Help on amplifying current in a circuit?

Question

Yes I'm a noob so be nice. There's no general electronics class at my high school so I've had to each myself. Right now I'm trying to build a circuit that makes a group of LEDs light up to spell a word. I have a 555 timer attached to an NTE4017b decade counter so the high current sycles through 10 outputs which are linked to a letter in the word. Everything works fine except that there isn't enough power that goes out of the decade counters output to light the bunch of LEDs of a letter. How could I fix this so the letter stays off until the counter output is high? I know I need to amplify the output from the counter but I need to know how. A specific schematic would be nice. Thank you for reading this far and I would apreciate any help you can offer!

Answer 1

The circuit in the first link below is a driver, but also provides current regulation, so acts as the ballast for the leds. This means the current used by the leds is independent of the supply voltage, so long as it is sufficient. A typical value of R would be 47 ohms, which allows almost 15mA.



When the input is high, the transistor connected to the leds is turned on. Current flows through the battery, leds, collector emitter path, and then R. At around 15mA the voltage across R is 0.7V, just sufficient to start turning the second transistor on. This robs some of the base current, which tends to prevent the current increasing further, so equilibrium is established when 15mA flows.



The second link has info about using leds, and shows a simple driver, where R is the ballast. See the 'transistor driver circuit'. TR1 can be a BC547 or 2N2222 etc (NPN with 100mA collector current or a little more). The value of Rb can be 4.7K. The ballast resistor R takes up the difference between Vbattery (Vcc) and the led forward voltage Vf, at the desired current. R = (V_cc - V_led) / I_led. I_led is 0.015A. Leds can be in series with a single ballast resistor. In parallel, each series string requires a ballast resistor to suit. This could be around 470 ohms for a red led and a 9V battery or 390 ohms for a blue or white led.. With 9V there can only be 3 red leds or 1 or 2 blue or white leds in series.

Answer 2

The 4017 will only put out a milliamp or so of current, which is barely enough to light up an LED. The problem with turning on several LED's is that each one needs to have about 2 volts across it. You don't say what voltage you are using, or how many LED's you want to turn on at each output. The IC will work from about 3 to 18 volts, and it will be easier to operate LED's if you are at the higher end of the range (12-volt power supply or 8 to 10 batteries in series). You can connect each of the 4017 outputs to a resistor going to the base of an NPN transistor, connect the emitters to ground, and connect the LED's in series between the collector and the positive power supply, with a series resistor to limit the current. If you have more than 4 or 5 LED's per letter, you will need two or more resistors, with a resistor in series with each group of five. The resistor should be around 1000 ohms. Experiment to see how many LED's you can put in series across the power supply -- the voltage across each one is different for different colors. Make sure to have the resistor in series, or you'll burn out the whole string. The resistor from the 4017 to the base of the transistors can be 10000 to 20000 ohms.

Answer 3

An amplifier produces an output it quite is a picture of the enter. It does no longer ~strengthen contained in the enter. The enter is a separate circuit from the output. No amplifier operates absolutely on modern-day (or voltage). truly the impedance is amazingly low for a modern-day operation or very intense for voltage operation. The adverse enter of an op-amp with the functional grounded is a modern-day enter because the output voltage will bypass to furnish a cancellation modern-day via the comments resistor. in case you want a modern-day output, you should rearrange the comments voltage to come back from a sensing resistor, ie the output is self sufficient of the surprising load impedance, as a lot because the decision of the availability rail voltages. after all, no amplifier quite operates on AC. truly it operates from both functional and adverse DC resources so as that the output can swing in both instructions. Audio amplifies commonly do no longer strengthen DC because you won't be able to listen DC and it basically motives warmth and burns out loudspeakers. The low frequency reaction is decrease off utilizing coupling capacitors and/or transformers.

Answer 4

If your LED's cathode is grounded, then your load requires current to be sourced by the driving device, that is output of 4017. If that cannot handle the required current, connect a transistor with base connected to 4017 output, and load (i.e LED's) connected to emitter. Connect collector to +ve supply. This transistor will increase driving capability at least 50 times. The draw back is that the voltage available at emitter will be about 0.7V lesser. If that is not adequate, increase the power supply by 0.7V.

Answer 5

1) Are the LEDs wired in series or parallel, for each letter?

2) 4000 series chips are CMOS so they are capable of handling higher voltages than TTL, what is the voltage of your power supply?





I suggest using a transistor (2N3904, or 2N2222) to amplify the current out of 4017.

Answer 6

Put an emitter follower circuit for each output to drive each LED. Read this....



http://en.wikipedia.org/wiki/Common_collector



Replace Re shown in circuit with LED.