I am filled with contrition. Please forgive my lapse.
hgmjr
Ha !
Wish I had half your knowledge.
OK, will be off-line for awhile doing the mechanics stuff, etc.............
Tx's everybody,
Oxbo
I am filled with contrition. Please forgive my lapse.
hgmjr
Your expertise and skill are elsewhere. I have never welded a thing. I would do myself a serious injury were I to attempt such a dangerous activity.Ha !
Wish I had half your knowledge.
That was "Across" R3 Sarge.You're showing 2.9v above R3. Does that mean you're measuring 2.9v across R3 to R9, or are you measuring 2.9v from the high side of R3 to ground?
If you are measuring 2.9v across R3 to R9, then since I=E/R, 2.9/120 = 24.16mA. This really isn't too bad because they are multiplexed.
If you are measuring 2.9v to ground, then you also need to subtract the Vce of the ULN2804, which should be around 0.65v, from the 2.9v - that leaves 2.25v, which means 18.75mA current through the LED segments - which is just fine.
The 360 Ohm resistor needs to be 1/4W.
Cheeez;The Simpson 260 is a good piece of equipment, but you have to keep in mind the Ohms per volt. In the case of the Simpson 260, it's 10k Ohms per volt. If you're dealing with a low-impedance circuit, the impact will be insignificant - but if you're measuring across, say, a 50k resistor and reading 5v, you must keep in mind that the Simpson will have an impedances of 10k Ohms/volt, so in parallel with the 50k resistor, you'll have a total of 25k - which can upset the balance of the circuit.
Digital multimeters have an extremely high input impedance (tens of megohms) so they have a minimal loading effect when measuring voltages in a circuit.
And for the next example ??OK, let's say you used the Simpson to measure the voltage across R3, a 120 Ohm resistor, and you measured 2.9v.
2.9v x 10k = 29k Ohms in parallel with 120 Ohms. The loading effect in this case is negligible; as 120 Ohms is about 0.4% of 29k Ohms.
120 Ohms is a pretty low impedance.
Well,Do you expect me to go through every possible combination of resistance and voltage?
I ain't gonna do dat for ya...
Hi Sarge;Hi Oxbo,
Have a look at the attached. This timer won't start out with the output high when you first power it up. That's because the timing cap, C1, is connected to Vcc rather than ground.
When triggered by 5v to 12v being placed on the left side of R5, a negative pulse is coupled through C3, which triggers the start of the timer. The output of the timer stays high for 1 second, and then returns low.
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