I'm looking forward to seeing the responses here -- in fact I'm just about to stroll over into the next bay and tell Ivan (the creator of the All About Batteries series) about this puzzle -- he's brilliant at this sort of thing.
My guess is the circuit is a 400 Hz inverter. The load is a transformer. The 555 is generating 800 hz signal. That gets divided down by 2 by the flip-flops. The two transistors are run in sequence 10 00 01 00 10 00 01 00 ... The transistors are tied to the ends of the primary. Power goes to the center tap of the primary. The secondary has a 400Hz AC signal on it.
Like Wnderer I think it's an inverter of sorts....arranged so there is no overlap on the Mosfets drive....so the load would probably be a transformer, centre tap primary, the outers connected to the transistors and the centre tap connected to the 48V supply. Who knows what voltage the secondary would be..??? 400 Hz is frequently used in aircraft systems, but I'm not sure what voltage (110V?) But 48V is a common telecommunications supply voltage.....
But 48V is a common telecommunications cupply voltage.....
Isn't it -48V?
I do agree that it appears to be switching two loads in anti phase. The outputs appear to have a belt and braces approach- it appears to be a zener for back emf protection, but also has some form of snubber. Perhaps that is some clue as to the load, but for the moment it escapes me.
The alarm circuit also gives some clues. Assuming the CMOS will switch state at 5 V the circuit looks for a smothed signal of above about 20V (assuming 400Hz) to trigger an alarm. But the alarm would not be open or closed circuit since a DC signal would not propogate across the capacitors C1/C3.
CR4 and CR6 in the alarm circuit would seem to imply the the signal could go negative. Are they merely good design practice or there for a real reason? And so perhaps CR8 and CR9 are more than back emf protection but aslo bypass for the transistor when (and if) the output signal is negative?
Diode CR2 and its unmarked partner in the other drive give directionality on the snubber, if it is indeed a snubber. There is a DC trickle current draw through R6 and R7, but what sort of device needs to have this low current to stay active?
The MOSFET drivers are not buffered so the output has no requirement for sharp edges.
Maybe it's my jet lag, but none of this leads me to any conclusion. I need to sleep on it.
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.
@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.
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.
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.
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