Transistor Timer With Darlington Pair.

Discussion in 'Homework Help' started by Babar Firasat, Apr 20, 2013.

1. Babar Firasat Thread Starter New Member

Feb 24, 2013
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The purpose of this is project to adjust the timing of when a bulb or a buzzer should go off. This is a dual transistor timer with Darlington pair circuit. In the picture the SW1 (switch one) is the trigger. That switch is already in a state where the circuit is active, (that is what is shown in the picture). can someone explain to me how this circuit works. this is my first as electronics year in as level and i know not much of electronics. thankyou.

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2. Jony130 AAC Fanatic!

Feb 17, 2009
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To understand this circuit, you need to understand RC time constant circuit.
http://www.electronics-tutorials.ws/rc/rc_1.html
Next you have a RS flip-flop.
http://www.play-hookey.com/digital/sequential/rs_nand_latch.html
And the BJT Darlington pair.

When the switch SW1 is in the trigger position IC1b NAND gate input see logic 0. A 0 level will force that output of a IC1b to a logic 1. This well turn ON the BJT Darlington pair. And the bulb is ON (lights).
But at the same time C1 is charging via VR1 resistor. And when voltage across C1 reach IC1B gate trigger voltage (NAND gate will see a logic 1 at the input). IC1B output force a logic 0. And the Darlington pair and the bulb is switch to off state.

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3. ScottWang Moderator

Aug 23, 2012
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The attached circuit diagram that you shown are two circuits, the 555 circuit is a normal oscillator, you can find it easily from this site or other site by google.

7805 is a 5V fixed voltage regulator.

The function of the IC1a of CD4011 is a pull high for the unused pin of the IC1b of CD4011, it is to replace the resistor pull high function, it means that normally we using a resistor to pull the unused input pin to high voltage for a AND gate or NAND Gate, so the function of IC1b is a inverter at here, it's function seems like a Seesaw, when the input is low and the output will be high, when the input is high and the output will be Low.

When the middle pin of SW1 switch to Ground, the C1 will be discharged and connected to Ground, the input of IC1b will be low, and the output of IC1b will be high, it will also active the Q1 and Q2 through R1, and the BZ1 will be working.

When the middle pin of SW1 switch to VR1, the C1 will be charge through VR1, when the C1 charge to about 3V, then the output of IC1b will turn to Low, and then the Q1 and Q2 will be turn off, the BZ1 will be off too.

When you turn the Sw1 in different position then the BZ1 will be working and off, the things will be goes on.

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4. Jony130 AAC Fanatic!

Feb 17, 2009
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Logic gate don't like slow rising edge of the input signal (voltage across C1).
So they add IC1a to the circuit to add positive feedback to the circuit.

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5. ScottWang Moderator

Aug 23, 2012
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Whatever a AND gate or NAND Gate, when any of one input pin is low, then the output is fixed on a stable state forever, so any other input pin can't generate any affect for the output.

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6. Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, of course you are right about IC1a. IC1a does not do anything in this circuit. So this circuit will not working properly.

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7. ScottWang Moderator

Aug 23, 2012
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No, it's working fine.
You can see what I described from the third paragraphs on #3.

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8. Jony130 AAC Fanatic!

Feb 17, 2009
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I full agree with your description. But what I don't like about this circuit is that we connect a slow rising edge to gate input. And this can cause a potential problem. Slow rising edge + noise on input can cause false gate triggers.
In this case it's better a Schmitt Trigger gate such as the 4093.

Last edited: Apr 20, 2013
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9. ScottWang Moderator

Aug 23, 2012
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Yes, you are right.

Because the function is a inverter, so I will using 74HC14 with +5V power, if the power voltage is 12V, then I will using MC14584.

You can see the schmitt trigger that I used with +5V power as below:

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