• posted

In the diagram below:

3V | ---------- NPN LED --- -

from my electronics learning kit, the NPN is in parallel with the led, with its base connected to a switch. When you press the switch, the LED, previously off, suddenly turns on. I've left out the resistance stuff.

My question if, since the voltage is in parallel with the LED, why can't the current just go through the LED and not the transistor. Why does the lack of current in the transistor keep the LED off until the base is activated?

• posted

Is the collector-emitter of the transistor really in parallel with the led or is it in series?

If the transistor is really in parallel with the led then the led will be OFF when the transistor is ON. This will happen because when the transistor is ON, the voltage from the emitter to collector (and thus also across the led) will be below the level that the led will draw any appreciable current. When the transistor is off then voltage across the led will be allowed to seek its own level (as if the transistor weren't even in the circuit).

If the transistor is in series with the led then the led will be ON when the transistor is ON.

Bob

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Leaving out the resistors makes it a lot harder to understand the circuit, because the voltage drops across those resistors are a big part of what makes this work.

LEDs require some minimum voltage across them before they conduct enough current to show a visible output. The voltage varies with the color (energy) of the photons emitted. Red LEDs take about 1.5 volts, blue require about

3 volts. Note that this circuit runs of a very low voltage, so there is barely enough to light an LED.

The problem with switching the LED by putting a transistor in its current path is that this circuit has very little extra voltage available to waste while that current passes through the transistor. So, instead, from your description, it seems that the designer has chosen to turn the LED off by detouring its supply of current around the LED with the transistor. As long as the transistor can carry this current while dropping less than the voltage it takes to light the LED, it works. The disadvantage of this method of control is that the current through the LED supply resistor is higher when the LED is off (and the transistor is on) than when the LED is on (and the transistor is off).

• posted

It makes no sense without the resistors.

Graham

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Thanks for the responses. I don't have the schematic with me (I'm at work). I gather that it all has to do with the voltage and current equations for the particular circuit. When you say it makes no sense without the resistors, I gather that there is some threshold of current and/or voltage that makes the LED come on, even though it is connected to the battery. The problem with electronics learning kits is they don't fully explain why there is a resistor here, a capacitor in the corner there, just a bit about the main capacitor and such.

• posted

Was the circuit something like this (view using fixed-spaced font):

If so, then pushing the button to turn on the switch will turn off the transistor. When the transistor is off, the LED will come on. That's because the transistor, when on, bypasses the LED and doesn't allow current sufficient to light it to pass through it. When the switch S1 is not engaged and is off, then current in R1 supplies the needed base drive for Q1 to turn Q1 on. When the switch S1 is pressed and is on, then S1 forces Q1's base to the negative side of the battery and this causes Q1 to be off. When Q1 is off, you can just imagine removing Q1 entirely from the circuit to see what is happening. When Q1 is on, think of it as a switch that is turned on and tying the (+) side of the LED to its negative side and the negative terminal of the battery

-- forcing the LED off.

Jon

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Thank you very much for the diagram, which is what I had in mind.

Actually, what would be really helpful would be, in case a and b, do draw arrows in the circuit. If I am a positive charge, starting at the + side of the battery, and I want to go home to the negative side, what is my route when the switch is off? Do I go through the NPN, do I go through the LED and THEN to the NPN? What about in case b? For example:

,--------------------+----------, | |/c Q1 |\/ --- +---------| NPN | - | |>e | + | end | | --- | | | \ / LED | ^ o | --- | | / S1 | | | | / push | | | | o button | | | | | | | | | | | | \/ | '-----------+--------+----------'

• posted

I the case where the button is NOT PRESSED, you have:

I the case where the button is PRESSED, you have:

Hope that helps.

Jon

• posted

With S1 open: When a small current from the (+) of the battery and through R1 enters the base of the transistor, it flows out the emitter and back to the (-) side of the battery.

That small current enables the transistor to conduct a much bigger current through the transistor as follows: from the (+) side of the battery, through R2 to the collector, from the collector to the emitter and back to the battery.

The current through R2 causes a voltage drop (V=I*R) which makes the voltage at the top of the LED in your diagram lower than the voltage needed to make the LED conduct, so there is no current through the LED. Therefore, it does not glow.

When S1 is closed: No current can enter the base of the transistor, because the bottom of the resistor (and therefore the base) is connected to the (-) of the battery, so there is no (+) to go into the base.

Since there is no current from the base to the emitter (there cannot be, because there is no potential difference between them), there can be no current from the collector to the base. Therefore, the transistor cannot cause a voltage drop across R2. Since the transistor causes no voltage drop across R2, the bottom of it is (+) and current can flow from the battery (+) through R2, through the LED and back to the battery (-). There will be a voltage drop across R2, but it will be smaller than the voltage drop caused by the transistor because the transistor, when conducting, draws more current than the LED does. So the voltage at the top of the LED will be sufficient for current to flow through it, and it will glow.

Ed

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I came across this post and was curious why someone would want to illuminate an LED this way. Is there an inherent advantage to running the load in parallel with the collector and emitter, as opposed to in series? I've seen some (simple) audio amplifier circuits that do the same thing, too.

Andy

• posted

I read the original post and came up with the schematic by "reading between the lines" and accepting the OP's ability to write accurately about whether or not the switch was pressed when the LED was on or off. It seemed the simpler circuit that would not be inconsistent with the OPs rather vague writing about it.

One of the insights that helped me accept the OP's writing was the phrase, "electronics learning kit." I don't think someone would actually do this as a practical matter, but since it was in some learning kit I had no problem accepting the circuit idea that seemed to fit, as a possible explanation for what he wrote.

Jon

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That makes sense. His wording does seem to lend itself to the diagram you came up with. I guess the schematic seems viable from an educational standpoint, as you pointed out. Thanks!

• posted

Jonathan Kirwan wrote in news: snipped-for-privacy@4ax.com:

Jon,

Actually I did this as a "practical" matter. I thought this schematic looked a little familiar, so I went back looking at some of my circuits from the past. This was a method I used of pulsing a laser diode fed by a switchable constant current source.

Ken

.----. | | +10v +------+------+-|7805|-+--------------+---+---+ | | '----' | 18 | 33| 82| | 0.1 -+- | --- .-. .-. .-. .-. --- | --- 0.1 | | | | | | | | | | | | | | | | | | | +---+----+ '-' '-' '-' '-' | | | | 390 | | .--+ | | | | | o o /o | | | / | | | / | | | o | | | | | | . | | | +----------------+------+ - | | | | | | | | | | | |/ | | | +----------------------+-+| TIP29 V | | | |> - | | 4.7K 4.7K | | | Laser Diode -- --- ___ ___ |/ | | Signal +--------+-|___|+--+-+-|___|--| | | | |> | | - | | | 1N4148 ^ 2N3904 | | | | | | | COM +------------------+------------+----+------+ Pulsed Constant Current Laser Diode created by Andy´s ASCII-Circuit v1.24.140803 Beta

• posted

Makes sense. I actually have thought about doing just about that, as well. Just that it didn't come to mind when thinking about a push button and a simple LED. In general, though, you are right to point out that there are some uses for "current steering."

Thanks, Jon

• posted

Maybe for an Off indicator? Something like this comes to mind:

+12 -----+-----+ | | [D1] [Rly] | | +-----+--{{------+ | | /c [1K] ---[R]---| | \\e [LED] Off | | Gnd ---------+--{{-------+

You might have an existing circuit like the relay circuit above, where you could add a resistor and LED to indicate that the relay is not energized.

Ed

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I see how the LED will be off when the transistor is on (and Vce is lower than the voltage required by the LED)... but when the transistor is off, why won't the current passing through the LED cause the relay coil to become energized?

Andy

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If you decided to use that approach, you would select the resistor that is in series with the LED and relay coil so that the current is too low for the relay coil to hold the relay energized.

For example, say you have an automotive relay with a 90 ohm coil that is rated to hold until the voltage across it drops to < 10% of nominal - that is, less than 1.2 volts. So pick 1 volt as a starting point, and figure the current that will produce 1 volt across 90 ohms: 1/90 = ~11.11 mA. Now choose a resistor that will limit the current to no more than 11.11 mA when placed in series with the coil and the LED. The LED drops the voltage by ~1.8 volts, leaving

10.2 volts. The total resistance needed is found by R = E/I, so 10.2/.001111 = ~918. A standard 1k is a near value, and reduces the current a bit more, to 10.2/1090 = ~9.36 mA, plenty to light the LED, but not enough to hold the relay energized.

Ed

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Yup. I've used it in constant current NiCd chargers. When the voltage reaches the design spec, you steal current from the constant source and steer it through a red LED and a parallel resistance to reduce the rate to C/50 (or whatever). That way, the NiCds can be left on the charger for extended periods with no damage and no self-discharge, and the red LED tells you the charging has stopped.

Ed

• posted

dear i think this circuit is like that

1 powersupply -ve one terminal to base and another terminal is connected to led 2emmiter connected to led' collector is with led's ans of ur Q , why

ur ckt is not complete without npn trans. if u connect the led terminal directly to power supply's terminal which is also connected to base of npn simply logic is that base is work as a switch if it doesnt has the voltage then the emmiter will not connect to collector and ur npn transistor will work as open switch and power will not Xfer to led and led will not be on

• posted

transistor.

Thank you for your responses, though some are a bit over my head. I've been particularly interested in the circuit below (a transistorized RS Flip-Flop):

+3V o----------------o | | | | 1K 1K | | ------ | 4.7K -----|------ | c ---\\ / /--- c | | b |----/ | b | | e e ~ RED | | \\ | | | | | R | | | | | | | ----------- | --- --- | --- | - - | - | | -------4.7K ---- -- | \\ S | --- _

R and S are pushbuttons; when the circuit starts, the red led is off. When I push S, the led goes on, but when I release S, the led stays on! I don't know how to work this out with the V=IR and closed-loop equations and such. When I push R (reset), and release it, the LED is off again until I push S. If someone has time to show how each state works and why it keeps ("remembers") that state, I'd appreciate it.

Thanks again.

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