the schematic is located at http://tinyurl.com/kk7zs
2006-08-06
14:26:09
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9 answers
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asked by
↓ImWithStupid ░░▒▒▓▓
4
in
Science & Mathematics
➔ Engineering
PIC with +5v @ 25mA output for each pin is attach to each base resistor.
2006-08-06
15:47:06 ·
update #1
I simulated this circuit in CircuitMaker 2000 Pro. The sim says it "works" LOL
2006-08-06
15:48:13 ·
update #2
Grumpy: Yes and Yes
2006-08-06
16:01:37 ·
update #3
RLY3 allows for Power ON/OFF, both Q3 and Q4 control RLY3. Q4, however, only controls RLY4. This lets both Q3 and Q4 to manipulate RLY4 as a reversing relay.
2006-08-06
16:04:04 ·
update #4
D8 provides current to RLY3 when Q3 is energized and D3 provides current to RLY3 when Q4 is energized. Q4 by itself should only energize RLY3. To stabilize operation, a resistor of less than 10k (4.7K or so) should be connected from each base to ground. This would help eliminate the possibility of false firing of Q3 and Q4 due to leakage of diodes or other transistor junctions. Inductive kick back of the relay coils could also be a problem even though it is limited to 0.7V by the diodes across them.
2006-08-07 05:08:06
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answer #1
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answered by Buffertest 3
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I now think the below quite wrong!!
You appear to be making RLY3 the priority. in that this will turn on and off depending on weather Q4 base is energised or not.
However if RLY4 is already energised or an attempt to energise is madewhile RLY3 is energised or RLY3 is energised while RLY4 is energised then RLY4 will turn off.
The only issue I can see here is that if there is enough voltage drop across the coil of RLY3 there might be enough difference for RLY4 to stay in if already in or it might be unpredictable. Why not do this with gates?
boolean
RLY4=Q3 saturated AND NOT Q4 saturated
Perhaps this
D8 provides a backfeed of FV to the return of RLY4. This ensures RLY4 stays off until RLY3 is energised.
This should all work fine.
Sorry to be a ding bat!
2006-08-06 15:32:52
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answer #2
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answered by slatibartfast 3
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I don't see a problem. It was difficult for me to read the numbers.
However, it appears Q3 will bring both relays on, and Q4 only relay 3.
Are you reversing a field, what are you doing with the circuit?
Yours: Grumpy
2006-08-06 15:58:47
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answer #3
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answered by Grumpy 6
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I would strongly recommend simulating this circuit. What is driving the base of the transistors? Might want to use 10k pull-down resistors on R8 and R9 just to make sure your transistors are OFF if nothing's driving the bases high
2006-08-06 15:40:03
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answer #4
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answered by Wattanabe 2
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seem for a 555 oscillator circuit with a variable duty cycle (adj. on time vs off time). then you certainly'll favor to have the 555 bias a transistor that can cope with providing the coil modern-day the relay will require.
2016-10-15 11:21:47
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answer #5
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answered by ? 4
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Here is what I see:
Q3 turned on, Q4 turned off will energize RLY3 and 4, and will turn on LED D5.
Q3 turned off, Q4 turned on will energize RLY3, and turn on LED D4.
Q3 turned on, and Q4 turned on will anergize RLY3 and 4, and will turn on LED D5, and D4.
My concern would be if Q3 can handle the current of both relays and LED D5.
2006-08-07 03:16:32
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answer #6
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answered by justme 7
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I don't get it. I think you just created a "buzzer". The coils are getting + and - from both directions
2006-08-06 14:37:50
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answer #7
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answered by civicsound 3
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Ouch that's a lot of diodes. My brain hurts.
2006-08-06 14:30:29
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answer #8
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answered by Anonymous
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looks like the charging system of most american cars..
2006-08-06 14:31:24
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answer #9
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answered by me too 6
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