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  • LED flashers, Line powered LEDs, LED Traffic lights
    LEDs The circuit below illustrates powering a LED or two from the 120 volt AC line using a capacitor to drop the voltage and a small resistor to limit the inrush current Since the capacitor must pass current in both directions a small diode is connected in parallel with the LED to provide a path for the negative half cycle and also to limit the reverse voltage across the LED A second LED with the polarity reversed may be subsituted for the diode or a tri color LED could be used which would appear orange with alternating current The circuit is fairly efficient and draws only about a half watt from the line The resistor value 1K half watt was chosen to limit the worst case inrush current to about 150 mA which will drop to less than 30 mA in a millisecond as the capacitor charges This appears to be a safe value I have switched the circuit on and off many times without damage to the LED The 0 47 uF capacitor has a reactance of 5600 ohms at 60 cycles so the LED current is about 20 mA half wave or 10 mA average A larger capacitor will increase the current and a smaller one will reduce it The capacitor must be a non polarized type with a voltage rating of 200 volts or more The lower circuit is an example of obtaining a low regulated voltage from the AC line The zener diode serves as a regulator and also provides a path for the negative half cycle current when it conducts in the forward direction In this example the output voltage is about 5 volts and will provide over 30 milliamps with about 300 millivolts of ripple Use caution when operating any circuits connected directly to the AC line Menu Line Powered White LEDs The LED circuit below is an example of using 25 white LEDs in series connected to the 120VAC line It can be modified for more or less LEDs by adjusting the resistor value The exact resistance will depend on the particular LEDs used But working out the resistor value is a bit complicated since current will not continously flow through the resistor In operation the output of the bridge rectifier will be about 120 DC RMS or 170 volts peak If we use 25 white LEDs with a forward voltage of 3 volts each the total LED voltage will be 75 volts The peak resistor voltage will be 170 75 or 95 volts but the resistor voltage will not be continous since the input must rise above 75 before any current flows This dead time represents about 26 degrees of the 90 degree half wave rectified cycle asin 75 170 asin 44 26 degrees This means the resistor will conduct during 90 26 64 degrees or about 71 percent of the time Next we can work out the peak LED current to determine the resistor value If the LED current is 20mA

    Original URL path: http://bowdenshobbycircuits.info/page10.htm (2016-04-26)
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  • Fading Red Eyes
    895 0465 1 24 Transistor 2N3904 568 8253 1 1 22uF Capacitor 852 6516 1 07 Solderless Breadboard 237 0015 1 6 99 Red Light Emitting Diode LED 670 1224 2 0 50 Note The LED listed has a narrow viewing angle of 30 degrees and appears brightest when looking directly at it It s not a pure red color and a little on the orange side but should be brighter compared to other selections For a wider viewing angle at reduced intensity try part number 670 1257 which is viewable at 60 degrees and has a red diffused lens Construction details Layout of the solderless breadboard Refer to the drawing below the schematic diagram and note the solderless breadboard is arranged in rows labeled A J and columns numbered 1 to 65 Each group of 5 holes in the same column are the same connection so that holes A1 B1 C1 D1 and E1 are all connected together Likewise holes F1 G1 H1 I1 and J1 are all the same connection The outer rows along the length of the board are also connected together and are normally used for power supply connections However there is a break in the mid section of the outer rows so a short jumper wire connecting the mid section of the outer rows should be installed to connect the entire outer row together If you have a DMM use the low ohms range and probe the various holes to get familiar with the board layout Installing the components Orientate the LM1458 so the nook or punch mark on one edge is near column 30 and the opposite edge is near column 33 Install the LM1458 on the breadboard so the pins straddle the center section of the board and pin 1 of the IC is occupying hole E30 and pin 8 is in hole F30 The pins are numbered counter clockwise so pin 4 will be occupying F33 and pin 5 will be in E33 Possible connections for the LM1458 9 volt battery and a couple other parts is illustrated in the lower drawing of the solderless breadboard but it is not complete with all parts Refer to the schematic diagram and install the various other components so they connect to the appropriate pins of the LM1458 Use whatever connection holes are convenient For example the 22uF capacitor connects between pins 1 and 2 of the IC which occupy holes F30 F31 so it could be placed in the holes H30 H31 or J30 J31 or I30 I31 But not all parts will conveniently fit so you may have to use a short jumper wire 22 preferred to connect parts from one side of the chip to the other The board I assembled was connected this way LM1458 F30 to F33 and E30 to E33 22uF capacitor H30 to H31 47K resistor I30 to I35 47K resistor C27 to C31 47K resistor F25 to Positive battery row 47K resistor J25 to Negative Battery row

    Original URL path: http://bowdenshobbycircuits.info/fade_led.htm (2016-04-26)
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  • Discrete LED Fading Circuit Using 5 Transistors
    using a 555 timer The circuit is basically the same as the Schmitt Trigger Oscillator in the misc section with the addition of two NPN transistors in a darlington arrangement to buffer the ramp output at the top of the capacitor The transistors can be almost any general purpose NPN PNP types such as 2N3904 2N3906 Ramp time up or down is approximately 3 4 RC so that a 100uF cap and 20K resistor yields about 1 5 seconds The LED peak current is about 20mA This is worked out from the peak ramp voltage of 6 volts minus the 600mV E B drop on each of the transistors Q1 and Q2 and also minus the LED voltage of 1 5 This leaves 6 0 6 0 6 1 5 3 3 across the resistor in series with the lower LED The resistor value will be R 3 3 0 02 165 ohms or 150 standard value Since the collector current of Q1 is very close to the emitter current the second LED can be placed in series with the collector of Q1 without any additional resistor Adjustments to points where the LEDs just extinquish and reach maximum brightness can

    Original URL path: http://bowdenshobbycircuits.info/desfade.htm (2016-04-26)
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  • Page 13
    R5 and the combination in series with R2 When the output is low the base voltage is set by R4 in parallel with R2 and the combination in series with R5 This assumes R3 is a small value compared to R2 The switching levels will be about 1 3 and 2 3 of the supply voltage if the three resistors are equal R2 R4 R5 There are many different combinations of resistor values that can be used R3 should low enough to pull the output signal down as far as needed when the circuit is connected to a load So if the load draws 1mA and the low voltage needed is 0 5 volts R3 would be 0 5 001 500 ohms 510 standard When the output is high Q3 will supply current to the load and also current through R3 If 10 mA is needed for the load and the supply voltage is 12 the transistor current will be 24 mA for R3 plus 10 mA to the load 34 mA total Assuming a minimum transistor gain of 20 the collector current for Q2 and base current for Q3 will be 34 20 1 7 mA If the switching levels are 1 3 and 2 3 of the supply 12 volts then the high level emitter voltage for Q1 and Q2 will be about 7 volts so the emitter resistor R1 will be 7 0 0017 3 9K standard A lower value 1 or 2K would also work and provide a little more base drive to Q3 than needed The remaining resistors R2 R4 R5 can be about 10 times the value of R1 or something around 39K The combination of the capacitor and the feedback resistor Rf determines the frequency If the switching levels are 1 3 and 2 3 of the supply the half cycle time interval will be about 0 693 Rf C which is similar to the 555 timer formula The unit I assembled uses a 56K and 0 1 uF cap for a positive time interval of about 3 5 mS An additional 22K resistor and diode were used in parallel with the 56K to reduce the negative time interval to about 1 mS In the diagram T1 represents the time at which the capacitor voltage has fallen to the lower trigger potential 4 volts at the base of Q2 and caused Q1 to switch off and Q2 and Q3 to switch on T2 represents the next event when the capacitor voltage has risen to 8 volts causing Q2 an Q3 to turn off and Q1 to conduct T3 represents the same condition as T1 where the cycle begins to repeat Now if you look close on a scope you will notice the duty cycle is not exactly 50 This is due to the small base current of Q1 which is supplied by the capacitor As the capacitor charges the E B of Q1 is reverse biased and the base does not draw any

    Original URL path: http://bowdenshobbycircuits.info/page13.htm (2016-04-26)
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  • IR Remote Control Key Test Circuit
    off to 3900 since I had one laying around The gain of the transistor was assumed to be around 100 so the base current will be about 10 microamps The voltage from collector to base is about 4 volts so the base resistor is about 4 00001 400K I rounded that off to 330K since I had one of those laying around The actual collector voltage measured 4 6 The second stage is a narrow band tuned stage using a parallel LC circuit tuned to 36KHz and operates at about 3mA The resonant frequency can be found with formula Fo 2 pi sqrt LC The actual resonant frequency of the LC parts 6 8mH and 2700pF works out to 37144 The capacitor could be a little higher at 2874pF to move the frequency to 36K but I had a 2700pF laying around and used that I did add another 200pF in parallel but it didn t have much effect Other values of inductance and capacitane could be used but the rule of thumb is for the reactance to be about 1K to 2K Inductive reactance is 2 pi F L or in this case 6 28 36000 0068 1537 ohms Most IR remotes operate at 36KHz where each bit is transmitted with maybe 30 cycles of the 36KHz carrier followed by some off time to separate the bits Using a tuned LC circuit allowes the receiver to reject most of the noise and only respond to the desired data within a narrow band The bandwidth is the range of frequencies that the LC circuit will respond while maintaining at least 70 of the max amplitude So if the tuned center frequency is 36KHz and the amplitude falls off to 70 at 34KHz and 38KHz the bandwidth will be 38 34 4KHz The quality factor or Q is the center frequency divided by the bandwidth or Q 36 4 9 There is a compromise between the Q of the circuit and the response time The LC circuit needs about the same number of cycles as the Q to charge to full amplitude so if the Q were 100 and the transmitter only sent 30 cycles the circuit would not work since it needs more than 100 cycles to charge up to full amplitude Since we only have about 30 cycles to work with the Q should be quite a bit less than 30 The unloaded Q of the LC circuit itself when not connected to anything can be found from the ratio of inductive reactance to the resistance of the inductor wire or Q XL R 1538 60 25 6 The actual Q will be less when the circuit is loaded and is fairly complicated to figure out It depends on the transistor parameters bias current and other things I found a Q calculator that works out the Q based on LC values inductor resistance and generator resistance The actual Q worked out to about 9 with the assembled circuit Using

    Original URL path: http://bowdenshobbycircuits.info/irtest.htm (2016-04-26)
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  • Window Comparator"
    10 degrees the total resistance falls 3750 ohms the current will be 12 66K 3750 193uA and the thermistor voltage will be 193u 33k 3750 5 65 volts This represents a voltage change of 6 5 65 350 millivolts for a 10 degree change The center resistor of the window voltage divider must then drop 350 millivolts Using 20K resistors on the top and bottom of the window voltage divider produces a current of 6 350 2 20K 291uA and the center resistor is 350 291u 1 2K When the temperature is in the center of the window range the voltage at pins 5 and 6 will be 1 2 the supply voltage or 6 volts in this case The voltage divider 20K 1 2K 20K produces a voltage of around 5 8 at pin 4 and 6 2 at pin 7 Since the voltage at pin 5 6 volts is more positive than the voltage at pin 4 5 8 volts the output at pin 2 will be a high level At the same time the voltage at pin 7 input is higher than pin 6 input causing pin 1 to also be a high level This condition produces a high 12 volt level at pin 10 input which produces a low level at pin 13 lighting the window LED indicating the temperature is in the window range As the thermistor voltage moves above the upper 6 2 limit pin 1 will switch low extinguishing the window LED and illuminating the Low Temp LED Similar action happens as the thermistor voltage moves below the lower 5 8 limit causing pin 2 to switch low Over Temp LED while the other two LEDs remain off Light Level Indicator Using a Window Comparator The second example below uses a LDR light

    Original URL path: http://bowdenshobbycircuits.info/window.htm (2016-04-26)
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  • Quartz Crystal Controller for Mechanical Wall Pendulum Clock"
    is close to 53 4 complete swings per minute Can t figure out why that particular period was used I have seen similar movements on ebay The clock pendulum was made using a 10 inch 3 16 wooden dowel with a strong magnet attached to the bottom and a small weight near the top to adjust the period close to 53 beats per minute The circuit board and electromagnet to drive the pendulum are located on a small shelf not shown and positioned so the pendulum magnet swings close to the stationary electromagnet and receives a small pulse on each swing to sustain oscillation The pulse duration is about 5 of the pendulum period and a LED is used to indicate the pulse output The clock starts fairly easily by releasing the pendulum near the magnet when the LED flash is observed Quartz Crystal Synchronizing Circuit The synchronizing circuit that produces a short magnetic pulse to keep the pendulum in near perfect time was made using a crystal oscillator and binary counters to generate a 60mS pulse at the required rate of 53 4241 PPM I didn t know the exact rate but it appeared to be close to 53 5 cycles per minute by just adjusting the pendulum length and monitoring the error over a several hour period The oscillator circuit uses an old 20KHz quartz crystal but other low frequency crystals can be used A standard watch crystal of 32 768 KHz is probably the best and easy to obtain The idea is allow the counters to count to the desired number and then reset the counters generate the desired pulse and repeat the cycle In this case I needed a time of 60 53 4241 or 1 1231 seconds At a frequency of 20KHz this is about

    Original URL path: http://bowdenshobbycircuits.info/pendulum.htm (2016-04-26)
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  • TRF Broadcast Receiver"
    better if the loop is rotated 45 degrees so the frame looks like a cross instead of an X That way the top and bottom sides will not be parallel to the ground and the antenna will have more height The antenna signal is buffered by the first emitter follower stage which presents about 150K input impedance The buffer stage avoids losing much voltage from the antenna when connected to the circuit The buffered RF voltage at the emitter of the transistor is rectified by the diode and the RF component removed by the capacitor at the base of the second transistor This leaves only the audio signal at the base with about 5X higher amplitude at the collector The 3 remaining transistors form an audio amp to drive the 8 ohm speaker The transistors used are 2N3906 PNP and 2N4123 NPN however 2N3904 should work as well The 100K variable resistor shown is only used to reduce the volume of very strong stations such as KFI in my case The speaker is a 4 inch model with a heavy 4 inch magnet inside a 6 inch box Seems to be more efficient than speakers with smaller magnets Headphones work better and I can hear 8 stations in the Los Angeles area at low volume The complete circuit draws about 10mA from a 9 volt battery Loop Antennas A loop antenna can greatly improve medium wave reception Loop antennas are directional and receive signals along the plane of the windings The directional quality improves signal to noise ratio of the desired signal while rejecting signals perpendicular to the plane of the windings Larger loops are better than smaller ones but good results can be obtained from moderate sizes of one or two feet on a side The shape doesn t

    Original URL path: http://bowdenshobbycircuits.info/trf.htm (2016-04-26)
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