Hi All

I have a 12v circuit using Eagle Timer A103-006 - Eagle Signal - Panel Mount Timer, Elapsed Time Indicators, 115 VAC (newark.com)

I plan to use the OUT/unwind signal to the winch as my trigger to begin the Eagle timer counting.
I timed 1 revolution equals 1 ft of rope release. I plan to use the timer to count how many feet I have out. Therefore, when off I should have 0 volts.

To my dismay there is actually 0.0009 Volt DC and this is enough to start A103-006 to start counting....ugh don't want it too!
When I operate the winch OUT/unwind position, I measure 0.100V DC. When the winch operates IN/wind-up position, I measure 10+ V DC.

I learned of zener diode. I am not sure this will help with such a low voltage.

I plead to you for help/solution of diode or???. Where can I obtain a component that only allows 0.1 volt signal to pass through so Eagle timer can be used?
Please advise of my options. thank you

 

Welcome to AAC!

Try putting a 1kΩ resistor across the input. If this does not work, try lower resistances.

 

Welcome to AAC!

Try putting a 1kΩ resistor across the input. If this does not work, try lower resistances.

I made quick drawing of the back of Eagle Timer. Is this the location of resistor you recommend?

 

I made quick drawing of the back of Eagle Timer. Is this the location of resistor you recommend?

No. The resistor goes across the input and GND (or COMMON).

 

 

View attachment 303307

thanks for this update! What I am concerned is (I turn off the 0.0009VDC with resister) but when I operate the winch the 0.1V+ signal will go through to the Eagle timer to turn-on and start counting?

 

As MrChips has said; experiment with resistance values. Seems to me the 1KΩ resistor should be a pretty good starting point.

Here's how the numbers look based on information you provided: 10 volts with a 1KΩ resistor should draw 10mA (milli-amps) of current. When your wench is powered there should be no interruption to your signal sense (pin 3). Also, this should be plenty to draw away 0.0009V from the signal sense and hold sense at ground. Higher resistance will further reduce the current when active but should still be sufficient to hold the signal sense at ground potential when inactive. But again, experimentation is the key to finding what will work using the lowest resistance possible, then the highest resistance that does not affect the functioning's.

 

As MrChips has said; experiment with resistance values. Seems to me the 1KΩ resistor should be a pretty good starting point.

Here's how the numbers look based on information you provided: 10 volts with a 1KΩ resistor should draw 10mA (milli-amps) of current. When your wench is powered there should be no interruption to your signal sense (pin 3). Also, this should be plenty to draw away 0.0009V from the signal sense and hold sense at ground. Higher resistance will further reduce the current when active but should still be sufficient to hold the signal sense at ground potential when inactive. But again, experimentation is the key to finding what will work using the lowest resistance possible, then the highest resistance that does not affect the functioning's.

Great! Thanks for further explanation. Appreciate your help and earlier from Mr Chips

 

Give thanks to @MrChips. He was the one who gave the initial direction. I just clarified it a little for you.

 

Oh Wow! I just saw this thread. What I get is that the TS wants to monitor the amount of rope run out of a winch based on the run time signal from the winch. I do not predict a great deal of accuracy because of two variables: First, the winch speed is dependent on the exact supply voltage and the tension load, and second, the amount of rope fed per revolution also depends on how tightly the rope was wound, which is affected by the load on the rope.
Besides that, the length per revolution will change with each layer on the drum of the winch.

 

He wants to measure extend time and convert that to some rough estimate of how many feet have extended. Yes, we all know that as the drum spools out the diameter gets smaller and smaller. With each layer the diameter decreases by the cross section of the wire. All he wants is what he's asked for. It's good you pointed this out - but it doesn't seem he's too worried about it. This isn't like he's building a space shuttle and needs measurements down to 2 ten-thousandths of an inch.

We answered his question concerning how to get rid of that pesky 900µV trigger.

 

He wants to measure extend time and convert that to some rough estimate of how many feet have extended. Yes, we all know that as the drum spools out the diameter gets smaller and smaller. With each layer the diameter decreases by the cross section of the wire. All he wants is what he's asked for. It's good you pointed this out - but it doesn't seem he's too worried about it. This isn't like he's building a space shuttle and needs measurements down to 2 ten-thousandths of an inch.

We answered his question concerning how to get rid of that pesky 900µV trigger.

Need to revisit this chat. I re-read the Eagle A103-006 spec. sheet and found
SPECIFICATIONS Start/Stop Input: NPN, Contact Closure; Accumulates time when connected to common;
Low State: < 1.0 VDC,
High State: > 2.0 VDC (28VDC max)
see attach drawing for now please ignore the yellow lead going into the A103. Also, ignore the 3k Ohm resistor. (these are a proposed solution?)
Thus on Thursday, I had green-wire straight into contact#3 of Eagle A103-006 and thought well the motor is "off" right now so I should have "off" state on my A103-006. Alas I did not, it was counting. Thus I learned that, per spec, Low State: < 1.0 VDC turns on so I measured a 0.0009 V on the green wire confirming why A103 was counting already. I have discovered too late that the counter counts only with <1 VDC. Sadly I might need to get a different timer with one that counts only when signal is greater than 0.1V. Can I sell someone my lightly-used Eagle A103-006 timer Or are there any more ideas to keep use of the A103-006 in my setup?

 

Adding a single silicon diode in series with the voltage being used to trigger the time counter will add a 0.7 V voltage drop under all conditions. The resistor across the timer input will assure that when the signal is not on, that the voltage is pulled down. If that somehow is not enough, another diode in series with the negative power connection of the timer will raise the effective required input trigger voltage by about one diode drop at the current draw of the timer. A simple cheap trick that will work.

 

Simple. Invert the logic signal with a transistor. If you don’t know how to do that we can show you how.

 

OK, if I understand correctly. See attached JPG. I am to use a transistor wired into the A103-006 contact#3. Also, use one with base-emitter voltage of 0.7V. This way contact#3 will only receive a voltage into it once the transistor is energized. Do I have this correct? I no longer need a resistor. Thanks again to ALL who have invested time to answer. I await further feedback.

 

Here is an example of a logic inverter circuit.

 

OK, if I understand correctly. See attached JPG. I am to use a transistor wired into the A103-006 contact#3. Also, use one with base-emitter voltage of 0.7V. This way contact#3 will only receive a voltage into it once the transistor is energized. Do I have this correct? I no longer need a resistor. Thanks again to ALL who have invested time to answer. I await further feedback.

This circuit is wrong in multiple ways.

Why is the run signal going to the motor? I thought it was to enable the counter.

And you need the transistor and two resistors as shown by @MrChips

 

This circuit is wrong in multiple ways.

Why is the run signal going to the motor? I thought it was to enable the counter.

And you need the transistor and two resistors as shown by @MrChips

The goal by splicing the signal from the motor into the A103-006 is so when motor starts(triggered elsewhere) then I pickyback signal to be used to start my timer to count in seconds. The challenge hence why I'm on this forum is
SPECIFICATIONS of Low State: < 1.0 VDC, I need a low voltage to start my counter.

 

I measured a 0.0009 V

STOP! WAIT A MINUTE! You're measuring "Nine Tenthousandths of a volt"? With what? Every meter I have on its best scale measures down to .001. MOST of what I have only go down to .01. And the actual reading has to take into account line resistance of the meter leads. You're claiming to measure 0.9mV. You must have a very expensive meter for that kind of resolution.

 

STOP! WAIT A MINUTE! You're measuring "Nine Tenthousandths of a volt"? With what? Every meter I have on its best scale measures down to .001. MOST of what I have only go down to .01. And the actual reading has to take into account line resistance of the meter leads. You're claiming to measure 0.9mV. You must have a very expensive meter for that kind of resolution.

Oh wait perhaps I misspoke, here is image from my voltmeter, I was reading this voltage (see todays drawing its the voltage from the purple wire before I plugged it into contact#3) when nothing was operating but it was enough to start A103 to count. After this weekend chat guidance by you ALL, I'll be heading over to make measurements again now that I have a better understanding of my options.