Tech Forum Questions

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]

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.

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.

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.

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)

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.

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.

When I read the question, I remembered that RadioShack sold a blinking LED but in red (part 276-0036). I bought the LED, and testing the current flow while it was blinking, I discovered that the IC inside reduces the current to less than 0.03 mA (30 µA) when off, and conducts over 40 mA when on. So you can buy the blinking LED and add it in series to your LED flashlight. When you turn the flashlight on, the red LED will blink, and make the main LED blink, too.

You will have to open the bicycle flashlight to wire the blinking LED in series to the white LED. This LED can handle up to 5 volts and up to 80 mA (as printed on the package) with no additional components. If you do a web search on the part number, you can find other bicycle projects using the same blinking red LED to flash an array of nine red LEDs for the rear light.

There are a number of different multi-mode LED flashlight driver modules available, many of which include flashing modes. Check out the selection at dx.com. Make sure to get one that supplies a current appropriate for the LED in your light. They also carry complete lights based around similar modules which might be easier than trying to modify your existing flashlight.

I had one of those (30 GB Classic iPod) and the best thing you can do is simply buy a new battery for $5 - $10 on eBay. You can even get a kit with tools to help you take it apart (that's the $10 end). It's easy to take apart that iPod, there are several YouTube videos to help you do it.

The first iPods were famous for the irreplaceable battery, and they are extremely hard to open. IFixit considers the iPod Classic battery replacement "Very difficult", and you can find some YouTube videos on opening and replacing the battery if you dare to do it yourself.

If you wish to continue using your son's iPod Classic as a portable music player but don't want to risk damaging it, consider attaching a portable external battery. If your iPod Classic uses a USB cable to charge, then buy an external USB battery which can supply 5 volts and at least 0.85 Amps (850 milliAmps). Some are as large as the iPod, so just tape it back-to-back, and use the shortest USB cable between them. They are rechargeable from any other USB port, and will be useable with smartphones if you desire to retire that iPod.

Forget the old battery and buy a new Lenmar Replacement Battery for iPod Classic 5 G (30 Gb) This battery replaces 616-0230 and EC008-2. Amazon carries them for around $15. The Portable Rechargeable Battery Association (PRBA) highly recommends against using reconditioned lithium ion batteries and I would assume this also applies to lithium polymer batteries due to their propensity for catching fire. I found one web site that advocated a repeated freeze and charge process but for $15 i would buy a new, known battery versus trying to resurrect a possibly defective unit.

You have an RB5X from RB Robotics. These have been around since the early 80's and were designed for use in the classroom. They do sell parts for your robot at their website - Check out http://www.rbrobotics.com/parts_dept.htm I also have one of this same model, which I want to re-fit with updated microcontroller hardware, as the original control system is very dated. With a few new body parts you could have your RB5X looking like it should and then begin incorporating some of the great projects in Nuts & Volts or Servo to get it animated and doing something fun! Best of luck to you as you resurrect this classic robot!

It appears that you have the bottom portion of the RB Robotics RB5X Robot. The domed top, top portion of the body cover and programmer are missing (check eBay for parts for sale). See the web site [url=http://www.theoldrobots.com/rb5x.html]http://www.theoldrobots.com/rb5x.html[/url] for more information.

I am SO jealous! That is the RB5X robot made by RB Robotics in 1982. You can read more about this awesome robot on my web site here - www.robotsandcomputers.com/robots/rb5x.htm - or this other great site here - www.theoldrobots.com/rb5x.html ... I hope you enjoy your very impressive find! I'm still looking for one of my own...

There are many transformers, less than $15 that are available to match low impedance microphones to a high impedance input. Some can be permanently wired while others like the Audio Technica  CP8201 are adapters that simply plug in.

Google for 'microphone transformer' to see what's available. You might also raid old audio gear. Many times, depending on your needs , exact values are not so critical and there's no risk in trying them!

I am going to go against the two prior answers. I have directly wired devices with XLR connectors to ones with phone plugs. And have wired balanced to unbalanced and vice versa. I have done this in the professional environment of broadcast TV stations where any technical problems would be instantly noticed and can not be tolerated. You just have to know what you are doing.

First, the type of circuit used, balanced or unbalanced, and the impedance of the circuit do not necessarily directly correlate to each other. Nor do either of these directly correlate to the type of connector used. Although the XLR style of connectors are generally used for balanced, low impedance (600 Ohm) circuits and phone plugs are generally used for unbalanced circuits, I have seen the opposite successfully used in many cases. Connectors are connectors and do not care about impedance or balanced/unbalanced. If they have the correct number of pins and the required quality level, they will work.

OK, now your question. Your XLR terminated microphones have an equivalent circuit something like this:

It functions as an AC Voltage source in series with a resistance. The shield connection is not connected to anything in the microphone cable or in the microphone itself except the outer case. It is for shielding only. You did not say what the impedance or signal level of the microphone is so there are two main possibilities, high impedance or low impedance (Rs). Either of these can have a number of different signal levels (Vac). This is the ideal way of wiring any microphone as it can be used with any type of input, as I will show below. If it were wired as an unbalanced device, then it would be OK for an unbalanced input but not well suited for a balanced one.

The other side of your circuit is the amplifier input. The only thing you say about this is that it is a high impedance connection that uses a telephone style jack. This could be either a two pin or a three pin telephone jack. You should look inside the amplifier to determine which it is. As for the "high" input impedance, the most common value used is 10 Kohms, but some amplifiers may differ. But it really makes little difference. 

If it is a two pin (T&S - Tip and Sleeve), then it is definitely an unbalanced input and the equivalent circuit will look like this:

Your microphone and amplifier could then be connected as follows:

Notice that the ground is not connected to the XLR pin 3 at the microphone. The two resistors form a Voltage divider so the signal delivered to the input stage of the amplifier will be a simple fraction of Vac. If Rs is low, almost the full value of Vac will be delivered to the input stage. If Rs is higher, then this will be reduced, depending on the exact values. There should not be a lot of reactance in the amplifier input stage so there should not be any high frequency roll off. In any case, the connection should work with no expected problems unless the amplifier has too little gain. If gain is a problem, you will need to use another microphone or a pre-amplifier.

If the telephone jack on the amplifier is a three pin type (RTS - Ring, Tip, and Sleeve), then the circuit will most likely look like this:

I said "most likely" because I have seen microphone inputs that used the three pin telephone jacks for other purposes, like selecting between a line and a microphone input. Also, it could still be an unbalanced input, just with the extra pin. In that case, the R and S contacts would probably be connected together and you are just back to the unbalanced situation above and you still use the hookup shown below. But let's assume that your amplifier's input is strictly a balanced, microphone input and the above circuit is correct.

For a three pin telephone jack your connection should look like this:

If you look at it carefully, you will see that there is no difference in the signal loop. Only that the ground is not connected to it so both sides are floating, as opposed to one side being grounded in the unbalanced connection. This configuration is less susceptible to picking up noise from the environment, but does not operate any differently as far as the signal is concerned. So all my comments for the signal in the unbalanced connection still apply to this balanced one.

One more thing, if a three pin telephone jack is used with an unbalanced amplifier input, the above circuit is still valid except that the bottom end of Ri is connected to ground at that amplifier connector and the lower arrow to the input stage is not present.

In short you should simply use a shielded pair for your microphone cable. Connect one wire of that pair to the T (Tip) connection at the amplifier. Connect the shield to the S (Sleeve) connection and the other wire of the pair is connected to the R (Ring) terminal if it exists and to the S (Sleeve) if it does not. This covers all cases.

A transformer may help if the microphone level is too low for the amplifier input. But that will introduce inductance and it will have it's own frequency response curve which may not be very good. Good audio transformers need a lot of steel and copper and are expensive. As I said above, a far better way to do this would be with a pre-amplifier. Even an inexpensive pre-amp will have better specs. than almost any transformer, hands down. Or get higher level microphones, but that can also be expensive.

Andre, yes you can connect an XLR microphone to a 1/4" TS (or TRS) jack. I'm 67 and am a retired studio chief engineer for a major studio in Hollywood, and it works no matter what some people might tell you.

Connect the XLR pin 1 (ground) and pin 3 (low) to the 1/4" sleeve (TS jack) or ring and sleeve (TRS jack), and connect the XLR pin 2 (high) to the 1/4" tip. Those who haven't done this get confused because, theoretically, you then have "unbalanced" the microphone, but the signal still exists. The less obvious problem is that an impedance matching transformer gives you 10-20 dB of noiseless gain so your signal-to-noise ratio is a bit better (the noise includes hum, of course, plus electronic thermal noise with a floor of -131dBm).

Finally, good transformers are EXPENSIVE (I like Jensens, jensen-transformers.com). All transformers need RC compensation on the secondary; send a square wave through one if you don't believe me and look at the ringing (ask if you want more info on this). HERE'S THE BOTTOM LINE: Connect the XLR to the 1/4" jack (use a NO-transformer adapter or wire one yourself) and see if you're happy; if not, then (and only then) think about using a transformer (and don't use a cheap one, they have more distortion). Heck, even a lot of today's semipro equipment runs XLR "unbalanced" (look at the schematics). Call me at 818-951-8646 (desk line, daytime Los Angeles) if you or anybody else wants further info.

Yes Andre, you need an adapter to mate that "Unbalanced" (Hi-Z) microphone input to a "Balanced" (Low-Z) microphone. Here are three ideas:

RadioShack sells item #274-016 (you'll need cables to connect the two ends). Here's the link: www.radioshack.com/product/index.jsp?productId=2062443.  Chances are your local Shack has it in stock.

Parts Express (www.partsexpress.com) sells a similar adapter: www.parts-express.com/pyle-ppmjl15-1-4-male-to-xlr-female-mic-cable-15-ft--248-4614.

If you're handy with a soldering iron, here's a DIY idea: www.mediacollege.com/audio/connection/xlr-jack-mono.html (all necessary parts are available at RadioShack).
All three methods will do what you need — just pick the option that's easiest for you to implement.

For starters you CANNOT use a 1/4" male plug wired to a XLR jack. The 1/4" line is unbalanced meaning there is one signal line and a ground. The XLR line has two signal lines with opposite and equal signals plus the ground. Use a RadioShack Model; 274-016 (or equivalent) A3F XLR jack to 1/4" plug adapter. This adapter contains a balance-to-unbalance (balun) transformer to match the two lines properly. The balanced line and High Impedance (HiZ) are both used to reduce interference on low level signal lines like microphones.

Telemarketers are harassing millions of people every day which is a major problem. In 2012 the FTC offered a $50,000 reward for a solution. A couple of solutions are available:

#1 Some telephone carriers will put a telemarketer blocker on your phone service. Check with your provider who may have a charge for the service.

#2 The Telebouncer Blocker TB1000 is advertised to block 100 percent of robo-calls and 99 percent of solicitor calls. I have never tried this and it’s pricey at around $120.

#3 I just hang up, but if you have caller ID you may just not answer calls with numbers you do not know. This still does not help the telemarketers loading up your voice mail.