ---A lot of the smaller switching transformers I've seen over the years, in things like DVD players and TV's, are just general parts, with extra pins for windings that are missing. Sometimes they would still have a winding, and just short both ends to GND.
--- End quote ---
There appears, as introduced above, to be a winding (primary side) both ends of which are tied together. I just wonder if that can possibly be done without blowing the transformer?
MathWizard:
Maybe I'm mixing that up, since an open winding, would have no current, and a closed loop being cut by magnetic flux would have max current.
max.wwwang:
--- Quote from: max.wwwang on December 25, 2024, 10:06:48 pm ---(It's puzzling what the function the relative independent part, top right around U3B, is.)
--- End quote ---
Figured it out – the subcircuit around U3B works as an oscillator and square wave (between +12V and 0V) generator, output on U3-1 (i.e. U3B-1). Its frequency will be dependent on the values of the resistors and the (only) capacitor. This will shape the final output on U3-14 (P190-6), in a way that whenever it is LO, the final output will be HI (it's HI cycles will not have any effect).
The relevant part reproduced in the attached.
Such application is explained here: https://www.electronics-tutorial.net/analog-integrated-circuits/op-amp-comparators/comparator-as-a-function-generator/
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version