标签: transistors

What are the missing loads? This circuit board plugs into something, but we know not what.   Click here for a larger image.   From the clues (there are at least four of them) hidden in the schematic above, what would you say are the most likely loads for the drains of transistors Q1 and Q2 connected to EdgeConn1 and EdgeConn2, respectively? Also, please describe what led you to your conclusions.   Glen Chenier Engineer

A week ago, I posted this blog in which I posed a puzzle involving the following circuit:   Click here for a larger image.   As I noted, these are a number of clues hidden in the schematic. From these clues, your mission was to describe the most likely loads for the drains of transistors Q1 and Q2 connected to EdgeConn1 and EdgeConn2, respectively.   Now, all will be revealed. Let's start by performing some simulations, and use the results to try to figure out what this circuit is all about.   Clue No. 1 The NE555 is wired as an 800Hz astable with a highly offset duty cycle. Its inverted output waveform, shown below, is at Pin 2 of the 4069 IC. The vertical scale is normalized to 1 for a 10-volt swing.     Clue No. 2 The D-type flip-flops and NAND gates produce the gate drive of heat-sunk power FETs Q1 and Q2 as shown below. Because the two flip-flop clocks are inverted, Q1 switches on the rising edge of the above clock and Q2 switches on the falling edge. The low duty cycle of the clock causes a brief interval where neither Q1 nor Q2 is on at the same time. The frequency is 400Hz, which is a standard power frequency (hint, hint).     Clue No. 3 Knowing that the main power input is 48 volts, let's assume the FETs are switching resistive loads that pull up to +48 volts. The resulting DC levels at the NAND 4011 inputs (pins 8 and 9) are generated by AC-coupled diode peak detector networks, but after startup transients stabilize the level is wrong for a CMOS gate with a Vdd of 10 volts. Not only is it too close to the switching threshold, but is always going to be seen as a logic 0 whether the FETs are switching or not.     Clue No. 4 To get the peak-detected DC level at the NAND inputs to a proper logic level of at least 7 volts, let's assume that the FETs' peak drain voltage is actually doubled or 96 volts. Now the NAND gets valid HI input logic levels, either of which will pull down to LO if either FET stops switching and gets stuck on (shorted) or off (open).     Yes, of course we all know that -- in the real world -- the NAND input clamp diodes would not allow the initial transient to rise above 10.5 volts, but the simulator doesn't know that.   Clue No. 5 The FETs are shunted by zener diodes, which clamp any positive spikes to 110 volts. They are also shunted by RCD networks that look suspiciously like snubbers. From all of the clues above, can you now deduce the load?   The answer to life, the universe, and everything Zener clamps and snubbers imply inductive loads. The FETs' on times do not overlap, so they probably have a common load. The alarm monitor NAND wants double the 48V supply voltage, and the frequency is a power standard 400Hz. If you deduced that the FETs are connected to the ends of a center-tapped power transformer primary winding, you'd be absolutely right. Such a load most certainly would not be happy if both FETs were to be on simultaneously, and the push-pull drive causes the opposite ends of the winding to pivot about the 48 volts at the center tap. So when one side of the winding is pulled to ground by its FET turning on, the other end of the winding rises to 96 volts as shown in the simulation below.     Thanks to everyone who took part in solving this puzzler. Please post any comments and questions below.   Glen Chenier Engineer

Can you identify the missing loads? This circuit board plugs into something, but we know not what.   Click here for a larger image.   From the clues (there are at least four of them) hidden in the schematic above, what would you say are the most likely loads for the drains of transistors Q1 and Q2 connected to EdgeConn1 and EdgeConn2, respectively? Also, please describe what led you to your conclusions.   Glen Chenier Engineer

Several days ago, I posted this blog in which I presented a puzzle involving the following circuit:   Click here for a larger image.   As I noted, these are a number of clues hidden in the schematic. From these clues, your mission was to describe the most likely loads for the drains of transistors Q1 and Q2 connected to EdgeConn1 and EdgeConn2, respectively.   Now, all will be revealed. Let's start by performing some simulations, and use the results to try to figure out what this circuit is all about.   Clue No. 1 The NE555 is wired as an 800Hz astable with a highly offset duty cycle. Its inverted output waveform, shown below, is at Pin 2 of the 4069 IC. The vertical scale is normalized to 1 for a 10-volt swing.     Clue No. 2 The D-type flip-flops and NAND gates produce the gate drive of heat-sunk power FETs Q1 and Q2 as shown below. Because the two flip-flop clocks are inverted, Q1 switches on the rising edge of the above clock and Q2 switches on the falling edge. The low duty cycle of the clock causes a brief interval where neither Q1 nor Q2 is on at the same time. The frequency is 400Hz, which is a standard power frequency (hint, hint).     Clue No. 3 Knowing that the main power input is 48 volts, let's assume the FETs are switching resistive loads that pull up to +48 volts. The resulting DC levels at the NAND 4011 inputs (pins 8 and 9) are generated by AC-coupled diode peak detector networks, but after startup transients stabilize the level is wrong for a CMOS gate with a Vdd of 10 volts. Not only is it too close to the switching threshold, but is always going to be seen as a logic 0 whether the FETs are switching or not.     Clue No. 4 To get the peak-detected DC level at the NAND inputs to a proper logic level of at least 7 volts, let's assume that the FETs' peak drain voltage is actually doubled or 96 volts. Now the NAND gets valid HI input logic levels, either of which will pull down to LO if either FET stops switching and gets stuck on (shorted) or off (open).     Yes, of course we all know that -- in the real world -- the NAND input clamp diodes would not allow the initial transient to rise above 10.5 volts, but the simulator doesn't know that.   Clue No. 5 The FETs are shunted by zener diodes, which clamp any positive spikes to 110 volts. They are also shunted by RCD networks that look suspiciously like snubbers. From all of the clues above, can you now deduce the load?   The answer to life, the universe, and everything Zener clamps and snubbers imply inductive loads. The FETs' on times do not overlap, so they probably have a common load. The alarm monitor NAND wants double the 48V supply voltage, and the frequency is a power standard 400Hz. If you deduced that the FETs are connected to the ends of a center-tapped power transformer primary winding, you'd be absolutely right. Such a load most certainly would not be happy if both FETs were to be on simultaneously, and the push-pull drive causes the opposite ends of the winding to pivot about the 48 volts at the center tap. So when one side of the winding is pulled to ground by its FET turning on, the other end of the winding rises to 96 volts as shown in the simulation below.     Thanks to everyone who took part in solving this puzzler. Please post any comments and questions below.   Glen Chenier Engineer

In the latter part of the 80s, I was on a bike ride with my two sons, one about eight was on his own bike and the other, who was maybe two, was strapped into a kid seat over the rear wheel of my bike. On our usual ride, which runs through a small neighbourhood along the river, something electronic looking caught the corner of my eye. Funny how that works, but I'm sure many of you know exactly what I mean. It was the evening before trash day and partway down a small side alley I saw something sticking up out of a trash can with just the back panel visible. We rode on over for a closer look and I pulled on it. Other bits of garbage slid back into the can as a Pioneer SX-535 receiver revealed itself—this might be a real score!   Even spending time in the trash, it only had a small scar on the top back edge of the wood cabinet. Not a problem as far as I was concerned. The lady of the house spotted us, and after a short conversation summarised, "It doesn't work anymore," and granted that I could have it. I tried riding with it under one arm, but the terrain here was rough and it didn't take long to conclude I might either drop it or dump my bike with my young son strapped in. Rationally, I set the receiver back down and we rode for home; driving back to pick up the unit. Analogue wasn't my strength, and my test equipment at the time was a really cheap analogue voltmeter and a similarly priced soldering iron, but that wasn't going to stop me from trying to impress my son. I borrowed speakers from the living room and confirmed "it doesn't work anymore." It lit up, and the radio signal strength worked, but there was no sound, no clicks, no pops, no radio hiss, no nuthin'. And dare I need to say—no schematic.   With the cover off. We looked inside and saw cracks and bulges in the power transistors. The rest of the PCB and parts looked great, and a little probing said the power supply was good (reading plus and minus 30v). There was a pattern to the layout that let me easily check both channels and the voltages "seemed" OK.   Here's an undamaged view of the PCB and power transistors.   I thought it was a shame the fuses didn't protect anything, they were all good. I'm pondering the failure? short circuit? weak design? operator error? overheating? Perhaps a stray strand on one speaker lead shorted the channel? Maybe, but could that blow both channels? I'm thinking it was likely that the design couldn't stand up to heavy use, and the owners were likely running it hard. We pulled the power transistors out, but with bubbled and burnt cases, it took quite a study to extract what I thought to be the part numbers. Off we went to the local electronics supply house to see if they had the parts. They had the "Sam's" schematics for most everything—except this receiver. But I had pulled a good-enough part number for the NPN, and they recommended the complementary part. As I recall it was about $20 for a pair. I was willing to risk $20 on this dumpster salvage, but I wasn't ready to go "all in" for $40 since I didn't know if I had the fix. Under a watchful eye my son soldered in the parts. I put my voltmeter on the output, and gave the power switch a flip. I saw just a tiny pulse on the needle of the voltmeter as the power supplies came to life and it settled back down to about zero. As I probed about, all looked pretty good. I made a small adjustment to the bias drive for the power transistors, mostly because the control was there since I didn't see any effect on the meter. I powered down once more and attached the speaker.   Part of the schematic I didn't have back then – showing the power amp for one channel. Another flip of the power switch, and sound came out! I did basic voltage checks again—I sure wish I would have had a scope back then. I assumed the bias I was adjusting would affect the zero-crossing and quiescent drive, and I gently cranked up the volume. We're thinking—this is a pretty decent amp for the money! Off to the supply store for another pair of parts. I soldered those parts in, switched it on, was probing the circuit, making the small adjustments, and "poof"—the smoke erupted out of the parts I had just put into play. Argh! Had my son not been a witness, I would've let out a few choice words. Clumsy me, with a dull-tipped voltmeter probe, I slipped, and paid the price. OK, now I was into this dumpster salvage for $40; do I cut my losses, or run it on up for another $20. I made one more trip to the electronics supply, tried not to establish much eye contact for fear the guy might ask, "What takes three pairs of parts?" Successfully installed and this time I sharpened the probe tips before the next power check—all turned out well. That amp served the basement family room in the house for nearly 15 years. Then one day, when I came home from work, I learned "it doesn't work anymore..." That's when it joined the dead device shelf, taking up residence with an old Hi-Fi stereo VHS deck, an old cassette deck, and more...   What do I miss most about that receiver? I have to say, it's the high-contrast black labels on the silver panel—very easy to read! For some reason, everything new seems to be dark lettering on a black panel and these eyes don't appreciate it. About the author My dad was quite a mechanic and co-owner in a local machine shop. We kids got to play with tools beyond our years since we could stand—drill presses, sheet metal benders, metal lathes, and more. In the garage one day, I must've been about six (it was the mid 60s), I was studying an extension cord. On the end of the cord was a light bulb socket with a 2-prong outlet adapter screwed in. After unscrewing the adapter I asked dad, "What will happen if I put my thumb in there?" He said "try it"—so I did. After that experience, I was hooked on electrical things. I would tear apart motors and old tube radios. In my early teens I managed to "fix" an old BW tube tv, but also had some "experience" with the big lead on the side of the tube. After that I tried to avoid high voltage, but electronics became both my profession and one of my favourite hobbies. David Smart submitted this article as part of Frankenstein's Fix, a design contest hosted by EE Times (US).