Thanks iONic. Could you give more details. When you refer to Power Requirements, are you referring to the Watts. I see that there are several different wattages available starting at 1/4 watt and going up to 1 watt. What does IsqR stand for and is that the formula after that (.02x.02)/R.

Thanks.

Yes, the power requirements I am referring to is the resistor wattage rating.
"IsqR" was my way of giving you the equation for determining the power consumption for the resistor... It's the current squared x the resistance as in the following image:

 

Thanks Ionic. That is a neat little chart you have there. Just printed it off.

 

Here is the circuit I have designed. I built it from ideas I got from you guys from the forum, and I certainly appreciate the help. I started pretty basic and from there have added to a nice little circuit that I have learned a bunch from this little circuit. It was in fact the first circuit I have built.

Just wanted to show the completed schematic, and I have some final questions that have started to be answered in this post. And here are the last questions concerning my first circuit.

The power supply is 9 volts. As you can see from the schematic there are 4 LED's (2 of which are lit at 1 time) and 1 zener diode. When the power is to the circuit the 2 Amber LED's are lit, and the 2 Green LED's are off. But when the tilt switches are closed, the Green LED's are lit and the Amber LED's are off. So when the Amber LED's are lit I have 4 volts there plus 3.3 volts to the Zener Diode giving me a total of 7.3 volts (2+2+3.3).

When the Green LED's are lit I have a total of 7.5 volts total for the circuit (2.1+2.1+3.3).

Finding the resistance for the resistor, I came up with (9 - 7.3) = 1.7 / .02 = 85 ohms. I don't think that was right, because after doing some readings on the LED's there was too much voltage and current. About 2.5V and a little over 70 mA. So I kept adding "bigger" resistors. I eventually got up to a 200 ohm resistor and the readings are where I wanted them. For the Amber LEDS 1.8 with about 18mA. The Green LED's were around 2.1 but they had a higher current rating of 33mA. (This confuses me a little bit, but after checking the datasheet it shows max amperage of 95mA with average of 25mA and for DC it shows 30mA. They are manufacturing #HLMP-3962)

So after a long post here is my question: Should I have included the Zener Diode in figuring my resistance for my LED's. Including it I come to a rating of 85 ohm. Not including it I come to a rating of 250 ohms which is much closer to the 200 ohm resistor I have in there now. Looking at my schematic you will see that there is a 96" cable between the connectors. I figure that the extra 50 ohms would be lost in there somewhere. Would that be correct? Or is my readings and figuring the ratings just completely wrong?

Thanks for your help thus far and any further help you can provide.

 

You are still missing a key point. Each LED drops a fixed voltage. These voltages add. If you have two LEDs with a Vf of 2.5 volts in series, then you need over 5V (2.5V + 2.5V) to create current through them. LEDs depend on current, the voltage they drop is a quantum effect of how they create light, but they are current controlled devices.

Another thread has opened with similar questions to yours. You are not alone.

Question regarding ohms law

 

I'm not sure I understand your use of the zener diode at all. Is this a circuit you intend to build? Can you explain its function. From what I see, when the tilt switch is closed you will be attempting to light all 4 LED's and not just the two green LED's.

For reference to others, here is the OP's Circuit.

 

I certainly appreciate the replies.

The purpose of this circuit is (sorry I didn't explain it) is that when the 2 tilt switches are closed, it should light only the Green LED's. It is a simple case of 2 LED's are lit until the switches close and then 2 other LED's light. The Zener Diode (from what I understand, I actually got the idea from this forum) is for when the Zener voltage reaches close to 3.3 it allows voltage to reverse. Again I could be wrong so please correct me if I am wrong, is that voltage goes through the circuit to the Amber LEDs, but when the 2 switches are closed, there is in effect 2 circuits it can travel, so the Zener (Breakdown) Voltage which I think from my figures is around 2.7V from my testing of the circuit closes the Amber LED circuit. Hopefully that is the correct explanation. Again please correct me if I am wrong.

 

The zener will increase the voltage at which the red LEDs turn on. 2X Red LEDs and a 6.3V zener means you need around 11 Voltage before they start lighting up. I've done something similar in the past.

Green LEDs need more voltage than red LEDs do. Where are you getting the dropping voltages you are using?

 

Thanks Bill. That makes sense now. I like your example. Sometimes the easiest way to explain something is by use of an example.

Now I understand it. So I need about 7.5volts to turn the amber LED's on but when the green LED's turn on, the Amber ones don't get the voltage they need to they turn off.

Thanks again.

 

Sorry I didn't answer this in my last post.

I bought the green LED's from newark.com. They are hlmp-3962. The datasheet shows 2.1v and 20mA.

 

So if ~7.5V is dropped on the two amber LED's and the Zener, the resistor I calculated for 18mA would be ~83 ohms. But for the Green LED's you would need a resistance of 270 ohms. Each resistor ought to be placed in the two parallel branched as apposed to the single resistor. CORRECT ME IF I'M WRONG!!

The Zener properties talked about are actually new to me and can offer an interesting circuit option.

 

Sorry Ionic

I am new to circuit design and really couldn't "correct" you if you were wrong. One of the more experienced will have to answer the question.

This circuit has gone through a lot of major changes as a learning experience for me.

 

Sorry Ionic

I am new to circuit design and really couldn't "correct" you if you were wrong. One of the more experienced will have to answer the question.

This circuit has gone through a lot of major changes as a learning experience for me.

No problem, beeson76, The question was more directed towards others such as Bill.

 

Green LEDs work out to 22ma on my calculator I keep on my desk using 2.1 Vf.

The amber work out to around 17ma. Looks good.

Low voltage zeners are slightly mushy, that is the voltage drop isn't quite a crisp over the range currents, but it will work OK.

Even a low voltages the LEDs may glow dimly, though they are off (but drawing µA). Nothings perfect.