[ARCHIVED THREAD] - Electrical Circuits (Page 1 of 3)
Posted: 10/21/2011 1:57:27 PM EDT
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Anyone know a thing or two about circuit design, specifically using logic IC's? I ask because I've drawn out a schematic that doesn't work as I expect, but I can't figure out why...  T1 is a 24VAC transformer B1 & 2 are bridge rectifiers IC1 & 2 are 7805 regulators IC3 is a standard 74xx AND gate Not shown are the pull-down resistors for pin 1 & 2 on IC3 My problem is: Circuit works as intended (SW1 is closed and LED lights); but if SW1 is open it's STILL lit... Voltages are as follows: SW1 =Closed B1 input = 27.45VAC | output = 25.3Vdc B2 input = 27.45VAC | output = 25.3Vdc IC1 output = 4.73Vdc IC2 output = 4.73Vdc SW1 = OPEN B1 input = 0.452VAC | output = 15.3ish Vdc B2 input = 27.45VAC | output = 25.3Vdc IC1 output = 2.53Vdc IC2 output = 4.73Vdc ETA: DC ground is shared between IC 1, 2 & 3 and IC3 +V is from IC2 |
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Quoted: Put a resistor parallel to the LED. Capacitance is providing the energy for the LED. What value? Try a 50k ohm or so, all depends on the capacitor value. Or pull-down resistors parallel to the caps. Is it really that simple? That's what's causing a back-fed appearance? |
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Try using this:
http://www.falstad.com/circuit/ (The absolute most educating FREE circuit simulator I've ever found!) Right click to add components. Or try logisim for logic stuff, it ignores power supply things though. ––ETA: Your problem is the IC is at a floating ground. the Power for the NAND Gate needs to be common to both inputs, and the difference or sum of inputs cannot be greater than the supply voltage (typ 5V). You are putting two 5V signals to a logic gate, but the logic gate isn't grounded with reference to both 5V signals, add grounds after the 7805s and you'll be fine with a higher current power supply at 5V from a quick glance. Also, add a capacitor about 10x the value of the 7805 input cap to the output side. You are missing 7805 output caps, which are needed badly for load stabilization. Do not make the caps the same size, as this can make the 7805 become an oscillator. |
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The regulators are full wave.
When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... |
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Quoted: The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Use a half-wave rectifier (diode). I don't know enough about your circuit to be specific, but that should work. |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Can you put the switch between B1 output and IC1 input? |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Tie the commons together. Right now, the entire bottom half of the schematic is "Floating". Connect both "-" outputs of the bridges together if nothing else (assuming the input is from a transformer, and not 2 different transformers). Put the circuit into the visual simulator above. Also, why not just use 1 rectifier? Not enough current? |
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AZ-sky beat me. What he said is true. You are still supplying half a wave to the top rectifier.
Your switch needs to be after the transformer, but before the branch going to the rectifiers. As it is, when you open the switch, you are still feeding half a wave into the top rectifier, which is why you have the voltage drop when it is open. |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Tie the commons together. Right now, the entire bottom half of the schematic is "Floating". Connect both "-" outputs of the bridges together if nothing else (assuming the input is from a transformer, and not 2 different transformers). Put the circuit into the visual simulator above. Also, why not just use 1 rectifier? Not enough current? I am assuming he is monitoring something, since he is using an AND gate. You could replace the AND gate, with a NAND gate and have an LED light up for an application where you wanted to run duel rectifiers, and be notified if one died. |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Tie the commons together. Right now, the entire bottom half of the schematic is "Floating". Connect both "-" outputs of the bridges together if nothing else (assuming the input is from a transformer, and not 2 different transformers). Put the circuit into the visual simulator above. Also, why not just use 1 rectifier? Not enough current? I am assuming he is monitoring something, since he is using an AND gate. If he cannot use a common ground, he will need to use an optocoupler on one so that he does have a common ground for the logic. The bottom will measure 5V between out and ground, as will the top, when power is applied. Measuring voltage from GND tab of the bottom rectifier to the OUT pin of the top will yield 0V or unknown/floating voltage with no ability to source current. |
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Quoted: Quoted: Quoted: Quoted: The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Tie the commons together. Right now, the entire bottom half of the schematic is "Floating". Connect both "-" outputs of the bridges together if nothing else (assuming the input is from a transformer, and not 2 different transformers). Put the circuit into the visual simulator above. Also, why not just use 1 rectifier? Not enough current? I am assuming he is monitoring something, since he is using an AND gate. I am. I'm tapping an existing 24VAC transformer to supply constant power to my circuit. My circuit (B2) is monitoring a different system and SW1 is actually a relay controlled by the other device. When the relay closes, the AND gate logic goes 'high' and the LED lights. I'm using the 5Vdc output of IC2 to do some other stuff as well, but that all works. It's only the B1/IC1 circuit that's problematic. To be clear, B1 & B2 '-' (grounds) are connected |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. Tie the commons together. Right now, the entire bottom half of the schematic is "Floating". Connect both "-" outputs of the bridges together if nothing else (assuming the input is from a transformer, and not 2 different transformers). Put the circuit into the visual simulator above. Also, why not just use 1 rectifier? Not enough current? I am assuming he is monitoring something, since he is using an AND gate. If he cannot use a common ground, he will need to use an optocoupler on one so that he does have a common ground for the logic. The bottom will measure 5V between out and ground, as will the top, when power is applied. Measuring voltage from GND tab of the bottom rectifier to the OUT pin of the top will yield 0V or unknown/floating voltage with no ability to source current. The last sentence in his OP states that he does have a common ground. He just doesn't have it drawn that way.... |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... This is what I was guessing at, but couldn't confirm beyond my guess... Do you have any thoughts on a remedy? I can't switch the common to the rectifier. You can not tie the two rectifier inputs to the switch? The switch needs to be before the branch to the rectifiers. If you can not change that, I guess you could throw a comparator in between ic1 and ic2 and use that to trigger the logic input 1 of ic3.' ETA: or throw a resistor between ic1 and ic3 so that when the voltage drops to 2.53 volts, the resistor will drop it even further so that it falls below the logic level for IC3 |
 This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. |
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http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. |
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Quoted: Quoted: http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. |
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http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. Put a potentiometer between the output of IC1, and the input of IC3. Open the switch, and adjust the pot until the light goes off, Then close the switch, the light should come back on. If this just for a project, you can then remove the pot, and measure its resistance, then replace with a similar resistor. If this is for home work, you are going to need to find what the lowest voltage is that IC3 will read as a high input, then calculate a resistor that will drop the voltage below that from the 2.53 or whatever you get out of the regulator when the switch is open. When the switch is closed, the voltage will rise within the high level threshold, |
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Quoted: Problem with this approach ( I just tried it)Quoted: Quoted: Quoted: http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. Put a potentiometer between the output of IC1, and the input of IC3. Open the switch, and adjust the pot until the light goes off, Then close the switch, the light should come back on. If this just for a project, you can then remove the pot, and measure its resistance, then replace with a similar resistor. If this is for home work, you are going to need to find what the lowest voltage is that IC3 will read as a high input, then calculate a resistor that will drop the voltage below that from the 2.53 or whatever you get out of the regulator when the switch is open. When the switch is closed, the voltage will rise within the high level threshold, 22M Ohms drops the voltage value of IC1 to 0.6ish volts with SW1 open. HOWEVER, with SW1 closed IC1 is only putting out 1.7 volts and I need a minimum of 2 volts to pull IC3 'high'. |
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Problem with this approach ( I just tried it)
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http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. Put a potentiometer between the output of IC1, and the input of IC3. Open the switch, and adjust the pot until the light goes off, Then close the switch, the light should come back on. If this just for a project, you can then remove the pot, and measure its resistance, then replace with a similar resistor. If this is for home work, you are going to need to find what the lowest voltage is that IC3 will read as a high input, then calculate a resistor that will drop the voltage below that from the 2.53 or whatever you get out of the regulator when the switch is open. When the switch is closed, the voltage will rise within the high level threshold, 22M Ohms drops the voltage value of IC1 to 0.6ish volts with SW1 open. HOWEVER, with SW1 closed IC1 is only putting out 1.7 volts and I need a minimum of 2 volts to pull IC3 'high'. 22 mega ohms is the smallest resistor your have? Use a small potentiometer, so that you can set the threshold. Or use a diode and use the voltage drop it supplies (usually around 1 volt) |
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Quoted: Quoted: Quoted: Problem with this approach ( I just tried it)Quoted: Quoted: Quoted: http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. Put a potentiometer between the output of IC1, and the input of IC3. Open the switch, and adjust the pot until the light goes off, Then close the switch, the light should come back on. If this just for a project, you can then remove the pot, and measure its resistance, then replace with a similar resistor. If this is for home work, you are going to need to find what the lowest voltage is that IC3 will read as a high input, then calculate a resistor that will drop the voltage below that from the 2.53 or whatever you get out of the regulator when the switch is open. When the switch is closed, the voltage will rise within the high level threshold, 22M Ohms drops the voltage value of IC1 to 0.6ish volts with SW1 open. HOWEVER, with SW1 closed IC1 is only putting out 1.7 volts and I need a minimum of 2 volts to pull IC3 'high'. 22 mega ohms is the smallest resistor your have? Use a small potentiometer, so that you can set the threshold. Or use a diode and use the voltage drop it supplies (usually around 1 volt) For IC3's inputs I need to be below 0.8volts for 'low' and above 2volts for 'high'. I suppose the diode idea might work... |
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http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. I think the green box is the switch and cannot be moved, if I read the OP correctly. Brute force, kill the 2.5 volt with 4 or 5 1N914s in forward bias series between IC1 out and AND gate. |
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Why is B2 after the green box?
If the greenbox needs exactly what you're currently sending it, maybe consider having it activate a relay? I think the problem is definitely that at least a quarter of your transformer output is still going to IC1. You've gotta run that through the switch as well, or separate it from the switch with a relay or something. |
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Problem with this approach ( I just tried it)
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http://i54.tinypic.com/30ueoea.png This is the circuit (minus some other controls powered by IC2 that function) as I've tested it. This does NOT work. The 24VAC signal from the "Green Box" is switch on/off. When Off, IC1 floats at 2.5VDC since the fullwave rectifier is rectifying half the wave I think. I am saying put the switch at one end of the transformers output, before it goes to the rectifiers. Ok, i get what you're saying. However that won't work since the "Green Box" must always stay powered. Put a potentiometer between the output of IC1, and the input of IC3. Open the switch, and adjust the pot until the light goes off, Then close the switch, the light should come back on. If this just for a project, you can then remove the pot, and measure its resistance, then replace with a similar resistor. If this is for home work, you are going to need to find what the lowest voltage is that IC3 will read as a high input, then calculate a resistor that will drop the voltage below that from the 2.53 or whatever you get out of the regulator when the switch is open. When the switch is closed, the voltage will rise within the high level threshold, 22M Ohms drops the voltage value of IC1 to 0.6ish volts with SW1 open. HOWEVER, with SW1 closed IC1 is only putting out 1.7 volts and I need a minimum of 2 volts to pull IC3 'high'. 22 mega ohms is the smallest resistor your have? Use a small potentiometer, so that you can set the threshold. Or use a diode and use the voltage drop it supplies (usually around 1 volt) For IC3's inputs I need to be below 0.8volts for 'low' and above 2volts for 'high'. I suppose the diode idea might work... If the diode does not work, you can adjust the supply voltage of your AND gate to adjust the threshold. If that fails, I would use a comparator to drive IC3. If this is a redundant supply monitor, I figure I would have designed it so that two comparators monitor the rectifiers against each other, and their outputs would drive the AND gate. |
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Quoted: Why is B2 after the green box? If the greenbox needs exactly what you're currently sending it, maybe consider having it activate a relay? I think the problem is definitely that at least a quarter of your transformer output is still going to IC1. You've gotta run that through the switch as well, or separate it from the switch with a relay or something. A relay would work, but I'm limited on PCB space and would rather use what I have for other components... |
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Why is B2 after the green box? If the greenbox needs exactly what you're currently sending it, maybe consider having it activate a relay? I think the problem is definitely that at least a quarter of your transformer output is still going to IC1. You've gotta run that through the switch as well, or separate it from the switch with a relay or something. He stated he can not do this. This looks like a monitoring circuit for redundant DC power. If the light is off, one of your rectifiers is gone. |
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Quoted: The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... That was my take on it immediately when I saw it. The relay should be replaced with a double-pole/single-throw device. |
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The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... That was my take on it immediately when I saw it. The relay should be replaced with a double-pole/single-throw device. I think the switch is just for testing. |
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Also: Why in the world do you have a logic AND gate in there in the first place? If the intent is to have the LED light up when the switch closes, then just ditch the bottom part of the circuit, the AND gate, and connect the LED straight to the voltage regulator. |
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Also: Why in the world do you have a logic AND gate in there in the first place? If the intent is to have the LED light up when the switch closes, then just ditch the bottom part of the circuit, the AND gate, and connect the LED straight to the voltage regulator. He is monitoring both rectifiers, if one dies, the light goes off. Redundant power supply it looks like. Not sure why the circuit was designed this way, comparators are designed for this type of thing. |
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Quoted: Quoted: Quoted: The regulators are full wave. When you open the switch the B1 regulator is now operating half wave, not as efficient but still producing voltage to the regulator. Both regulators will be satisfied with the switch open. You need to totally remove the input from both sides of the full wave bridge..... That was my take on it immediately when I saw it. The relay should be replaced with a double-pole/single-throw device. I think the switch is just for testing. I just looked at the diagram with the green box and that makes sense. |
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Brute force idea: Just use a simple half-wave rectifier (single in-line diode). Not pretty, but you can then ignore the negative half-wave that still exists when the switch is opened. OP: Is this a school assignment and are limited to what you can use? He needs two full wave rectifiers, and this circuit is to monitor them. I would replace the AND gate with a quad comparator, and use two of the comparators to drive two seperate LEDs. The inputs will be cross referenced against each other. |
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What are your design priorities here? There are a number of methods to take here, ranging from brute force to expensive to elegant. What your resources, constraints and priorities are will dictate which solution is 'best'.
If you really want to win some nerd points you're going to need to find an excuse to stick an arduino in there somewhere... |
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What are your design priorities here? There are a number of methods to take here, ranging from brute force to expensive to elegant. What your resources, constraints and priorities are will dictate which solution is 'best'. If you really want to win some nerd points you're going to need to find an excuse to stick an arduino in there somewhere... Naa, no nerd points for arduino anymore. He needs to tie this sumbitch to an rs485 to modbus Ethernet bridge and set this bitch up on SCADA. 
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This is just a hobby project. I've built it this way since this is what I had laying around; if there are better ways to do this then I can source additional components. I'm not at my desktop at the moment or I'd revise the schematic to show the full circuit since it may remove some confusion... What's happening here: - Existing control circuit does it's thing and has a 24VAC output that is powered during specific events - I'm pulling power from the same 24VAC transformer as the 'Green Box' circuit. - My circuit has 24VAC input and rectified/regulated to 5Vdc fulltime to power future components - IC1/B1 (switched) is only supposed to be active when the 'Green Box' is outputting 24VAC - When the 'Green Box' is outputting 24VAC, the idea was to have the AND gate go 'high' ONLY if my circuit is still powered (should be, but may not be) - When the AND gate is 'high', the output would drive a transistor (not the led shown, that's just for testing) to pull in the contacts on relay to switch a completely different 24VAC source. I realize this is a bit confusing... Hopefully I can get back to my desktop and just post the full schematic. |
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Quoted: Quoted: - IC1/B1 (switched) is only supposed to be active when the 'Green Box' is outputting 24VAC Except that it won't work correctly that way. Drop the bridge rectifier for a single diode and you should be GTG. You can keep the bridge on the bottom circuit. I think someone had suggested that earlier and I got sidetracked before trying that out... |
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Ignoring your schematic for a moment, it looks like your goal is to illuminate an LED if both AC circuits are on. If this is true, your circuit should work. There can be no current flow if one of the legs of a full wave rectifier is not connected, so I don't think that is your problem.
The filter capacitor used to smooth the full wave rectified signal to DC acts as energy storage. That capacitance, depending on how large it is, could continue to power your 5V regulator. Your logic gates have a very high input impedance, and will draw very little current, which would cause the input to the logic gate to stay high. You can try adding a 330 or so resister from each gate input to ground. This will pull the input down and not allow it to float. Depending on your capacitor size, this could still take some time to turn off. If you are just trying to monitor the AC with a switch in it though, I'd just use one branch to power your regulator which in turn would turn on your LED through its current limiting resistor. Even simpler, you could half wave rectify it, add in the appropriate current limiting resistance (it will dissipate a significant amount of power), and illuminate your LED that way. |