I know this is just a general question and I will get get more specific later after I make a schematic and post it.

Here it is:

Is there a way use use just one Zener Diode to control a bi color LED with 2 post? I burned out the only Bi Color LED I have so I am using 2 LEDs in place of the one, but I cannot get my mind wrapped around it. My simple circuit is build upon a switch, resistor(s), zener diode and LED(s).

I would like the RED Led to stay lit until I press the switch, and then I would like the Red LED to turn off and the Green LED to light up. Very simple. My voltage source can either be a 4 A batteries or 1 9 volt battery. I can do both.

Thanks for any help provided. Its been a while and I am getting back into basic electronics again and I want to start with this simple circuit.

Thanks again.

 

Can we have a schematic of this circuit?

Ramesh

 

What kind of switch do you have?

Can you use a DPDT? If not, can you use a SPDT?

When you say, "press the switch," that makes me wonder if this is a momentary switch. Is it.

 

Here is the schematic so far. I got the circuit to work with 2 LEDS. The Red LED shuts off completely when switch is pressed.

I have several questions about this circuit which you guys helped me with about a year ago. I will ask as I get things sorted out.

Can anyone recommend a very good read on Zener Diodes. I am having a horrible time understanding them My one question is: this is a very simple circuit...how does it work? I understand the concept of Zener Diodes, the forward Voltage being about .7 volts, but the breakdown voltage is hard for me to understand. How do I get the Breakdown Voltage for the Zener Diode to work?

I would still like to get this down to a Bi Color LED if possible.

Thanks for the help guys.

 

Forget about electron flow and use conventional current. At the end of the day they are the same thing and trying to mix and match only leads to mishmash like you have here.

A diode conducts in the forward direction when its anode (the base of the arrow) is at a more positive voltage than the cathode (the line at the tip of the arrow). Notice that I didn't have to say anything about electron flow or conventional flow.

In your diagram you have the negative terminal of the battery connected only to the anode of a diode (an LED). Thus, in this circuit, no current can ever flow (unless you exceed the reverse breakdown voltage of an LED, which since these are commonly in the 5V to 7V range, is very possible with a 9V battery.

Also, saying that you understand the concept of zener diodes and then saying that the breakdown voltage is hard for you to understand and that you don't know how to get it to work, means that you don't understand the concept of zener diodes since the whole point of a zener diode is to operate in the reverse breakdown mode.

 

Your circuit can't light up the LEDs whatever the red or green one.

The direction of the LED is wrong, you could treat the LED similar as the diode.

If you want to know more about the Zener diode, you can see the following diagram.

 

I guess Im trying to ask is how do you know what value of Zener you should pick. Do you pick a Zener based on what voltage you need after the Zener--such as the circuit above. I need about 2 volts for the LED so I pick a Zener based on that, or do I pick a Zener based on how much voltage I need to use? Again the circuit above--The Zener has a breakdown voltage of 3.3, but the voltage across the Zener is about 2.6 volts when measured with a multimeter. I am in the process of the making a new schematic and will post it here as soon as possible.

I really do appreciate all the help and suggestions. Im still new to electronics, but Im learning gradually

 

I'm not sure why you want to using zener diode, there are some other ways to do, but you have to explain what's the purpose of those LEDs, different request has different solution.

The circuit haven't tested, you may need to adjust the values of R2.

 

This is similar to Scott's circuit, and it did work.

 

This is similar to Scott's circuit, and it did work.

If you are using two different colored LEDs, then you need to be sure to put the one with the higher forward voltage in the leg with the normal diode. This is because if you don't and if the forward voltages differ by more than a diode drop, pushing the button will have little or no effect.

 

I will be sending an updated drawing soon.

First of all to answer some questions. I am doing this project simply to learn. I could use other components for the project but I would like to start with these basic components and work up from there.

@ScottWang. I appreciate the drawings you sent and I have a couple of questions about them. So I Zener Diode is in affect a diode that is has 2 leads that the forward Voltage is .7 volts, but the reverse voltage (breakdown voltage is .7 times however many diodes are in the place.) For example, for drawing 5, you have one leg with one diode (.7) and one leg with 3 diodes (.7 x 3 = 2.1). Is that correct?? If that is correct, that is something I never knew and its starting to make sense now, because the breakdown voltage has to overcome the 3 diodes at .7 volts each.

I will stop there for now because I would like to understand this concept clearly before moving on.

 

Here is the updated drawing. I changed some simple things.

 

Here is the updated drawing. I changed some simple things.

What happened to your switch?

In your diagram the top branch might as well not be there. The voltage across the middle branch will be 2.1V which is insufficient to overcome the 5.1V needed to start conduction down the top branch.

Hence you will have (9V-2.1V)/195Ω, or 35mA of current in the green LED, which is nearly twice what you have indicated that it is rated for.

If you remove the green LED (either physically or by way of a switch), the the current in it would be (9V-1.8V-3.3V)/195Ω = 20mA.

BTW, where are you going to get a 195Ω resistor? Just call it 200Ω and be done.

Try replacing the zener with a normal diode, but turned the other way. Now you need about 2.4V to turn on the red LED and when the green LED is in the circuit it will only have about 2.1V. That 300mV difference should keep the current in the red LED down to less than 0.1% of the current in the green LED. If you then size the resistor for 20mA in the green LED you would need

R = (9V-2.1V)/20mA = 345Ω

Round that up to 360Ω (the next larger 5% value), which would give you a current of.

I_green = (9V-2.1V)/360Ω = 19.2mA

When the green LED is switched out, the current in the red LED would then be

I_red = (9V-1.8V-0.7V)/360Ω = 18.1mA

In theory, you could omit the diode altogether and put the switch in series with the red LED relying on the 300mV difference between the two forward voltages to keep the green LED off when the red LED is in the circuit. It's trivial to see if this will work -- just put the two in parallel connect them to your battery through a 330Ω to 470Ω resistor and see if only the red one lights up.

 

I will be sending an updated drawing soon.

First of all to answer some questions. I am doing this project simply to learn. I could use other components for the project but I would like to start with these basic components and work up from there.

@ScottWang. I appreciate the drawings you sent and I have a couple of questions about them. So I Zener Diode is in affect a diode that is has 2 leads that the forward Voltage is .7 volts, but the reverse voltage (breakdown voltage is .7 times however many diodes are in the place.) For example, for drawing 5, you have one leg with one diode (.7) and one leg with 3 diodes (.7 x 3 = 2.1). Is that correct??

That's right.
What voltage you want to using the Zener diode, it depends on the direction, the circuit diagram will make you to understand the Zener diode easier.

And what's going on with the switch?

Your circuit will make red LED won't work, because the voltage of red LED and Zener diode already limited at 2.1V by the green LED.

 

I really appreciate your post Wbahn. It gives me a lot to think about. I will certainly try your suggestions...this was an excellent post because it makes me think Thanks again.

When you speak of induction on the the top branch, what are you referring to? Can you give me a reference on where I can read about it in basic terms or better yet by examples of circuits.

I really liked this reply a lot.

A new drawing will be posted shortly with the switch in place.

 

Not "induction", "conduction". You won't get any current flowing if the applied voltage is less than the required voltage drops across the diodes. Once you exceed this voltage, then current starts flowing -- the branch starts conducting electrical current, hence conduction has started.

 

Here is the schematic so far. I got the circuit to work with 2 LEDS. The Red LED shuts off completely when switch is pressed.

I have several questions about this circuit which you guys helped me with about a year ago. I will ask as I get things sorted out.

Can anyone recommend a very good read on Zener Diodes. I am having a horrible time understanding them My one question is: this is a very simple circuit...how does it work? I understand the concept of Zener Diodes, the forward Voltage being about .7 volts, but the breakdown voltage is hard for me to understand. How do I get the Breakdown Voltage for the Zener Diode to work?

I would still like to get this down to a Bi Color LED if possible.

Thanks for the help guys.

I'd forget about the zener and put a 1N4148 in series with the green LED, put an appropriate current limiting resistor in series with that. Then put a pushbutton & red LED across the green & diode - when yoy press the button; the red LED will have lower Vf than the green + diode and cut it off.

If you want the green on when the button is pressed, you'll need extra diodes in series with the red LED to make the combined Vf greater than the green LED.

There's an elegant circuit floating about somewhere based on a 2 transistor Schmidt-trigger that will transition smoothly from 1 LED to the other with varying input voltage - or with the addition of the positive feedback resistor will snap-switch.

 

This is what has been suggested in Post #11 and Post #13. The OP is already trying these.