# using current to measure a time

Discussion in 'Electronic Basics' started by Panther, Dec 1, 2005.

1. ### PantherGuest

Hi,

I am making something to measure how much time two objects are in contact
when one of them is accelerated towards the other. To do this, I'm going to
put a wire on one object and another on the other, and make the objects
complete the circuit when they touch obviously. What happens with this is is
these wires link to a stopclock/watch (i can give the model name if you
want) and this stopwatch starts timing when it receives a current and stops
when it receives ANOTHER current. I mean, you give it a current. Starts
time. You "break the circuit." It still times. You give it another current,
it stops. This isn't useful because I need it to start timing when it
receives a current and stop when that current is taken away (as the objects
are no longer in contact; no more current).

What can I do to make this work? Is there anyway I would generate/make a
current come when the circuit is broken? How does a potential divider fit
into this? I am grateful for help. Thank you.

2. ### Dan HollandsGuest

Conceptually it is simple

Sense the current that starts the watch

When the current goes away, fire a one shot that will generate a current
pulse to stop the watch

OR

generate a current that is normally connected to the Stop input.

When current is detected switch that current to the Start input until the
current goes away

Other Thoughts

If I had this problem I would simple use an oscilloscope to monitor the time
that the objects touch. If the objects are metalic do you think they will
touch long enough to run a clock?

Dan

--
Dan Hollands
1120 S Creek Dr
Webster NY 14580
585-872-2606

www.QuickScoreRace.com

3. ### Andy BaxterGuest

Panther said:
I don't know if this would work, but you could try doing it like this:

Build a current generator circuit with a fast response rate that didn't
mind trying to drive an open loop.

Connect the output of this to a capacitor through the circuit made by the
two objects.

When they collided, a pulse of current would flow between, and the voltage
on the capacitor at the end would measure the impact time.

4. ### Nick.Guest

Blimey... I remember doing those experiments at school on equipment using
Nixies tubes! (Kicking a tinfoil covered football down the corridoor IIRC)

Forget the toy stopwatch. Use (or get) a frequency counter with a "count"
object to the osc, the other to the counter input and you have a timer
with 1uS resolution...

http://uk.farnell.com/jsp/endecaSearch/partDetail.jsp?SKU=112331&N=401

5. ### PantherGuest

Thanks, but I'm afraid this too complex for me

6. ### PantherGuest

Thank you for the assistance, but I am not allowed to use that (even if I
was, I wouldn't know how as I'm quite new to this stuff). But thanks anyway.

7. ### PantherGuest

What does a one shot mean? Sorry I'm new.

This looks like it would lead to inaccuracies because of slowness. Either
way, I don't think there is a Stop input or anything of that sort.
Unfortunately I can't do that anymore

Thanks anyway.

8. ### Nick.Guest

You can't use a counter or a 'scope. An analogue (capacitor) solutiom is
too complex. What answer was you hoping for?

9. ### PantherGuest

How I could use a potential divider to set a pulse when the circuit breaks?
Or a more complicated circuit design?

10. ### John FieldsGuest

---
I'm assuming you've got two balls hanging from the ends of wires and
you're going to pull on eway from the other and the let it go so
that it'll hit the other one, and what you want to measure is how
long the collision lasts.

Try this: (view in Courier)

GND +5V HFCK
| | |
| [100R] | +-------------+
| | +--|> |
+ +-----A |_ COUNT _| A---+
/ | OR Y---+-O|E Q0 R|O--Y NOR |
/ | +--B | +---------+---+ B---+
/ | | | | |
/ | | A--+ +--A |
/ | +---Y AND NOR Y---+ |
/ | B---------------+--B | | +5v
/ \ / \ | | | |
\_/ \_/ | A---+ | O |
+--Y NOR | |<-ARM
B--------+--O |
|
[1KR]
|
GND

11. ### PantherGuest

I'm sorry but this is way too complex for my level. Thanks anyway.

12. ### Rich GriseGuest

You should also learn to bottom-post.

Anyway, I think I have an idea what you want - a current source
that when the balls are touching, lets current flow for time T.

OK, take a power supply, a resistor, a capacitor, and a voltmeter.

PS+ ---- R ----- (B1) (B2) ----+---- Voltmeter +
|
=== capacitor
|
PS- -----------------------------+---- Voltmeter -

Use a high-impedance voltmeter, like a DVM.

Record the voltage while the balls are apart. It will very probably be
very close to zero. You can assure this by momentarily shorting the
capacitor terminals.

Drop the ball. While the balls are in contact, current will flow and
start to charge the cap through the resistor, with a time constant
of T = RC.

The voltage across the capacitor at the exact moment that the balls
separate will tell you T by using that exponential equation that I
can't remember now, but since you're in school you should look it up
anyway.

Good Luck!
Rich

13. ### John FieldsGuest

---
It seems that no matter how simple we make it, you can't quite get
there.

How about specifying what you want/need in a little more detail, if
you can.

Otherwise, well... we've offed trolls here before.

BTW, here in Rome we bottom post so, when in Rome...

15. ### Nick.Guest

You cannnot. A potential divider divides potential.
Some simple designs (e.g. a resister and a capacitor) have already been
suggested but are too complex (?)

If you really must use the stopwatch you mentioned what you really need is
an edge-triggered monostable. You may get away with a simple inverter,
depending on the input logic of the stopwatch. Either way, you will
probably run into problems with "switchbounce".

16. ### PantherGuest

Thanks but I think the poster above you's suggestion is simple enough to do.

17. ### PantherGuest

Thank you!!!!!!!!!!!!!! This looks very simple so I will give it a go. But
could you please explain what the time constant T = RC means? I've no idea.
So you're saying, for example, if I were to do the practical now and record
the voltages, I would be able to work out the time section using some
formula? IE I don't need the equipment? As that would very handy.

Thanks

18. ### Rich GriseGuest

T = RC is the "time constant". It's the amount of time it takes for the
charge to reach 63% (or something) of its final value, R is in ohms, and
C is in farads. With T on the horizontal axis, the cap voltage rises at
some exponential rate...
Maybe one of these sites will be more helpful:

Good Luck!
Rich

19. ### John FieldsGuest

---
It seems to me that by your not cross-posting, by your off-handed
rejection of several solutions which have been presented to you,
and by not using in-line references, you're not really looking for a