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Brushless Fans and Solar Panels Answered November 2014

I have three 12 VDC brushless fans and am considering running them from a 12V 30W solar panel. Two of the fans are rated at 5.4W and one at 7.6W. Unfortunately, the brushless motors can tolerate a maximum voltage of 13.8V and the solar panel has an open circuit voltage of over 17V. I am afraid I could fry the electronics in the fans with this solar panel but I can't find a 12V solar panel that outputs no more than 12V. I am sure this is not the first time this problem has occurred. What options do I have?

#11142
Terry Tanber
San Diego, CA



Answers

My suggestion would be to put an LM7812ACT linear three-pin regulator (about $0.69 each, plus shipping, from Digi-Key.com) between the solar panel and each fan. Be SURE to read the spec sheet carefully and use bypass capacitors on BOTH the input and output. (Guess how I know to warn about that!) You'll also need some sort of heat sink on each, though some scrap of aluminum should suffice -- my guess would be four square inches or more should do the trick. (If necessary, cut up an empty aluminum can, though fold the edges over to avoid sharp edge hazards.)


Using a "low dropout" regulator would mean more efficiency, but at a noticeably higher cost.


P.S.  I notice that you're in San Diego.  You can probably get the parts at Fry's Electronics, though for a little bit higher cost. The spec sheet can still be had at Digi-Key.com. (Wish Fry's would provide scooters for their mobility-limited customers!)

Clark Jones
Gilbert, AZ

It’s not likely that the voltage from the solar collector, under load, will be enough to damage the fans. However, to be safe, you could put a 13 volt, 15 watt zener across the collector output. The fans will be drawing 1.53 amps while the collector is trying (best case) to put out 2.5 amps, so the zener will have to sink 0.97 amps. I could not find such a zener in the Mouser catalog, so the alternative is to use a power transistor amplifier, see diagram (the part numbers are Mouser).


The zener current is just the base current, about 33 mA. It is DO-41 (1 watt) and costs 22 cents. The transistor sinks 0.97 amps and dissipates 13 watts.  It costs 99 cents. A heat sink will be needed; Mouser part number 588-FA-T220-51E is suitable and costs $1.56. The heat sink is mounted vertical. It has two 0.091 solderable posts, diameter spaced one inch.

Russell Kincaid
Milford, NH

A good solution to this issue is to use a switching regulator. It has very high efficiency, close to 90%, and will take the 20+ VDC from your panel and even increase the current output some!! Go to "sul-tech" on YouTube, and you will find the solar optimizer schematic....hope it helps.

 

 

Peter Sul
Hendersonville, NC

The 30W rating on your solar panel is very optimistic and requires a LOT of direct sunlight to reach that rating. You might need more than one panel to do what you want to do. I think you will find that you have the opposite problem of what you are concerned about. Solar panels have a very large output impedance, which means that the voltage will drop to a low value when you connect a load like an electric motor that has a low input impedance.


Also note that electric motors are rated at their RUNNING power requirements and they need much more current to START. You will need a buck-boost voltage regulator to do what you need to do. It raises the voltage when the light is dim and lowers the voltage when the light is bright.


Since your motors will require more than one amp you will need to find a regulator that is capable of supplying at least 2 amps and I recommend one rated at 3 amps or more.

Cliff Harris
Anaheim, CA

Your solar panels only show that high of voltage when they are unloaded. Once you put the fans on them, the voltage will adjust to their rated voltage, providing you have enough sun light to drive them properly.


The other good news is that the fans will not burn up when more voltage is applied to them. They will run faster and wear out sooner, but to fry them, you would need much more than a single solar panel can produce.


The solar panel you have says 30 watts, but what they actually mean, is that under ideal conditions, it can produce 30 watts. Two panels would assure that the fan would run in less than ideal sun light.

Joseph Massimino
Jensen Beach, FL

I would use a simple LM7812 voltage regulator for each fan. The 7812 can take up to 18V in and as low as 5V and has a current rating of 1 amp. Based on your specs, you will only be drawing a little over half an amp on the 7.6 W fan and a little under a half amp on the smaller fans.  If you find that they get hot, you can easily add a heat sink.


I will also mention that I did something similar, running a boom box radio at the beach, and found that unless the panel is pointed at the sun constantly, the output will drift when fair weather clouds cross overhead. So, I connected 3 1000000 20V caps in series, which slowed down the drifting effect caused by clouds.


Another option is to use a solar charger controller like the one I picked up at harbor freight tools. You would hook up the solar panel to the controller, then the controller to the battery (car/marine), and run the fans off the battery. I would still use the LM7812 voltage regulator just to protect the fans.

Dave Litle
Lynn, MA


D Cell or Gell Cell Adapter for Nikon Cameras Answered November 2014

I have the following three Nikon cameras: D3200, D5100, and D5300 that I want to use in my black bear wildlife research work. There are no electrical outlets to recharge the small lithium-ion batteries that come with the cameras in the desolate areas that I frequent. A car inverter to a regular lithium charger is out of the question because four-wheel drives can't get back to the remote areas that are accessible only by horse and by
packing on foot.


Furthermore, the cameras need to be in the "on" mode for several days at a time and left unattended in cave areas, where continually re-entering the caves would not be particularly safe were I to repeatedly reinstall the little batteries. There is some flash power being consumed as well with every exposure, so the batteries drain fairly rapidly.


I would like to run a bipolar cable from a couple of parallel connected 12 volt/18 amp-hour batteries into a drilled hole in the camera battery snap cover in the camera base. That probably will probably require cutting open and possibly destroying a battery pack and removing the lithium-ion cells so as to connect the voltage dropped wires to the battery terminals inside the battery pack which, in turn, will make contact with the camera's internal fixed contacts.


Can you please assist me with a "camera safe" regulated circuit to drop the 12 volt gell cells to a constant regulated output level for a Nikon EN-EL14 lithium-ion battery which is rated at 7.4 volts at 1,050 mAh (7.8 Wh)? The Nikon MH-24 battery charger that comes with the cameras has a charger output of 8.4 volts at 0.9 amps and charges each EN-EL14 in about 2.5 hours. I would greatly appreciate your help and advice with a voltage dropping circuit and battery disassembly details, or if disassembly is not required, how to proceed with the hook-up.

#11141
John Graff
via email



Answers

I want to offer some facts that may help you form your own solution.


  1 - 7.4 volts happens to be 2X of 3.7 volts, and 3.7 volts happens to be the exact rating of most all cell phone batteries. Cell phone batteries can be bought very cheaply via the internet.

  2 - The next fact is that the battery door on Nikon cameras can be removed. Open it to approx 45 degrees and gently twist it out.

  3 - The next fact I would share is that a battery grip is a way to attach, usually two batteries, to the bottom of your camera. Some of the grips come with interval timers for timed exposures, they are all made in China, and as such, eBay is full of second party battery grips for many popular cameras. The grip might be the better way to modify your cameras without damaging them.


If you were to use cell phone batteries to make your own battery pack, your next challenge would be to build a charger for them, since charging them one by one would be labor intensive. Making a battery pack of the cell batteries would require them to all be the same rated capacity. Connect them in series, in pairs, to bring them up to 7.4 volts and parallel as many as you like to increase capacity to meet your requirements.


Lastly, NEVER attempt to open a battery, especially a Lithium battery. If you were to expose the Lithium to air, it most likely will explode and seriously injure you or result in your death.

Joseph Massimino
Jensen Beach, FL


Chinese Capacitor Problem Answered October 2014

I have a Viewsonic model  VG2230WM LCD monitor that has quit. I read that these monitors have the "Chinese capacitor problem" but I can't figure out what exactly that means and what to do to fix it. Do I just replace all the caps on the PCB?

#10144
Jouko Koskela
Memphis, TN



Answers

Often a monitor goes dark because the back light power supply fails. Look for the power module which has the pink and blue wire back light voltage connectors. Usually there are two or four connectors for the backlight power. The capacitors on this module often fail due to heat over a period of several years, and they sometimes have bulging tops making them easy to spot. Replace all of the electrolytic capacitors on the power module, there may be five or six of them. In many cases this will restore operation of the monitor.

Bill Seabrook WETA-TV
via email

I have repaired several flat screen monitors, one of them was the VG2230WM. I was able to find the power supply on line and replace it after getting it open. Another one was fixed by buying the capacitor kit on line and replacing each of them. The hardest part of any flat screen repair is getting them open without damage.

Good Luck.

Bob Smith
Prescott, AZ


Carbon or Metal? Answered October 2014

Is there a rule of thumb for when it is better to use carbon film resistors over metal film resistors?

#10143
Brayden Lawlor
Norfolk, VA



Answers

Generally, if you require greater precision, better stability and/or less noise, you go with metal film resistors. In the olden days, this was an expensive option, but these days, precision metal film resistors have plummeted in cost, at least for the Asian imports. So use metal film if you're working with test equipment attenuators, voltage dividers and analog timing circuits.


Metal Film Is Better
  If you need a circuit to remain within operational specs for a long time without periodic adjustment.
  If you’re designing a low-noise audio preamp.
  If you need precise timing for your 555 circuit.


Some notes here.
  1 - The ancient trick of beginning with a carbon composition resistor of a lower value than needed and filing a notch in it until the value was right on, gave you a good, tight resistor for a few minutes. But even if that notch is sealed with clear fingernail polish, carbon resistors still have a lousy temperature coefficient. The resistor will change in value with any changes in temperature far beyond what you'd want with a precision part. It isn't the same as using a precision metal film resistor, no matter what that old timer tells you.
  2 - The second note concerns power dissipation. The old carbon comp resistors could take short spikes in power and survive even with short bursts of ten times the power rating, because the mass of that slug of a resistance element could absorb and dissipate that spike. Carbon film and metal film resistors can handle their specified power level, but can't handle spikes well, for they don't have that same mass in the resistance element. They can dissipate the average rated power, but not absorb the heat of a big power spike. So, that same spike in power can cause a carbon or metal film resistor to literally burn out.


Even with the problem of film over composition described above, carbon film resistors do seem to have tighter and more stable values than the comps. They still suffer from bad temperature coefficients and noise, but the typical new 5% carbon film resistor will often be well-within 2% of its marked value.

Dean Huster
Harviell, MO

As a "rule of thumb" I recommend: Choose carbon film resistors when low cost (at high quantity) is the main consideration. Use metal film resistors when low noise, low temperature coefficient, and/or greater precision are important.


All types of resistors generate thermal noise, also called "Johnson noise," which increases as resistance and temperature increase. This comes directly from the laws of physics which, as Engineer Scotty famously said on Star Trek, "you canna change." However, current flowing through a resistor causes additional noise, including "shot noise," and this is greater in carbon film than in metal film types. That is why the latter are preferred for critical analog circuitry, such as audio preamplifiers. Resistor noise hardly matters in digital applications.


For more information about noise, see Joe Geller's excellent resistor noise measurement project (JCan) in the July 2007 Nuts and Volts. After building the JCan kit, I could easily measure more noise in carbon composition types compared to metal film resistors of equal value when current was applied. There was no difference with zero current — just the thermal noise "background." The noise difference between carbon film and metal film resistors under power was smaller, but detectable.


The temperature coefficient tells you how much a resistor's value can change with temperature, in units of parts per million (PPM) per oC or oK. Common axial lead (through hole) ¼-watt metal film resistors are rated 50 or 100 PPM/ oC, but their carbon film cousins are 350 to 700 PPM/ oC. Sensitivity to temperature is a consideration in many sensor and measurement applications, but it does not matter in digital circuits.


Assuming the resistor offerings at Mouser Electronics (www.mouser.com) are typical of the industry, the vast majority of ¼-watt axial units are available only at 1-percent tolerance for metal film, and 5 percent for carbon film. Even if I wanted one, I could not buy a 5% metal film or a 1% carbon film resistor. Therefore, choice of type is related to precision, for practical purposes. The cost difference is negligible for small quantities; you are mainly paying for the labor to count them out and bag them.

Clark Huckaby
via email

I'm not sure about a rule of thumb but I would not use metal film resistors in an RF circuit. It would be my luck that I create a resonate circuit and have all kinds of strange oscillators going. Good Luck

Bob Smith
Prescott, AZ


555 for PWM Answered October 2014

Does anyone have a simple circuit/schematic to use a 555 timer to dim a 120 VAC incandescent bulb? In the little bit of research I've done, it seems like using the 555 for PWM should do the trick, but I don't understand "zero crossing" and why it’s important. Pointers welcome!

#10141
Chuck Pearson
Kansas City, MO



Answers

I have a 555 PWM circuit diagram on one of my YouTube videos here: [url=http://youtu.be/KrcqpGZQd9U?list=PLUsDZdokq6wxDP7L3sMoYFTA5WDoAtBu4]http://youtu.be/KrcqpGZQd9U?list=PLUsDZdokq6wxDP7L3sMoYFTA5WDoAtBu4[/url] The diagram is at about 2:18 in the video. I hope it helps for starters...

 

 

Freddy Cordero
via email


Noise Cancelling Answered September 2014

Is there such a thing as a noise cancelling technology that would allow me to play my guitar without disturbing the neighbors?


I get complaints about noise from my apartment when I play. I just can't use headphones and I've tried putting blankets on the wall for soundproofing, but I still get complaints. How about an electronic solution?

#9143
Brian Tate
Madison, WI



Answers

There is no practical noise cancelling technology on that scale. Building an anechoic chamber would be very expensive. Playing guitar in a vacuum would not allow you to hear yourself (since sound does not travel in a vacumn), and would suffocate you. That leaves two possible options:


Relax your ban on headphones and use a product such as the $40 Amplug [The headphone guitar amp that lets you enjoy serious guitar sound, fast.]  See www.voxamps.com/amplug or possibly the low tech solution of playing under a wool blanket.


More than 40 years ago we wanted to record short skits for radio brodcast in our college dorm room. Our input was not as loud as an accoustic guitar, and we were close mic-ed for voice and sound effects. We had such good sound isolation, that it effectively blocked the external noise of nearby stereos, running feet, shouting, and anti Vietnam War riots outside the building. Many people wondered where we found such a quiet place to record! This probably will not work for amplified electric guitar. You could use a self standing tent, with covering heavy bankets down to the floor. The level ouside the tent will be dramatically attenuated.

Barry Cole
Lacey, WA


UV LED VS UV Bulb Answered September 2014

I'm been working a lot with high power LEDs now that they're so affordable. I'm especially fond of the UV LEDs for visual effects at night, but I've heard that some UV LEDs can cause blindness, and that I should use a UV bulb to create the effects. Is this true? Can someone explain the difference?

#9142
Nick Robbins
Ft Wayne, IN



Answers

The safety difference between a UV light bulb and a UV LED stems from the difference in radiance between the two sources. Radiance is basically the density of light in the source.


The surface area of an incandescent-size lamp is approximately 100,000 sq-mm. The source area of a high power LED is perhaps 2 sq-mm. While its likely a single high power LED will produce less optical energy than the bulb, the radiance is still likely thousands of times greater for the LED and hence its greater potential for eye damage.


In practice, as long as you don't look directly at the light source but rather the light reflected from the objects of interest, there is little difference between the two sources.

Steve Paolini
via email


Lithium Charging Answered July 2014

I'm thinking of building a solar charger for my iPhone, but don't know how to handle the internal lithium battery in terms of taper current, etc. — especially when I have the phone on all day. What I've found online is information on charging disconnected lithium batteries, not ones under load. Any hints?

#7143
Anthony Suchek
White Plains, NY



Answers

First, under load shouldn't be that much different from being off as long as the circuit can handle the current needed to do both jobs. That you will need to measure and then find a pre-built circuit to do the trick.


I suggest pre-built since they are so cheap. I did the same kind of thing for a phone that refused to recognize its own charger as a valid charger. A good ol' Motorola phone, I'll never buy another phone that has some type of recognition routine in it to check the charger.


In any event, the first thing I noticed was that my camera charger was for the same type of battery, same voltage, same AHr rating, same Li Ion battery. I hooked up wires from the charger into the phone and it worked just fine.


Next I wanted a more universal Li-Ion battery charger & found more than one on-line. The one I settled on was from dx.com:

1A Lithium Battery Charging Module - Blue  $1.70/e
[url=http://dx.com/p/1a-lithium-battery-charging-module-blue-205188]http://dx.com/p/1a-lithium-battery-charging-module-blue-205188[/url]


There are others as well, a Red at 3A for under $9
[url=http://dx.com/p/1a-lithium-battery-charging-module-red-318740]http://dx.com/p/1a-lithium-battery-charging-module-red-318740[/url]


These are circuit cards usually with a USB power port to give it 5V 1A to work with, so you will need to do some soldering for connections to the battery from the board.


At 1A, expect to charge and run your phone without a problem, since most cell phones, even on transmit, are below 1W output. see: [url=https://en.wikipedia.org/wiki/Mobile_phone_radiation_and_health]https://en.wikipedia.org/wiki/Mobile_phone_radiation_and_health[/url]
and
[url=http://hypertextbook.com/facts/2006/EbruBek.shtml]http://hypertextbook.com/facts/2006/EbruBek.shtml[/url]

Phil Karras, KE3FL
Mount Airy, MD

Charging your iPhone should not require any special accomodations. The OEM charger from Apple supplies 5V at 1A via a USB connection with proper cable. Either the 30 pin connector or the new 5 pin "lightning" connector cable both deliver the same power.


The phone (as all phones now) has built in current limiting, temperature sensing and voltage sensing to protect the battery.


You just need to deliever 5V, regulated, to the cable with, of course, polarity protection for the solar charger. A simple LM7805 regulator IC is what most chargers use for the 5V.


It really is NOT a good practice to leave the phone on the charger continually, as most rechargable batteries 'like' to be excercised. Charge it up, let it run off battery until it shows recharge is needed then reconnect charger. You should get several years of service from it.

Rod Hogg
Scott City, KS


Photoresister Switcher Answered July 2014

I would like to know if I have correctly connected the photoresistors (CdS photocells) shown here in order to turn OFF during the day and to  turn ON during the night the two LED circuits attached to them. (If not, please indicate by a new diagram.)

Also, I would like to know:
  a) If any photoresistor would work?
  b) What would the optimum dark/light resistance values be for such a photoresistor?
  c) How would I calculate the values  (any formula?) from the transistor side (2N2222) that would best fit this ON/OFF photoresistor switcher?

#7141
Nate
via email



Answers

Depending on the current required by the different LED loads, the value of the Cds sensor is critical. In addition, you should include hysteresis to snap the load ON or OFF, rather than have it operate erratically at the dawn or dusk threshold. If the Cds sensor has a large value at the critical threshold, there will not be enough current to bias the transistor ON to it’s saturation state.

To remedy this problem, I have included two circuits which solve your issues. Figure 1 uses two transistors with positive feedback to create the needed hysteresis and gain to drive an output stage that provides 100 mA current limiting in case your load circuit shorts out.

Figure 2 circuit uses a 4093 quad-nand schmitt trigger (that has 1 volt hysteresis) and drives the same output stage as circuit one. The second circuit is probably the easiest one to implement, but requires the fourteen-pin 4093 IC. You will still need to select the a Cds cell that has around a megohm of resistance when dark and 10K or less resistance when light. Hope this helps you.

Ron Hoffman
Solon, OH

Power supply not specified.
Top circuit operating range using common 555 is from 5 to 15 volts by data sheet. IC 4017 data sheet specifies current drive limit about one milliamp. Some amplification required to drive LED at 10 to 20 milliamps


Bottom circuit requires at least 12 volts to operate series string of LEDs. Three LEDs in series add up to: G(3.2) + R(1.8) + G(3.2) = 8.2 volts. Requires at least 12 volt power to allow for current limit resistor voltage drop. LEDS will load down ring oscillator as shown. Need high impedance buffer to isolate LEDs from Resistor Capacitor time constant.
 

a) The logic of operation of CDS photocells is:
No Light = high resistance
Max Light = Low resistance
 

As shown, that logic will tend to turn on the transistor during the daytime and turn it off during the night. Simple solution is to swap 100K resistor with photoresistor in circuit. You will probably need to adjust resistor values to operate the way you require. You also need to limit current into the base of the transistor.


Not every photoresistor will work as the range of resistance variation must match the required transistor bias. You could measure the resistance of the photo resistor with an ohmmeter for both dark and light environment or look up the data sheet from Digikey, Mouser or the manufacturer.

See above circuit from Linear Technology Spice program.
 

Some CDS photo resistors are about 200K ohms in the dark and about 4K ohms in light. If the range is less, the transistors will always stay on. Measure the photoresistor in light with an ohmmeter. Cover the photo resistor to measure dark resistance.
 

b) Optimum bias would provide about 0.6 to 1.0 volts or more at the transistor base (relative to emitter) to turn on and less than that threshold to turn off. Depends on required current from collector to emitter.


You did not specify your power supply voltage. The 555 IC works best from about 5 to 15 volts. Three LEDS in series need at least 12 volts as shown.


Two red LEDS and 470 ohm minus (diode drops and IC drop)
(12 - (1.8 + 1.8)volt-1 -1 )= 6.4 volt
Max current = 6.4 volt / 470 ohm = 14 milliamps


The three LEDS with 1K resistor operating at 10 milliamps would require
VDC = (0.01 Amp * 1000 ohm)+(1.8)R+(3.2)G+(3.2)G   8.2
12 - 8.2 = 3.6 volts   3.6 volt/ 470 ohm = 7.6 milliamps
R8 is 330 ohm for 10 milliamps


Current limit for 4017 source about one milliamp out. need transistor drivers.
470 ohm drive about 14 milliamps


Ring oscillator may not work as LEDs are loading RC timing, transistor emitter followers may help. Linear Technology Spice program can be used to experiment with different values.
Calculation for transistor bias
R1 = base to power bus VCC
R2 = base to ground or common


Vbase = VCC * (R2 / (R1 + R2) )
This is first approximation as some current will go into base of transistor. Switch point is about in the range from 0.6 to 1.0 volts
High on , low off
Look up Ohm's law
 

See attached circuits
The .PDF extension is directly viewable. The .asc format is editable within Linear Technology Spice. LTSpice is free and downloadable from Linear Technology web site


Adjust the photo resistor value for the version you have at light and dark environment. Verify that the LOAD has a reasonable current (ON value) of about 10 to 100 milliamps or negligible current (OFF value)

Edward Wade
via email

The two circuits Nate submitted are configured to turn the LED sequencers ON during the day and OFF at night. This is because the resistance of the photocell decreases as light falling on the photocell increases, causing increased current flow from the base of the transistor to the positive power supply rail. The increase in base current will cause the transistor to conduct and supply current to the LED sequencer circuits in daylight. To modify the circuits so they turn OFF during the day and ON at night, swap the photocell and the resistor so that decreasing photocell  resistance will pull the base of the transistor toward ground, turning the transistor off in the daytime. The fixed resistor will turn the transistor on at night when the resistance of the photocell is high compared to the resistor. The circuit can be improved (see new diagram in Figure 1). Two issues are solved by the improved circuit.

Figure 1
1) Nate is using a 2N2222 transistor as a low side switch to control current to the light sequencer circuits. A MOSFET makes a better switch because it has a very high OFF resistance and a very low ON resistance, so more of the supply voltage is applied to the light sequencer circuit and less power is wasted in the transistor. For this reason, I replaced the 2N2222 transistor with a low cost IRF510 N-channel power MOSFET.


2) The other problem with Nate's circuits is they won't turn on and off quickly like a mechanical switch. At dawn and dusk — when the light changes gradually — the current through the transistor will also change gradually. The switching action should happen fast when a predetermined light threshold is crossed; a Schmitt trigger circuit is needed for that. The improved circuit contains four Schmitt trigger NAND gates in a single 4093 CMOS IC. Only one NAND gate is needed, so the other inputs should be tied to V- to prevent instability.


The 4093 has a hysteresis band which keeps the sequencer circuit from receiving rapid bursts of current at dawn and dusk when the control voltage hovers near the tripping point. The output of the 4093 drives the IRF510 MOSFET which switches the current to the LED light sequencer circuit on and off. The MOSFET is driven into saturation by the 4093 Schmitt trigger to provide clean switching and maximum current to the LED sequencer circuit. Next, the answers to questions a, b, and c.


a) Will any photoresistor work?
Most photocells will work with the improved circuit, but don't confuse a photocell (which is a photoresistor) with a phototransistor. I've included a pot wired as a variable resistor; this will adjust the voltage divider to match the photocell characteristics. To set the pot, go outside in the twilight near sunset and adjust the pot until the circuit turns ON, then back off slowly until it turns off. The LED sequencer circuit should then turn on at sunset and off at sunrise.


b) What are the optimum photoresistor light/dark resistance values?
The resistance of most photocells varies from a few hundred ohms (or less) in direct sunlight to a megohm (or more) in total darkness. The optimum resistance of R1 for a given photocell can be calculated using Ohm's Law, but the adjustment pot in the improved circuit makes it compatible with most CdS photocells.


c) What formula will calculate the optimum component values?
The 2N2222 transistor used in Nate's circuits is a current operated device which should be explained in terms of current. The improved circuit uses voltage controlled components. This explanation is only for the improved circuit.


Inputs 1 and 2 of the 4049 are tied together and connected to a voltage divider consisting of photocell PC1 and resistor R1, which are connected in series between V+ and V-. To predict circuit operation in different lighting conditions, calculate the voltage from V- to the point where R1 and PC1 join using Ohm's Law, and then compare this voltage to the upper and lower trip points of the 4093. The lower trip point will turn the LED sequencer circuit on because the 4093 is an inverter, which will drive the gate of the MOSFET to V+ when the 4093 input is low. When the MOSFET conducts, it will ground the LED sequencer circuit to V-. The lower trip point of the 4093 is +3.9 volts when V+ is 10 volts and the upper trip point is +5.9 volts.


EVENING CONTROL: The voltage across R1 must fall below 3.9 volts to turn the LED sequencer on at sunset; this example uses 3.5 volts to be well below the threshold. To calculate the optimum resistance of R1, it's necessary to know the resistance of PC1 in the twilight hours of sunset or dawn.


This example assumes the resistance of PC1 is 100K at sunset. The voltage drop across both resistors must add up to V+. So, if V+ is 10 volts, then 6.5 volts must be dropped across 100K (PC1) for 3.5 volts to be dropped across R1. Operation of the improved circuit is explained in four steps.


STEP 1: Calculate the current through the voltage divider at sunset.
If we know the resistance of PC1 and the voltage across it, then we can calculate the current through PC1 with Ohm's Law (I = E/R). Plugging in the known values, I = 6.5/100,000 = 65 µA. This is a series circuit, so 65 µA also passes through R1.


STEP 2: Calculate the resistance of R1.
We know the current through R1 is 65 µA and the voltage across R1 is 3.5 volts, so we can use another form of Ohm's Law (R = E/I) to calculate the resistance of R1. R1 = 3.5/.000065 = 53846 ohms. To check this, use E = IR, E = .000065 (53846) = 3.499 volts; 3.5 volts is under the lower trip point of the 4093, so output pin 3 of the NAND gate will go high, driving the gate of the IRF510 MOSFET to V+ through R2. This causes the MOSFET to conduct and provide a current path from V- to the LED sequencer circuit.


MORNING CONTROL: At the soft light near daybreak, a typical photocell has a resistance around 10K. If we use the value of 53.846K previously calculated for R1 and assume a PCI resistance of 10K at dawn, then the total resistance of PC1 + R1 is 10K + 53.846K = 63.846K.


STEP 3: Find the daytime current through the voltage divider.
If V+ is 10 volts, the current through the series resistances PC1 and R1 at sunrise can be calculated with I = E/R. I = 10/63846 = .0001566 A, or 157 µA.


STEP 4: Find the daytime voltage drop across R1.
Now that we know the morning current through R1 and the resistance of R1, we can calculate the morning voltage drop across R1 with E = I/R; E = .0001566 (53846) = 8.43. A daytime voltage drop across R1 of 8.43 volts is well above the 5.9 volt upper trip point of the 4093, so the output of the inverting 4093 will go low, grounding the gate of the MOSFET to V- and turning off current to the LED sequencer circuit. During the day, the input to the 4093 should vary between 8 and 10 volts, keeping the LED sequencer circuit turned off.

Ed Gore
Panama City, FL


Wireless Mirrors Answered July 2014

I have an old car that came with electric windows but has old fashioned side view mirrors which are almost useless. I would like to upgrade to more modern mirrors that are aimed by control of 12V electric motors. On the driver's side, that is no problem as you simply use a four-way switch scavenged off the vehicle that you got the mirrors from, using the 12V available to run the electric windows.


However, to use that same switch to control the right hand mirror, I would have to drill a hole in the edge of both doors and both front pillars, and snake a wire through from the left side to the right side. It occurred to me that the use of radio waves from the left hand door could be used to control four relays in the right hand door and obtain the required action.


Something like Ron Newton used in his “Super Detector” article in the October 2013 issue of NV. Any suggestions would be appreciated.

#7142
Dean Kaul
Kalamazoo, MI



Answers

There is a Chinese company that sells lots of off the wall stuff. Here is the link to a 4 channel remote control relay box. The unit will cost $17.50 with free shipment. (1 week to 1 month). I have bought a similar item and it worked fine. www.dhgate.com/product/4-channel-rf-wireless-remote-control-switch/179280704.html#s1-2-7|328444939 Here is a simplified wiring diagram.

Ray Perry
Jacksonville, FL

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