11
\$\begingroup\$

I have made the following circuit:

enter image description here

When power is applied, C1 charges up and the LED lights up. When power is disconnected, the LED gradually dims. It turns off when C1 reaches around 1.5V.

I'm trying to get another LED to turn on when the first LED turns off. So basically, I want a circuit to turn on when the capacitor has below 1.5V:

enter image description here

I am using a SS9014 Transistor (it says something else on the diagram). I'm not too familiar with reading transistor data sheets, so I have no idea of the conditions that make a transistor turn off. Is it a certain voltage or is it a certain amperage that makes it turn on?

If it's a certain amperage, then I could just adjust R2 to make LED2 turn on when C1 is under 1.5V right? What is that amperage?

If it's not amperage that makes Q1 turn on, then it must be voltage. I assume the voltage is less than 1.5V because 1.5V can turn it on. What are my options in this situation? Is there a way (or possible a 4000 series IC?) to reduce the 1.5V from C1 to the minimum voltage that makes the base turn on Q1? If so what is that voltage?

Sorry if my question is confusing. Feel free to ask questions in a comment.

\$\endgroup\$
  • 4
    \$\begingroup\$ Please get into the habit to draw your power supplies on a vertical line by themselves, with + up and - down. This will be easier to read and you'll avoid stupid mistakes like in your second diagram. \$\endgroup\$ – starblue Jun 17 '12 at 12:57
3
\$\begingroup\$

With bipolar transistors it's the current that controls the transistor, not voltage (though there is a minimal voltage). So, increasing the base resistor (R2) will make the transistor turn on at higher voltage than 1.5V).

If you want the LED to turn on when the voltage is below 1.5V *that is, invert what it is now) then you can do it like this:

schematic

Now when the transistor turns on, it shorts out the LED so the LED goes dark. When the base current drops below a certain level, the transistor will turn off and the LED will turn on.

You may have to find a suitable value for R1 so that the LED turns off completely when you want it to turn off. The power supply voltage (5V in my circuit) does not really matter, as long as R2 is suitable for the LED.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ How much current turns the transistor on? The datasheet isn't clear about that \$\endgroup\$ – blake305 Jun 17 '12 at 4:04
  • \$\begingroup\$ @blake - the current changes gradually with input voltage because of the base resistor. So you can vary it from a few micro-amperes to (in Russell's schematic) ten mA by changing the voltage. It's completely on if it tries to draw more than 10 mA. In that case R2's voltage drop of 5V will limit the current; that's all that's available, and the voltage across the transistor will go as low as a few hundred mV. \$\endgroup\$ – stevenvh Jun 17 '12 at 6:05
  • 2
    \$\begingroup\$ Note that this circuit consumes more power with the LED off than when it's on. \$\endgroup\$ – radagast Aug 20 '13 at 14:04
  • \$\begingroup\$ Use a comparator. \$\endgroup\$ – Chupacabras Feb 20 '18 at 12:58
2
\$\begingroup\$

As I understand it you want to go one LED brighter as the other goes dimmer.

I would use the following circuit:

enter image description here

The top part (Q1, D1, D2 and R1) are a constant current source. The diodes create a 1.4 V difference, 0.7 V per diode. The transistor's emitter-base junction also acts like a diode, and therefore also drops 0.7 V. Then the other 0.7 V from D1 is across R1, so according to Ohm's Law the current through it should be 0.7 V / 35 \$\Omega\$ = 20 mA. That's the sum of the current through the LEDs.

A transistor is current controlled, but that's not always handy, and therefore we will often make it voltage controlled by adding a resistor to the base. The resistor will turn a voltage difference into a current, again Ohm's Law. So by driving the CTRL input with a voltage that will create a base current, which will cause a current through LED D3. As D3's current increases D4's current will decrease, because the sum of the currents is constant.

Remember that the base-emitter junction has a 0.7 V voltage drop. CTRL has to go higher than that before there will be any current. Above 0.7 V the current will change linearly with the control voltage.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

You could use something like this:

Schmitt Trigger and Inverter

Your requirement of having one LED on while the other is off is achieved by Q5. This transistor inverts the signal coming from Q2's collector, so either Q3 or Q4 are on.

Looking at this part (Q3, Q5, Q4) only, you could also turn the whole thing upside down, using NPN transistors.

The reason I chose PNPs was that I have signal coming from a Schmitt-Trigger (around Q1, Q2) that is referred to VCC, and Q3, Q5, Q4 being also referred to VCC, this is just right for PNPs.

Using a supply of 5V, this circuit turns on when IN reaches ca. 4 V and it turns off when IN goes below ca. 1 V. The resistors around Q1 and Q2 determine these levels. If you don't need a Schmitt trigger or somewhat precise levels, you can omit Q1 and Q2 and start at the node at Q2's collector.

Notes: C1 will load the input quite heavily. Omit or use a smaller value. R17 and R18 are only required when you use the signals at Q3's and Q4's collectors as an input for more logic East of this diagram.

Enjoy simulating it (or even better: breadboarding it) and play with the values unit it does what you need. Exact transistor or LED types don't matter, pretty much anything will work.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.