-2
\$\begingroup\$

I have 2 led strips and I want to turn one ON when the other turns OFF. I have a 50V transistor from a atx psu, and a switch. When I assembled, the circuit works (not gate), but the strip that should light up stable, blinks. Can you please help with a schematic?

In this schematic, switching is not working at all. When the transistor is conducting, the strip that follows it lights, the other one just working independently.

\$\endgroup\$
10
  • 4
    \$\begingroup\$ Please add a schematic of what you have tested and clear up your formatting. \$\endgroup\$
    – winny
    Nov 28, 2016 at 12:47
  • \$\begingroup\$ How about you helping with a schematic of what you assembled? \$\endgroup\$
    – Andy aka
    Nov 28, 2016 at 12:47
  • \$\begingroup\$ Closing due to no schematic, just hand waving. \$\endgroup\$ Nov 28, 2016 at 13:19
  • 1
    \$\begingroup\$ Where did you put the switch and the transistor? Is it a MOSFET, BJT, JFET? Where is the NOT gate? We cannot help you without a schematic. \$\endgroup\$
    – 12Lappie
    Nov 28, 2016 at 13:23
  • 3
    \$\begingroup\$ You can use the schematic editor on the question form. Will be much more readable and practical than hand drawings. \$\endgroup\$
    – Wesley Lee
    Nov 28, 2016 at 15:16

1 Answer 1

1
\$\begingroup\$

Your schematic shows the LEDs connected backwards. The diode symbol should 'point' in the direction of classical current, that is from positive to negative voltage.

I was unable to find the transistor model you labeled, but if it is a p-channel transistor you may need an n-channel transistor to drive it (see Sziklai Pair).

If it is an n-channel transistor you may be able to get it to work. The voltage threshold will be different depending on your transistor. The following diagram will put 6V at the base of the transistor when the switch is open, so if a higher voltage is needed you can adjust the values of R3 and R4.

The way this circuit works is the transistor base is connected to the output of a resistor divider (R3 and R4). When the switch is open, the resistor divider outputs 6V at the transistor base. When the switch is closed, the switch provides a path of less resistance to 0V and pulls the transistor base down to 0V, causing the transistor to switch off. With this circuit, the first LED does make a complete circuit even when the switch is open. But as long as R3 and R1 are a high enough resistance, the current will be too limited for the LED to light up.

I added R1 and R2 since if you are using normal LEDs with 12V you will want current limiting resistors, otherwise the LEDs will break very quickly. Those resistor values should probably be somewhere around 510 ohms but this really depends on what LEDs you are using.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
4
  • \$\begingroup\$ I'm actually unsure how R1/D1 will affect the output of the resistor divider when the switch is open. This might not work like I had initially thought when I wrote this answer. \$\endgroup\$
    – Nathan
    Nov 28, 2016 at 17:10
  • \$\begingroup\$ The transistor is a npn NTE 293 & NTE394 equivalent, leds are strips that come with resistors and all the rest, just plug to 12v and play. I managed to find a simple diagram on dummies .basically 1st led drives off the 12v rail into the transistor base trough a switch and the second led is grounded from the rail trough the transistor. Will try both designs tomorrow. Thank you for your help. \$\endgroup\$ Nov 28, 2016 at 17:24
  • \$\begingroup\$ This circuit may blow out T1. When SW1 is open, the full D1 current will pass through T1's base, so so D1 will always be on, just slightly dimmer when the switch is open. A 2N3904 is rated for 200 mA collector current, but there is no such spec for base current. \$\endgroup\$ Nov 28, 2016 at 18:46
  • \$\begingroup\$ Could that be fixed with a resistor on T1 base? \$\endgroup\$
    – Nathan
    Nov 28, 2016 at 18:52

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