# Flip-Flop Latch Switch Circuit help

Discussion in 'The Projects Forum' started by timbudtwo, Mar 26, 2015.

1. ### timbudtwo Thread Starter New Member

Mar 26, 2015
3
0
Hello,
I am building a model for a product design class and I would like there to be a functional LED on it, controlled by a tact switch. There is not enough room for a toggle, so a latch will be necessary.

I purchased a flip-flop latch switch on ebay because I do not have enough experience with circuits to build one myself. I wasn't able to understand all the debouncing and other logic required.

Here are the schematics that were provided with the circuit:

What I did: Positive end of 3V button cell was wired to tact switch, VCC, and positive leg of LED. negative leg of LED was wired to Vo. Other end of tact switch was wired to Vi. Negative side of button cell was grounded. I did not use a resistor after the LED as the input voltage was already correct for the LED.

With this setup I was unable to get it to work. All help getting this working would be appreciated!

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
"I did not use a resistor after the LED as the input voltage was already correct for the LED."

There is your problem. When illuminating LEDs, there is no such thing as the voltage being "correct" for any LED. LEDs are never driven by connecting them directly to a voltage source (battery).

The correct reasoning would have been: I have a 3V LED. I need to drive it with xxmA. After allowing for a minimum voltage drop across a ballast resistor of say 1V, the minimum input battery voltage has to be greater than 1V+3V=4V. To get xxmA, the ballast resistor must be (by Ohms Law) R=E/I = (drop across the ballast resistor)/xxmA = ...

for a 20mA LED and a 4.5V battery, that would be (4.5-3)/0.02 = 1.5/0.02 = 0.75/0.01 = 75Ω.

The reason that the LED didn't light when connected to the switching device is because the switching device uses a transistor which has a saturation voltage of about o.25V when it is conducting. Subtracting o.25V from 3V doesn't leave enough to light the LED, and even if it did, you still need enough headroom for the voltage across the required ballast resistor.

You have two ways to proceed:

1. Start with a higher battery voltage, like 4.5V. Put a 75Ω resistor in series with the LED on the output side of the switching device as shown on the schematic. After testing the actual current draw from the battery with the LED lit, you might have to slightly adjust the 75Ω resistor value to set the final current...

2. Use a 3V battery, but temporarily replace the LED with a standard Red LED ( the kind that has a forward voltage of ~1.9 to 2.2V). Connect a ballast resistor of (3V-0.25V - 2V)/0.02 = 0.75/0.02 = 37.5Ω (use 33Ω or 36Ω standard values) in series with the LED and try the switching device.
If the LED lights, measure the drop across the switching device when the LED is on. If you get a voltage drop is substantially different than what I guessed (0.25V), you might have to repeat the calculations for the ballast resistor...

Your real problem is too low a battery voltage and trying to operate a LED without the required ballast resistor.

Last edited: Mar 26, 2015
3. ### timbudtwo Thread Starter New Member

Mar 26, 2015
3
0
I have never taken apart a flashlight and had it be anything other than LEDs driven by batteries.

I was using a red LED by the way. I didn't know they had different operating voltages.

I only have a 100ohm resistor to test it out with. The LED's I have are 3mm clear red.

I have some 18650 batteries I can try it out with.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Only super-cheap throw-away Chinese crap flashlights do that. They rely on the internal resistance of the battery to provide a modicum of ballasting. These flashlights dim out long before the battery is used up. They are greatly effected by ambient temperature (because a LED's Vf is strongly temperature dependent).

Here is a table of Vf for different color LEDs

5. ### timbudtwo Thread Starter New Member

Mar 26, 2015
3
0
I must have had a bridge or something wired wrong the first time... It works now, even without the resistor. It also works with the 3v cell.

However, I did notice that it was rather dim. There is only about 2v across the led when the device is on. This was when I was using an 18650 that was at 4.1v.

6. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Here is a comparison of two ways of driving the same LED as temperature varies. I made the LED voltage and current be the same at 25degC. Now compare what happens to the LED current (hence brightness) for LED1 and LED2.

7. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Here is a comparison of two ways of driving the same LED as batteries discharge. In this case, I am using a (mythical) nominal 2V battery that has some internal resistance (0.1Ω). The battery starts out at 2.1V and discharges toward 1.7V. Note what happens to the LED currents I(D1) and I(D2). Note that LED brightness is proportional to LED current.