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Take another picture.
- Warren
- Warren
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Twinfun2
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Looks good to me... Keep in mind that the horizontal stripes of holes at the top and bottom of the board do not always connect all the way across the board. Make sure that all of your VCC connections, and all of your GND connections, are continuous with your multimeter.
- Warren
- Warren
Twinfun2
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chroot said:
Ok, I will do that, and then insert my new timer afterward.
Wish me luck guys!
Twinfun2
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Still no dice guys...light stays on, and the timer is actually STILL getting hot. Could it simply be the breadboard? If so, I might make an attempt at soldering the circuit together instead.
Output of Pin is in this order:
Resistor, LED, Ground
Is there any connections missing? According to the schematic supplied by Bob S, there should be some extra connections that are not included in my circuit.
Output of Pin is in this order:
Resistor, LED, Ground
Is there any connections missing? According to the schematic supplied by Bob S, there should be some extra connections that are not included in my circuit.
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At this point I'm at a loss. Have you gone through all the nets in the design, measuring their resistance to all other nets? It is entirely possible that your breadboard is damaged. If it is, it'll show up in those continuity tests.
- Warren
- Warren
Twinfun2
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Could "Rechargeable Batteries" be the culprit?
I'm just looking through every detail now...
I'm just looking through every detail now...
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No, rechargeable batteries should not be an issue. Again, there's something going where a lot of current is flowing through your device. You should try:
1) Measuring the resistance between each net and all the others.
2) Using your multimeter to measure currents through the VCC and output pins.
The second test needs to be done with the part installed, of course, but you should be able to get a reading of the current in just a second or two.
- Warren
1) Measuring the resistance between each net and all the others.
2) Using your multimeter to measure currents through the VCC and output pins.
The second test needs to be done with the part installed, of course, but you should be able to get a reading of the current in just a second or two.
- Warren
That has been corrected:
Still at a loss what's going on. Weird that the LED lights whether it's connected to Vs or to ground.
Twinfun2 said:
Still at a loss what's going on. Weird that the LED lights whether it's connected to Vs or to ground.
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I compared each and every node on the breadboard circuit with the ones on the schematic in the OP and they all match. I'd suspect the board at this point.
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What does the writing on the 555 chip actually say? Could you read it please?
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One last silly question: Is that electrolytic cap connected with (-) lead to ground?
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Still no dice guys...light stays on, and the timer is actually STILL getting hot. Could it simply be the breadboard? If so, I might make an attempt at soldering the circuit together instead.
Just reading back through the posts, I saw this one (above).
If you can get away from the breadboard approach, it would be good. You don't need printed circuit boards. I find breadboards generate more problems than anything else, usually because of poor connections.
This is a simple and extremely reliable ciircuit and it shouldn't be causing so much grief.
You turn the chip so its legs are facing upwards. (this construction method is called "dead bug construction" for this reason).
Any legs that are to be grounded are bent downwards and soldered to the PC board. Also any capacitors etc that need to be grounded are soldered in.
Power leads etc are attached to tag strips and not directly to the IC. (I use header pins cut to shape as tag strips).
This way, you can lay out a circuit as it is in a diagram and quickly see if something is wrong. It is very safe and the IC is in no danger of being pulled apart. And you can easily change a component value if you want to.
What does the writing on the 555 chip actually say? Could you read it please?
.
Just reading back through the posts, I saw this one (above).
If you can get away from the breadboard approach, it would be good. You don't need printed circuit boards. I find breadboards generate more problems than anything else, usually because of poor connections.
This is a simple and extremely reliable ciircuit and it shouldn't be causing so much grief.
You turn the chip so its legs are facing upwards. (this construction method is called "dead bug construction" for this reason).
Any legs that are to be grounded are bent downwards and soldered to the PC board. Also any capacitors etc that need to be grounded are soldered in.
Power leads etc are attached to tag strips and not directly to the IC. (I use header pins cut to shape as tag strips).
This way, you can lay out a circuit as it is in a diagram and quickly see if something is wrong. It is very safe and the IC is in no danger of being pulled apart. And you can easily change a component value if you want to.
What does the writing on the 555 chip actually say? Could you read it please?
.
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vk6kro said:
LM393?
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LM393...Where did you get that?
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It's a joke. The 393 is also a common 8-pin PDIP part.
Twinfun2
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http://www.radioshack.com/product/index.jsp?productId=2062596
Would you guys consider Radio Shack to be a distributor of quality merchandise of this sort?
Yes.
At this point, all I there really is to blame is the breadboard, I completely agree.
During the past 3 or 4 days I have been posting here, has anyone been able to replicate my exact circuit?
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Twinfun2:
I repeat: measure the resistance between all of the nodes in the circuit, with the 555 installed, but without power. There are only a handful of nodes, so this should only take a few minutes. With this information, we can hopefully positively identify your problem as a bad breadboard.
- Warren
I repeat: measure the resistance between all of the nodes in the circuit, with the 555 installed, but without power. There are only a handful of nodes, so this should only take a few minutes. With this information, we can hopefully positively identify your problem as a bad breadboard.
- Warren
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I found a very old 555CN chip and made up your circuit.
AND I got the same result as you did, but without any heating. The LED was on all the time. I checked with an oscilloscope and there was no other oscillation going on.
It will not oscillate with a 100 K resistor unless the supply is less than 4.5 volts.
To oscillate at 6 volts the 100 K resistor has to be about 10 K but no more.
So, you might like to try that. Use just 3 alkaline AA cells to get 4.5 volts or get a 10 K resistor.
A data sheet I found gives 100 K as the upper limit on this resistor.
I'm sure I have used 555s that were a lot less fussy than this one. Guess it depends on the maker. Mine was obviously ***P quality.
If the Radio Shack is convenient, you could try one of these:
http://www.radioshack.com/product/index.jsp?productId=2062595
It is the CMOS version of the 555 and should let you use any resistor you like.
I was also going to suggest you pick up a cheap digital multimeter, but not from there. Overpriced and poor features.
This is the sort of meter I was thinking of:
http://www.altronics.com.au/index.asp?area=item&id=Q1053A
that is about US$7.50
but Australia is a long way from Philadelphia.
AND I got the same result as you did, but without any heating. The LED was on all the time. I checked with an oscilloscope and there was no other oscillation going on.
It will not oscillate with a 100 K resistor unless the supply is less than 4.5 volts.
To oscillate at 6 volts the 100 K resistor has to be about 10 K but no more.
So, you might like to try that. Use just 3 alkaline AA cells to get 4.5 volts or get a 10 K resistor.
A data sheet I found gives 100 K as the upper limit on this resistor.
I'm sure I have used 555s that were a lot less fussy than this one. Guess it depends on the maker. Mine was obviously ***P quality.
If the Radio Shack is convenient, you could try one of these:
http://www.radioshack.com/product/index.jsp?productId=2062595
It is the CMOS version of the 555 and should let you use any resistor you like.
I was also going to suggest you pick up a cheap digital multimeter, but not from there. Overpriced and poor features.
This is the sort of meter I was thinking of:
http://www.altronics.com.au/index.asp?area=item&id=Q1053A
that is about US$7.50
but Australia is a long way from Philadelphia.
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There's really nothing wrong with his meter. Nothing you're going to do at the board level is going to require more than 1% precision, so just a couple of significant figures is plenty. You can easily read a few sig figs off an analog meter. The only advantage of digital meters is that you can read them a little more quickly. They may appear to have greater precision, but their least significant figures are mostly noise anyway.
- Warren
- Warren
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The non zero voltage readings in the following chart should have all been 5 volts. Because of the low resistance of the analog meter, much lower readings were obtained.
This was the test where the chip was removed from the socket.
1: 0v
2: 2v
3: 0v
4: 5v
5: 0v
6: 2v
7: 4.5v
8: 5v
This was the test where the chip was removed from the socket.
1: 0v
2: 2v
3: 0v
4: 5v
5: 0v
6: 2v
7: 4.5v
8: 5v
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vk6kro:
The readings that were obtained have nothing to do with his meter.
The voltages on pins 2 and 6, etc. should not be 5V, and it's ridiculous of you to assert that they should have been. C1 had not been properly discharged, so it had some residual voltage on it -- nothing is driving that node.
There's nothing wrong with an analog meter -- certainly, nothing that's going to make his small 555 circuit fail to work. You're leading him down a frankly idiotic path.
I am a little concerned, though, about the voltages on pins 4 and 8 (RESET and VCC) being 5V, rather than 6V. If you're using 4 AA batteries and the voltage is sagging all the way down to 5V, you must be pulling a lot of current through them... what's the voltage of the battery pack when it's not connected to the circuit? What's the voltage when it is connected to the circuit? Are they substantially different?
- Warren
The readings that were obtained have nothing to do with his meter.
The voltages on pins 2 and 6, etc. should not be 5V, and it's ridiculous of you to assert that they should have been. C1 had not been properly discharged, so it had some residual voltage on it -- nothing is driving that node.
There's nothing wrong with an analog meter -- certainly, nothing that's going to make his small 555 circuit fail to work. You're leading him down a frankly idiotic path.
I am a little concerned, though, about the voltages on pins 4 and 8 (RESET and VCC) being 5V, rather than 6V. If you're using 4 AA batteries and the voltage is sagging all the way down to 5V, you must be pulling a lot of current through them... what's the voltage of the battery pack when it's not connected to the circuit? What's the voltage when it is connected to the circuit? Are they substantially different?
- Warren
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The readings were taken under power from 4 AA rechargeable cells.
This means that the capacitor was fully charged and that the voltage should have been about 4.8 volts (4 times 1.2 is 4.8 volts).
So, since there was no current flowing in the resistor chain (after a 1 second charging time), the supply voltage should have been across the capacitor.
So there should have been 5 volts there.
The only thing wrong with his analog meter is that it is a low impedance device trying to measure high impedances. Better digital meters have a 10 Megohm resistance even on the low voltage ranges, but even cheap ones have 1 megohm input resistance. The error in this case was up to 60%, quite enough to lead him to the wrong conclusions.
I actually prefer an analog meter, but I wouldn't use one in this case.
This means that the capacitor was fully charged and that the voltage should have been about 4.8 volts (4 times 1.2 is 4.8 volts).
So, since there was no current flowing in the resistor chain (after a 1 second charging time), the supply voltage should have been across the capacitor.
So there should have been 5 volts there.
The only thing wrong with his analog meter is that it is a low impedance device trying to measure high impedances. Better digital meters have a 10 Megohm resistance even on the low voltage ranges, but even cheap ones have 1 megohm input resistance. The error in this case was up to 60%, quite enough to lead him to the wrong conclusions.
I actually prefer an analog meter, but I wouldn't use one in this case.
chroot said:
Good idea to check these. Either the analog meter is somewhat off, or the unloaded battery voltages are that low, or the batteries are being drained when connected and that is pulling their voltage down.
I'll just add, try measuring the connected-battery-voltage both with and without the chip in the circuit. EDIT: but don't leave the chip in AND battery connected very long, due to the overheating chip issue.
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vk6kro said:
Well, I looked through the analog meter's manual, and am very surprised to see that its input resistance, when set to the 25V scale, is only 20kohm. As you said, that's definitely low enough to cause problems here. Please accept my apologies -- I just never expected any meter to be that poor.
It does achieve 2Mohm resistance, but only on the 1000V scale...
- Warren
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~10-20k is a typical low-scale input impedance for basic analog meters--even the venerated Simpson 260 only had a 20k input-Z. Better ones incorporate a FET to raise the effective impedance to the low tens of MOhms.
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Yeah, I made the (apparently really bad) assumption that all modern meters use the same kind of sensing circuitry (FET input stage, for example), and that the only difference is the display... and I was wrong!
- Warren
- Warren
Twinfun2
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chroot said:
Roger, I will do this when I get the chance, but I am away from home again (Been a busy few weeks).
Sorry for such a late update.
chroot said:
Wow.
Weird, I looked through the manual and could not find that info. Are you looking at something other than:
http://rsk.imageg.net/graphics/uc/rsk/Support/ProductManuals/2200223_PM_EN.pdf
?
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Very last page, bottom of right column: "Specifications."
- Warren
- Warren