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Lightning Detector Michael Bloem December 5, 2002 Engr 311 The Circuit 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •The purpose of the circuit is to set off an LED when there is a lightning strike in the area (even if too far away to be seen) •This will indicate the proximity and intensity of a storm Circuit Operation: Input Stage 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •A lightning strike is approximated by a small voltage pulse •The circled part of the circuit is resonant at 300 kHz, so that is the frequency passed through the circuit Circuit Operation: Input Stage 40nA 20nA SEL>> 0A I(Q1:b) 2.0mV 1.0mV 0V 0.1MHz 0.2MHz V(V2:+) 0.4MHz 0.6MHz 0.8MHz 1.0MHz 1.2MHz Frequency •This is the Fourier transform of the pulse current input and the input to the base of the first transistor •Note that the pulse has components at all frequencies •Also note that the input stage of the circuit picks out the 300 kHz part and sends it to the circuit Circuit Operation: Input Stage 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 C3 0 Q2N4403 Q4 D1N4148 R1 1000k C1 .001u R3 1000k 2 .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 120NQ045 D12 1 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •The first transistor provides amplification of the signal before it is sent to the flasher part of the circuit Circuit Operation: Flasher 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •The second transistor is biased by adjustment of the variable resistor so that it doesn’t allow current to flow until a radio burst pulls the base down Circuit Operation: Flasher 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •Once the second transistor’s base is pulled down, it turns full on and current flows as shown by the red arrows •This sends a relatively strong current to the third transistor •This current flows until the capacitor discharges to ground •Then, the current from the diode flows to the capacitor to charge it again to be ready for the next pulse (green arrows) Circuit Operation: Flasher 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •The third transistor should amplify the current and send it on to the LED driver Circuit Operation: LED Driver 0 V2 TD = 20u TF = 0 PW = 2.5u PER = 100000000 V1 = 0 TR = 0 V2 = .01 200K SET = .00001 33 4.7k R10 10k Q6 D8 1 2 120NQ045 0 Q2N4403 D12 C3 Q4 1 D1N4148 R1 1000k C1 .001u R3 1000k .001u Q2N4401 C5 1u Q2N4401 V1 3Vdc C9 10u 1 L2 C2 680p Q3 Q2N4403 R4 3.9k Q1 2 390uH R11 R2 10k R8 .000001 L1 10mH R6 R7 R9 1000k 2 0 •The LED driver sends a current pulse to the LED, lighting it up. Circuit Simulation Output 2.0nA 0A SEL>> -2.0nA I(D8:K) 10mV 5mV 0V 0s 5us 10us 15us 20us 25us 30us 35us 40us 45us 50us V(V2:+) Time •To simulate a lightning strike in Cadence, I just used a small (.01 V) and short (2.5us) voltage pulse. •Why the output is good: it does create a current pulse that starts when the lightning strike occurs and then goes away •Why the output is bad: •Sinusoidal response- the capacitor should discharge fast enough so that it is just a pulse •Too small of a current to set off an LED Physical Implementation •I have found all the parts for this circuit and put them together on a breadboard •Testing •To approximate a lightning strike •Van de Graaf machine •Square wave from function generator •Result of testing •Essentially the same results as from Cadence- the current that ends up at the LED is too small Problems Opportunities for Improvement •Get enough current to light up the LED •Change the biasing of the amplifying transistors (2nd and 3rd transistors) so they actually amplify •Add my own amplification component at the output •Opamp or transistor •Get the current to be a single pulse rather than a sinusoid •Play with the capacitor value •Not as important- still get an LED flash with the sinusoid
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