Lightning Detector

Michael Bloem

December 5, 2002

Engr 311

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

The Circuit

•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

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

Circuit Operation: Input Stage

•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

Frequency

0.2MHz 0.4MHz 0.6MHz 0.8MHz 1.0MHz 1.2MHz0.1MHzV(V2:+)

0V

1.0mV

2.0mVI(Q1:b)

0A

20nA

40nA

SEL>>

•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

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

Circuit Operation: Input Stage

•The first transistor provides amplification of the signal before it is sent to the flasher part of the circuit

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

Circuit Operation: Flasher

•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

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

Circuit Operation: Flasher

•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)

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

Circuit Operation: Flasher

•The third transistor should amplify the current and send it on to the LED driver

Circuit Operation: LED Driver

C51u

C910u

D8

120NQ045

1 2

D12

D1N4148

Q1

Q2N4401

L110mH

1

2

L2

390uH

1

2

C1

.001u

C2680p

C3

.001u

R11000k

R210k

R31000k

R4 3.9k

R6

4.7k

Q3

Q2N4401

Q4

Q2N4403

V13Vdc

R7

200KSET = .00001

V2TD = 20uTF = 0PW = 2.5uPER = 100000000V1 = 0TR = 0V2 = .01

R8.000001

R91000k

Q6

Q2N4403

R10

10k

R11

33

0

0

0

•The LED driver sends a current pulse to the LED, lighting it up.

Circuit Simulation Output

Time

0s 5us 10us 15us 20us 25us 30us 35us 40us 45us 50usV(V2:+)

0V

5mV

10mVI(D8:K)

-2.0nA

0A

2.0nA

SEL>>

•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