Hello, I am trying to make the Darlington pair circuit found on http://www.kpsec.freeuk.com/trancirc.htm (The one under the dotted green box)
I think the problem is simply that I am reading the schematic wrong. I have the +9v connected to the 470 and 100k resistors. Then the 100k connected to B. Then E to B. Then I cannot figure out what to do. I dont understand why the LED goes from +9v through the resistor and then after the +9v comes through the two transistors it is connected to the cathode because the voltage is still positive, right? Then what is the 0v for at the bottom? To me it looks like both positive and negative flow through the transistors. Could you clear this up for me? Thanks a lot!

 

The Circuit is basically correct.
The Zero volts is the negative of the battery.
The 470 ohm resistor is a Current limit for the LED.

The 100K from the Bast to the 9 Volts is just supplying a Temporary signal to the transistor to light the LED.

I would also put a resistor (somewhere between 100K to 1M) between the Input Base and the Zero volts. To pull the Base Low with no Signal.

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So both the positive and negative go into the transistor? The +9 into the base of transistor 1, and the negative into the emitter of transistor 2? Thanks.

 

No. The circuit doesn't have a negative 9V battery. It has only a single battery that supplies +9V and 0V. You need two 9V batteries to make + and - 9V.

No. If you connected +9V to the base of transistor 1 it would blow up.
The +9V connects to one of the touch contacts and the base connects to a 100k ohm current-limiting resistor. When you touch both contacts with something conductive then a small current flows through the 100k resistor into transistor 1, turning it on and it turns on transistor 2.

Attached Images

darlington.png (10.2 KB, 573 views)

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Uncle $crooge

 

If you want to experiment, just make sure you don't go lower than about 300 ohms for any resistor.

The maximum current the transistor can handle is very important. Audioguru said that "connecting 9V to the base of the NPN will blow up the transistor" is correct, because the resistance between +9V and the NPN's base is well under 1 ohm, and when you calculate the current, you get 9/1 = well over one amp. and because one amp exceeds most transistors, it blows up!

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aah, thanks it is starting to make sense. I'm still lost on the 0v. I think I'm making it harder than it needs to be! There are two connections on the 9v battery. One is +9v, and one is ground (negative) correct? Lets say you used a battery pack with a red positive and black negative wires. Which would connect to where on the circuit. I am lost on the LED because you need a positive and negative to make it light, then how is it connected to the circuit but there is no negative in the circuit? Thanks, I hope I didnt confuse you.

 

Hi Mike,
I showed how the battery connects to the circuit. One battery terminal is +9V and the other battery terminal is marked "-", but it is actually 0V.

The anode of the LED is connected to +9V through the 470 ohm current-limiting resistor. The cathode of the LED is connected to the collectors of the darlington transistor, which is connected to 0V. When the transistor turns on then it connects the cathode of the LED to 0V and battery current flows in the LED.

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Uncle $crooge

 

Batteries have a positive plate and a negative plate. If you connect say the negative plate to gnd, then the potential of that plate is 0V since gnd is assumed to be at zero potential. So the voltage (potential) of the positive plate, measured with respect to (wrt) gnd is +9V.

On the other hand, if you connect the positive plate to gnd, then the potential of that plate is 0V. So the voltage of the negative plate, measured with respect to (wrt) gnd is -9V. It all depends upon your chosen reference point.

Look at it this way. If you connect the black probe of your multimeter to gnd, and measure the other plate of the battery, it will read +9V in the first case and -9V in the second. In both cases, if you connect the red probe to gnd (leaving the black one connected to gnd), it will read 0V.

The LED needs a difference of potential (ie. a voltage) of sufficient amplitude to make a current flow through it in order to make it glow.

I hope this helps. If not ask more questions so we can clarify it further for you.

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Len

 

ok Thanks for the help! So 0v goes into the Emitter of transistor 2 and that is the only thing connected to the negative on the battery. Then the positive goes through the 470 ohm resistor to the anode of the LED then to the 'touch' switch? Thanks a lot, it is comming together now. Does anyone have any other easy transistor projects I can try out? Thanks!

 

According to audioguru's circuit, yes.

yes

I do. In fact, it is about the most simplest NPN transistor experiment yet. Here is what you do:

1. Connect the NPN's emitter to the anode of an LED.
2. Connect the cathode of the same LED to one end of a 1K resistor.
3. Connect the other end of the 1K to ground (which is 0V).
4. Connect the NPN's collector to +5V.

and try one of the following steps, but do NOT do them both at the same time.

5. Connect the NPN's base to +5V.
6. Connect the NPN's base to +0V.

you will notice that you have made yourself a simple transistor switch.
Also, you will realize that an NPN requires +ve voltage (I think at least 0.65) at it's base to turn on.

AND, you will see why an NPN transistor is two diodes with the anodes connected to each other.

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Cool, I built the circuit. When the only connections I have are Emitter to the anode of the LED, cathode to 1k to 0v, and +5v to the collector the LED lights. I thought for the power to flow through the transistor it needed +5v on both the collector and base. When I put a 270 ohm resistor on the base instead of +5v the LED dims. Is that how it is supposed to work? I appreciate all of the help!

 

The LED should glow when you connect the +5V to the base. It should not glow if you connect the base to 0V.

If you insert a resistor between +5V and B then the LED will glow less depending on the resistor value.

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Len

 

In an NPN, the base is in 99% of the cases, the controlling factor.

If you meant that you connected the 1K and a 270 ohm in a row from 5V to the base, then the LED should dim, because the resistance is now 1270 ohms instead of 1000 ohms. If you want a brighter LED, lower the resistance. Don't go too low, or the transistor can burn out.

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