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Old 16th May 2004, 03:21 AM   (permalink)
Default Pulse train to DC level:

Hi,

Im trying to change a pulse train into a DC level, and im failing in all my attempts.

The pulse on or off time can be anything from 0.5s -> 3secs.

When the pulse line goes high, the output should go high. The output should remain high until the pulse line has been low for more than 3 secs.


Code:
pulse:    __----__--_-_-_-------__---________________---_-

output:   __-----------------------------------______------
I tried to use a 555 monostable, but once the pulse has gone low, there is no way to restart the time period.


Help!
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Old 16th May 2004, 07:54 AM   (permalink)
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You need a retriggerable monostable, with longer output pulse as 3sec.
CD4528 or similar can do this.
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Old 21st May 2004, 05:41 PM   (permalink)
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Hi,

I now have an almost working circuit. I was hoping someone could figure out what is wrong with it.

Circuit operation:



I needed to change a pulse into a DC level, the INPUT led follows the input. The OUTPUT led turns on at the same time as the input turns on, and remains on for 4 seconds AFTER The input has been disconnected.

This works, except the output LED doesnt turn off correctly, it sort of fades down and i cannot work out why!


Any ideas?
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File Type: gif pulse_to_dc.gif (12.3 KB, 170 views)
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Old 21st May 2004, 06:36 PM   (permalink)
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Since You apply diodes for output transistor driving, it haven't pull-down.
Definitely need a resistor between base and GND. (10k)
But why need additional driving the output transistor directly from input?
Try this simplyer circuit..-
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Old 21st May 2004, 06:55 PM   (permalink)
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Quote:
But why need additional driving the output transistor directly from input?
The first transistor (after the opto-coupler) is required because the opto-coupler can only source/sink 50mA, the LED shown is actually for testing, in reallity, the LED is a 100mA hi-power device, i've used a 10mA device for testing as i dont want to be blinded during testing!

The final output transistor is required becase the output will be driving a relay (150mA) and the LE Din the diagram. Again, this LED is a 100mA device, but im using a 10mA LED for testing and have not placed the relay (and back EMF protection diode) into the circuit yet.

{edited...}
Also noticed you removed the diode mixer. This was required so the output transistor turns on when the input goes high, when the input goes low, the multivibrator is triggered (pin 6 goes high) and remains high for the 4 sec duration set by the RC on pins 1 and 2.
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Old 21st May 2004, 07:27 PM   (permalink)
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If the falling edge triggered input goes to low, the Q output immediately goes to high (in my sematic). I've don't protest against transistors...
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Old 21st May 2004, 08:44 PM   (permalink)
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Indeed, but the output will not go high until the input goes H to L - i need the output to go high as soon as the input goes high, then stay high for 4 secs after the input goes low
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Old 21st May 2004, 08:47 PM   (permalink)
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O.K. just see the difference between two circuit: You triggered with positive edge the falling edge input...
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Old 21st May 2004, 08:58 PM   (permalink)
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Yes i need to trigger the device on the falling edge, the 4 sec delay i require must begain on the falling edge of the input.


But i also require the output to be ON when ever the input is ON - hence i used a diode mixer (Wired-OR).

With your modifications, the output will ON time will begain when the input falls, which means the output will be OFF when the input is ON. This is wrong.
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