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.
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...
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.
I'm going to guess, based on the 110V zener diodes in the output, that the load is around 45mA (since that's what the zeners are going to be conducting when the JFETs are off and the opposite rail is active).
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.
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).
That makes much more sense. I missed that red bar over the ALARM.
I don't see why you would need to prevent the two transistors from turning on at the same time in a motor. I think a two phase motor would have its coils driven 90 degrees out of phase not 180 degrees.
If you look closely at the schematic, you'll see that the connection points on the four off-page connectors are red, which means they're not actually connected. [PCB designer rule #1: never trust the schematic. Always check the netlist!] This means there's no power in, no connection to loads be they inductive or otherwise, and no connection to ~Alarm.
Actually, I'm going to go with a couple of off-beat guesses. Since the input voltage is +48V, that sounds like four 12V batteries, which is a common configuration battery back-up of serious electronics. I'm guessing a really loud alarm bell, actually a pair of bells with a hammer between them which is driven by two electromagnets alternating at 400 Hz. I mean really loud, the kind of alarm that clears a building in case of fire.
My other off-beat guess is a flipper torpedo, with a single tail fin powered by alternating electromagnets. The reason for a flipper rather than a circular propeller is so that it has a passive sonar signature that sounds like a fish.
@betajet My other off-beat guess is a flipper torpedo, with a single tail fin powered by alternating electromagnets.
This would be great if used on one of those helium-filled air-floating sharks that Max wrote about a while back. The 400 Hz tone would be a bit annoying, but at least would give warning that the beast was approaching - from miles away! And could also double as a fire alarm bell if two gongs were located near the tailfin. :^}
As for the red connection points, thay are just part of the net naming symbol in TinyCAD schematic drawing software. One has to position the red connection point on the wire to give it a name, so there really is a connection.
As seen from the protection stuff on the outputs I think it is driving a small transformer, probably center tapped. This transformer is driving 2 nasty devices capable of giving a big burst of energy back to the primary drivers. It might be 2 enormous MOSFET's or IGBT's that -depending on the speed and load- are able to destroy direct logic driving. I think at least I should win half a pint of lukewarm beer....(Yikes! ;-)