## [misterhouse-users] OT: power meter watch circuit

 [misterhouse-users] OT: power meter watch circuit From: Alan Womack - 2002-11-28 06:39:45 Attachments: schematic.jpg ```I have built the circuit based on the schematic attached. I've double checked = my work twice, but am unable to make it function. Where do I start for troubleshooting? I have power 7.69v. The two leads from the Cadmium sulfide cell are connected to a pair of header = pins. One connected directly to ground, one goes to a "junction" where a 100uf = cap goes from + on the line side to ground (7.61 v shows). A 10K ohm resistor = goes from the line to 5V with the gold stripe to the line. I show 7.84v on the = brown end and 7.84 on the gold end. Then comes a 22uf cap I have 3.1v here. Then I hit a pair of 100K resistors, = one going to 5V (3.6v) the other to ground (4.2v) via a cross board ground. Then I go to pin 2 of the opamp (4.2v) Pin 1 from the opamp combines at solder point, a 1K (brown black red) resistor = going to power (7.4v) Pint 1 at the solder point also has a 1M resistor connecting from jumpers going = from Pin1(7.55v) to pin3(7.55v), jumpers from pin 3 to pin 6, and the middle = point on a 100K potentiometer (7.55v) The poteniometer lines have 470k resistors, one to ground (.04v) and 1 to 5V = (7.56v) pin1 also has a jumper to pin 5 (7.58v) Pin 7 I get (5.81v) If I apply a ground path for pin 7 I get the led to come on which comes from = pin 7 and goes to 5V. But playing with the Cadmium cell I get no difference in = pin 7's behavor. I know what the circuit is supposed to do, but not HOW it does it technically. = I think I have built it correctly, but it is not doing it's job. Alan=20```

 [misterhouse-users] OT: power meter watch circuit From: Alan Womack - 2002-11-28 06:39:45 Attachments: schematic.jpg ```I have built the circuit based on the schematic attached. I've double checked = my work twice, but am unable to make it function. Where do I start for troubleshooting? I have power 7.69v. The two leads from the Cadmium sulfide cell are connected to a pair of header = pins. One connected directly to ground, one goes to a "junction" where a 100uf = cap goes from + on the line side to ground (7.61 v shows). A 10K ohm resistor = goes from the line to 5V with the gold stripe to the line. I show 7.84v on the = brown end and 7.84 on the gold end. Then comes a 22uf cap I have 3.1v here. Then I hit a pair of 100K resistors, = one going to 5V (3.6v) the other to ground (4.2v) via a cross board ground. Then I go to pin 2 of the opamp (4.2v) Pin 1 from the opamp combines at solder point, a 1K (brown black red) resistor = going to power (7.4v) Pint 1 at the solder point also has a 1M resistor connecting from jumpers going = from Pin1(7.55v) to pin3(7.55v), jumpers from pin 3 to pin 6, and the middle = point on a 100K potentiometer (7.55v) The poteniometer lines have 470k resistors, one to ground (.04v) and 1 to 5V = (7.56v) pin1 also has a jumper to pin 5 (7.58v) Pin 7 I get (5.81v) If I apply a ground path for pin 7 I get the led to come on which comes from = pin 7 and goes to 5V. But playing with the Cadmium cell I get no difference in = pin 7's behavor. I know what the circuit is supposed to do, but not HOW it does it technically. = I think I have built it correctly, but it is not doing it's job. Alan=20```
 [misterhouse-users] OT: power meter watch circuit From: bazyle butcher - 2002-11-29 00:22:21 Attachments: Message as HTML ```Alan, You are right in needing +ve on pin 8, and 0v on pin 4. Also you do need = a ground to tthe computer to complete the circuit as you suggested. The NTE 943 is a comparator, and can run from 5v to 30v see data sheet = http://www.nteinc.com/specs/900to999/NTE943.html Your circuit is designed to give pulses when the light level changes; it = is not intended to show two states, light and dark. Under steady light conditions pin 2 is at half voltage so adjust the = potentiometer to make pin 3 just below half voltage. Pin 1 will be low = because pin 2 ( inverting input) is higher than 3, but pin 7 follows pin = 5 ( non inverting input to second stage), so pin 7 is low and LED is on. ( the reason you 'are always high' is probably because you have set the = potentiometer to make pin 3 higher than half voltage) Shine a light on the cell and it reduces resistance so pin 2 gets pulled = low, 3 goes high, 7 goes high and the LED is off, but only while the = 22uf capacitor is charging. So you only get the LED off for a while. If you want it to stay showing the light level, instead of a pulse, just = remove the 10k resistor and the 22uf capacitor. You didn't say which light cell you use but if you can get an ORP12, = it's been around for 30 years and still the best device around. (Also because this circuit is designed to only give high or low output, = not progressive, don't bother to try feeding it into an analogue = input.) Baz ```
 Re: [misterhouse-users] OT: power meter watch circuit From: Don Wilde - 2002-11-28 18:23:08 ```Alan, Your description leads me to suspect that you havent got power and ground to the chip. Those connections are not shown in typical schematics, only the signal lines. There should be connections to two ther chip pins, one power and one ground. Havent got a chip manual handy, so I cant tell you which pins, but often opposite corners. Don ----- Original Message ----- From: "Alan Womack" To: "Majordomo leben.com" Sent: Thursday, November 28, 2002 1:39 AM Subject: [misterhouse-users] OT: power meter watch circuit I have built the circuit based on the schematic attached. I've double checked my work twice, but am unable to make it function. Where do I start for troubleshooting? I have power 7.69v. The two leads from the Cadmium sulfide cell are connected to a pair of header pins. One connected directly to ground, one goes to a "junction" where a 100uf cap goes from + on the line side to ground (7.61 v shows). A 10K ohm resistor goes from the line to 5V with the gold stripe to the line. I show 7.84v on the brown end and 7.84 on the gold end. Then comes a 22uf cap I have 3.1v here. Then I hit a pair of 100K resistors, one going to 5V (3.6v) the other to ground (4.2v) via a cross board ground. Then I go to pin 2 of the opamp (4.2v) Pin 1 from the opamp combines at solder point, a 1K (brown black red) resistor going to power (7.4v) Pint 1 at the solder point also has a 1M resistor connecting from jumpers going from Pin1(7.55v) to pin3(7.55v), jumpers from pin 3 to pin 6, and the middle point on a 100K potentiometer (7.55v) The poteniometer lines have 470k resistors, one to ground (.04v) and 1 to 5V (7.56v) pin1 also has a jumper to pin 5 (7.58v) Pin 7 I get (5.81v) If I apply a ground path for pin 7 I get the led to come on which comes from pin 7 and goes to 5V. But playing with the Cadmium cell I get no difference in pin 7's behavor. I know what the circuit is supposed to do, but not HOW it does it technically. I think I have built it correctly, but it is not doing it's job. Alan ```