I haven't spent much time on this, but the deadband drive (90usec) and the 110v zeners and 250v c1/c3 mean that the loads are definately inductive, and the high voltage is intentional. The 510K bleed r6/r7 could simply be for safety, to discharge any capacitive load (in addition to the inductive load) when power is turned off. I believe that the 1N4003's 200v reverse rating is also proof of the high voltage, and that those diodes and c3/r5 c4/r8 are snubbers, to quench the ringing that happens from the q1/q2 Ldi/dt turn-off. The alarm goes low whenever the pulsing AC voltage quits on either EdgeConn1 and/or EdgeConn2. This means that if the 555 fails, or the external inductors blow open, or the external voltage being supplied from somewhere to the inductor-loads (may not be the same 48v+ supply) fails, then the alarm goes low (asserted). A 400Hz reluctance motor comes to mind. If the inductance is correct, the AC voltage swing on the EdgeConn1 and EdgeConn2 will depend on the proper spinning of the motor -- perhaps the chopped swing AC voltage drops somewhat when the motor stalls, producing insufficient rectified voltage for the AND gate, triggering the alarm. Another possibility could be a 4-wire stepper-motor driver: however, the 400Hz makes this not quite as attractive an answer. I agree that a 400Hz power inverter, via a center-tapped transformer is also possible, but not as likely an answer. Another possibility is a BTL-connected inductor/capacitor charge pump, such as might be used in an LED lighting application. I whipped up a spice run of the circuit to try some inductors out, but I'm not happy with the results, and I'm out of time. Even with small inductors, the output hits the 110v zener clamps, as I suspected.
Actually, looking at this again, I'm thinking that the alarm signal is a LOW true signal. In that case, the alarm would detect an open circuit on either of the outputs. If the alarm condition is high true, then both outputs would have to be affected.
The zener plus snubber on the output implies that the load is high inductance such as you would see on a transformer input so I'd be inclined to believe that there is a center tap transformer on the output except that the circuit seems to be deliberatly set up so that both outputs can't be on at the same time...
The 250V rating on C1 and C3 and the 110V zeners imply that voltage spikes could get that high... C1 and C3 will block the high DC but when the output switches the signal after the capacitor will go negative so the diodes prevent that from getting to the AND gate and also provide filtering...
Way back in my misspent youth I spent a lot of time with slot car racing (Scalextric track with Monogram cars (the Scalextric cars were much slower and had a higher centre of gravity)) and at the time I thought I understood how the DC motors worked. It was reinforced through my A-levels with Flemings Left and and/or Right Hand rules and all was right with the world. When I got to university they told me that everything I knew was wrong, but the new explanations did not make much sense and besides, the huge motors they used in the labs always smelled and gave me headaches. When it came to heavy current, unfortunately I "switched off" and merely blindly documented results. But I digress.
Brushing my teeth this morning, it occurred to me that this puzzle circuit would switch the current much like my old brushes and commutator, so I am going to say that this is an electronic commutator for a 2 pole DC motor. If the load increases, so the current in the coil increases (and also the voltage) and would trigger an alarm on overload. However it doesn't explain the trickle current through R6/R7. Some relays like a wetting current, but ~0.1mA seems very light for that purpose and I don't think it would have much affect on any magnetics either. The directionality (by CR2) also suggests that my guess is wide of the mark
If this guess leaves me with egg on my face, EE Times gives me the option to delete the comment completely. Or when the answer is revealed, I can edit to make me look like a genius.
Maybe by load they're talking about what is on the secondary of the transformer. I think it might be some sort of lamp. I also want to see the alarm signal as a ready signal and not an error signal. Like it is a flash bulb. The alarm signal comes on when it is charged to a certain point and the snubbers are there to deal with the effect of the lamp firing. If it was an error signal it should shut things down or at least there should be a method for the user to shut things down.
@Antedeluvian - "Just a few more thoughts before I go to bed"....half your luck, my post was before I went to work, hence somewhat perfunctory. I am at work now, so goofing off to answer this (don't tell the boss :-)
> Isn't it -48V?
-Yes, but they might be using the -48 as the local ground, not real ground.....
The alarm circuit....the left side of C1/3, by my theories, would have an AC voltage of +/- 48V on them. That would be reduced by the dividers of R1/R9 and R2/R? to around 12V AC. CR4/6 would short the negative half cycles but CR 5/7 and C11/12 would put +12V on the CMOS 4011 inputs - the CMOS appears to get its supply from the +10V line so it probably would be happy enough with that. So assuming the loads are being driven, the output of the 4011 would be low, the alarm output would be high. If either output went off the alarm output would go low. - ie if the 555 or any of the driver circuitry went bad.
I'll bow to your superior knowledge on the power side, it sounds good to me. If there is a centre-tapped transformer on there, then the side that is not being driven would go to +96V, any spikes would get caught by those Zeners.
Just a few more thoughts before I go to bed- the mosfets are, by my standards, hefty devices capable of ~10A operation. CR8 and 9 are 5W devices and in forward bias could handle (just vaguely) 5A, so it would count against negative voltages, I think. And CR2 (and CR?) are 1 A devices, so not much continuous current is expected to flow here.
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