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Voltage drop detection

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Sorry, but that was my typo. Of the four comparators in the LM339, I started with comparator #2, which indeed used pins 1,6, & 7. I then tried using pins 2, 4, & 5 which is for comparator #1, and got the same results in that 1) the relay is energized as soon as the power (6 volts) is applied, whether any input is attached or not, 2) the flash will only fire if flash ground is attached to input pin, and flash positive is attached to the circuit ground, and 3) when the input voltage goes to zero, the relay does not de-energize.

Short of trying a new chip, can you think of anything else why these items are happening?
 
Both circuits work the same. If you connect +6V power to the circuit with the input floating (input floating looks like the voltage is high), the relay will instantly be activated. It will be de-activated when a voltage less than about +2.7V relative to circuit ground is applied to the input, and activated if the input is greater than about +3.3V. If you need to invert the relay functionality so that
When Vin<2.7V, relay ON
When Vin>3.3V, relay OFF
let me know, and I will modify the design. I don't understand what is happening when you connect the input as you describe.
 
Ron-
It looks like I wil need to have the circuit modified and inverted for the opposite operation (thank you), but I am still trying to get the circuit to work as you have so patiently explained. I will also get some additional LM393 chips during the week as maybe the chip I am using is bad.

I rechecked the existing breadboard wiring, resistor values, and pin hookups again to make sure it is set up as in your schematic, but with no input attached at all, as soon as I connect the 6v supply, the relay energizes. If I connect (momentarily) a single AA cell as a test input (battery neg to circuit ground and battery pos to input pin), the relay stays energized. If I reverse the single AA cell polarity (battery pos to circuit ground, and battery neg to circuit input) the relay will de-energize upon momentary contact.

This is just what I was finding with the flash in that the trigger voltage before the flash occurs is 4.5 volts, and if the pos lead is connected to the input, and the negative lead is momentarily grounded, the flash doesn't fire and the relay stays energized. If I reverse the flash leads so the flash negative is connected to the input pin, and the flash positive is momentarily connected to the circuit ground, the flash will fire with each momentary contact, and the relay will also release with each momentary contact.

To me this sounds opposite to what you said should be happening, and the reverse polarity of the input doesn't seem correct either.

So, the question is, if the circuit on the breadboard is indeed wired correctly, might it be that the comparator is bad? I can take a picture of the breadboard and send it if that might be useful.
 
It sounds like R2 is not connected to the power supply, or is much too large in value (or R3 is much too low). Do you have a voltmeter? You need to measure the voltage at the junction of R2 and R3, relative to circuit ground.
 
Hi, JMB

I suggest you have a look at the following site:

**broken link removed**

for a concise tutorial on the three comps you are working with.

Hope this helps.
 
AllVol -- Thanks for the link, but it doesn't appear to come up correctly...the full URL didnt' come through, just the shortened version containing the "~"

Ron -- With only the 6 v supply connected, the junction of R2, R3, R4, and pin 2 (on your schematic), the voltage is 2.61 v. Connecting the single AA cell (1.34v) gives a reading of 2.85v, while connecting the flash leads (4.5v) gives a reading of 2.61v.

This might be obvious, but with the flash on and ready, putting a meter across the terminals gives the 4.5v, but naturally the terminals have to be shorted with a mimentary switch for the flash to fire. The electonics in the flash then take this voltage to zero after 90 sec. It is this voltage drop I am trying to capture and then have the relay energize at the moment, and then the reverse after this works. Thanks!
 
AllVol -- Thanks for the link, but it doesn't appear to come up correctly...the full URL didnt' come through, just the shortened version containing the "~"

Ron -- With only the 6 v supply connected, the junction of R2, R3, R4, and pin 2 (on your schematic), the voltage is 2.61 v. Connecting the single AA cell (1.34v) gives a reading of 2.85v, while connecting the flash leads (4.5v) gives a reading of 2.61v.

This might be obvious, but with the flash on and ready, putting a meter across the terminals gives the 4.5v, but naturally the terminals have to be shorted with a mimentary switch for the flash to fire. The electonics in the flash then take this voltage to zero after 90 sec. It is this voltage drop I am trying to capture and then have the relay energize at the moment, and then the reverse after this works. Thanks!
Forgive my ignorance about flash units. What if you want to use the flash again after more than 90 seconds has elapsed? The 4.5V is no longer there.
 
That's ok, I was trying not to confuse the issue, but the relay will trigger a pulse from the 6 v supply to another contact separate from all of this that will wake the flash back up and then the 4.5 v will be present again, and open the relay. This will all happen though an adjustable timer cicuit to be able to vary the periods of activity and inactivity.

Michael
 
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OK, so when the output goes to zero after 90 seconds of inactivity, does it go to zero by turning off the supply in the flash unit?
 
Well, the display blanks, a voltmeter across the terminals will read zero volts, and if the terminals are shorted, the flash will not fire. The main power swtich of the flash has to either be turned off and then turned back on, or the "Ready" button is pressed which send the wakeup signal and the flash will be ready to fire again. When the flash is sitting on a camera, the camera will send this signal to wake up the flash just before you are taking the picture. For this off camera usage, the relay will close the circuit allowing the same trigger switch that originally fired the flash to now send the 6 volt signal to a different contact to wake up the flash. After the flash is awake again, the trigger voltage across the flash terminals would be back to 4.5 volts, the relay would open, disconnecting the pulse of the 6 volts to the extra contact point, and allowing the same momentary contact switch to now fire the flash itself. If this 6 volt voltage were to be connected to both the extra contact and the main contact, the flash would not fire (and it wouldn't go to sleep either).

Michael
 
AllVol -- Thanks for the link, but it doesn't appear to come up correctly...the full URL didnt' come through, just the shortened version containing the "~"!

Sorry, my bad. Eyes are too old, I guess.

Let's try this: **broken link removed**

I really think this site will help you.
 
Great site! Should be required reading after (or before) a couple of questions!

It will be my reading for tonight. Thanks
Michael
 
Here is a timing diagram of my understanding of your input signal to the comparator. If you can add a line to show what you need the output signal or relay closure to do, it would be helpful.
The squiggly lines just mean the time interval is indeterminate.
 

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I added the timeline in the lower portion of your image showing that the relay gets energized when the flash goes into sleep mode. (I might have to also try it that the relay is de-energized when it goes into sleep mode.) The flash goes into sleep mode when there has been no activity to fire the flash within these 90 sec. The same momentary switch that is used to fire the flash is then be used via the energized relay to now connect the 6v supply to the extra wakeup contact. When the momentary switch is pressed and the flash is in sleep moed, the flash will first wake up and stay awake for another 90 sec. The relay will be de-energized connecting the momentary trigger switch so that the next time it is pressed, the flash will fire, as long as 90 sec have not elapsed, and the flash went back into sleep mode.

If this couldn't work because of something to do with the flash circuits, my next test is to use an optical sensor taped against the flash ready light. When the flash goes into sleep mode, the ready light (usually a red or green LED) will be off. The disadvantage here is that while the flash is powered on and charging the main capacitor, the ready light is off, until the flash is actually ready to fire, so the relay would cycle many more times this way.
 

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I'm still not getting the big picture. Can you describe what you are really trying to accomplish, independent of how you think it should be done? Do you want the flash to fire every 90 seconds, repetitively, with no external controls (I'm guessing not)?
I'm just confused by your description of what you want.
 
Here is the setup and what happens with the flash itself:

Turn the flash main power on and leave it on. The flash is not connected to a camera in any way, but the flash terminals are connected to a momentary contact firing switch.

The flash will charge the capacitor, and when it is ready to fire, the ready light will be lit. As soon as the main power switch is turned ON, the voltage across the terminals of the momentary switch is 4.5 volts. The momentary switch is external to the flash unit (2 wire connection). If you press the momentary switch (when the ready light is on) the flash will fire and then recharge the main capacitor. As long as you press the momentary switch at least every 90 seconds, the flash will remain awake. If the momentary switch is not pressed within 90 seconds, the flash will go into sleep mode, but the master power switch has not been physically moved, and the flash will not fire if the momentary switch is pressed. To wake up the flash and allow the momentary switch to again fire the flash, either you 1) physically turn the master switch OFF, and then ON again, or 2) physically hit a RESET button to wake the flash up, or 3) add a small voltage to a secondary contact.

Here is what I am trying to accomplish:

It is this item # 3) which is what the relay will allow me to do, using the same momentary switch that can normally fire the flash. If the flash is awake and powered on, this switch will connect through the NC contacts of the relay to the pos terminal of the flash with the ground of the flash connected to the other side of the momentary switch. Any press of this momentary switch will fire the flash if it is charged up (ready light on).

After the 90 sec of inactivity, when the flash goes into sleep mode, the relay will energize, connecting the NO contacts of the relay to the "wakeup terminal" through the 6v supply. The next press of the momentary switch will send the 6v on to the "wakeup terminal" of the flash, which will then wake up, energize the relay which will disconnect the connection to the "wakeup terminal" and then the next press of the momentary switch will again fire the flash.

There is the possiblity that the relay will need to be energized not when the flash is in sleep mode, but in awake mode. I attached a rough drawing showing the relay, flash, and momentary switch.

Thanks,
Michael
 

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In your timing diagram, you showed that the momentary contact closure would last about 10mS. How is this controlled? What if the relay switches faster than this? If this is a problem, we could probably add a little delay to the comparator.
 
Well, it might be a problem, in that when the flash fires, the voltage drops to zero, and the relay would try to energize briefly, which technically should be ignored, but the circuit isn't working yet, so I don't know if it is a problem. Right now the momentary switch is just a pushbutton.

Any other ideas of things to check here?

Michael
 
I just realized that making the comparator ignore the momentary closure will allow the comparator to trigger only on the transitions to sleep, which I believe is what you want. Otherwise, it will trigger whenever the input goes low. Unfortunately, you can't possibly make the contact closure as short as 10ms. It would be easier for the comparator if the contact closure time was consistently the same, but this would require putting some circuitry between the switch and the flash. Short of that, we could make the comparator ignore anything shorter than, say, 100ms, but you would have to be certain to make all your contact closures shorter than that.
 
I think I am understanding, but let me ask what might be an elementary question: Does this have to do with the momentary switch being open most of the time?

To trigger the flash, the momentary switch is pressed, normally just a brief push, by hand at maybe 0.1 sec, or by an auto trigger which would be a duration of about 0.01 sec, I think. When the switch is not pressed, the voltage across the flash terminals is the 4.5 volts.

From what I understand, the way the circuit is currently setup, comparator output is OFF (and should de-energize the relay) when the voltage at the PLUS input is greater than the reference voltage at the MINUS input. Measuring the voltage at the MINUS input as you asked (having a 4 AA battery supply) gave a reading of +2.61v. Since this is less than the +4.5v of the flash, the relay should NOT be energized, but unfortunately, as soon as the 6v supply is connected, the relay immediately energizes.

This seems like it should be so simple already, and the tutorial referenced from AllVol was very helpful in getting a better understanding of all this, but obviously, something still is amiss. Does it indeed have to do with the momentary switch? Sorry, but I am at a loss for what else to check to make this work correctly.
 
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