Tech Forum Questions

The non-dimmable variety use a largish electrolytic capacitor in the power supply, to generate a DC voltage, and then the LEDs are driven with a constant current circuit. So because of the constant current, the LED output is unaffected by supply voltage, so the brightness won't change with voltage(or dimming). Dimmable leds only have a small film capacitor, and the control circuit sets the current according to the input voltage (so they pulsate at mains frequency) and when dimmed the light output is chopped up similar to an incandescent bulb. The above applies to screw in bulbs, larger LED supplies may still use electrolytic capacitors, but within these another circuit measures the incoming duty cycle and adjusts the LED current to match.

Chad, a small headphone amplifier could be used. Look at the ones for amps and guitar accessories.

The easiest way to listen to your TV via headphones is by using a set of computer speakers connected to your TV using a stereo 1/8” (3.5mm) jack to RCA plugs y-adapter (see below).

Plug:

• The RCA plugs to the LINE OUT jacks on your TV
• The 1/8” stereo jack to the stereo plug of the computer speakers
• The speaker AC cord into the wall
• Your headphones into the HEADPHONE jack of the computer speaker.

Turn the computer speaker on and adjust the volume to your liking.  

Here are some sources for the y-adapter:
Parts Express
www.parts-express.com/dual-rca-male-to-35mm-stereo-female-y-adapter-audio-cable--240-070
MCM Electronics
www.mcmelectronics.com/product/24-13885
RadioShack
NOTE: RadioShack doesn’t have a “direct” adapter, you need to create one from the following parts - http://comingsoon.radioshack.com/3-ft-1-8-stereo-to-dual-phono-rca-plug-y-cable/4200494.html#.VaesjHWuPIo and http://comingsoon.radioshack.com/1-8-stereo-phone-coupler/2741555.html#.VaesxHWuPIo

Finally, depending on how far away from your TV you are, you may need the appropriate length male-female stereo extension cord to patch the adapter to the computer speakers.

I would use a external computer speaker set. It will provide a cheap small amplifier, volume control and an input for your headphones to plug into. If it comes with an external power supply you won’t need to buy batteries. The computer speaker system will most likely come with a stereo mini 1/8” phono plug so you will also need to get an RCA adapter cable. The computer speakers must have a place for you to plug in your headphones.
 

Attach the adapter to your TV and then plug the speakers into the adapter. Once the speakers are connected you should be able to play audio out of them. Then plug in your headphones and the speakers will shut off and all the sound will feed into your headphones. Adjust the volume on the speakers.

Done and your total cost should be less than $25 (15-$20 speakers and $5 for the adapter).

The tiebreaker circuit could also be implemented with relays, which wouldn't require any soldering or breadboarding. A relay would be required for each contestant, with as many poles per relay as there are contestants. The schematic I provided uses four 4-pole relays. For a higher number of contestants, relays could be ganged together (wire the coils in parallel) to obtain more poles. For each relay, pushing the corresponding button activates the coil; the normally-closed contacts of the other relays in series ensure that only one coil can be activated at once. A normally-open contact of the same relay placed in parallel with the button latches the coil so that it will stay on when the button is released. A normally-closed reset button (which could also be the on/off switch) will unlatch the relays.

Low cost multi-pole relays with screw-down wire terminal sockets can be found on eBay or at electronics surplus stores. Some even have indicator lights which would eliminate the need for separate lamps. Supply voltage would be selected to match the relay coil voltage. Since relays are mechanical devices, they have a finite switching time, which is usually on the order of milliseconds. It is theoretically possible that two button presses extremely close together could cause more than one relay to latch. To test this I put together a system like I described but with only two relays and buttons instead of four. The good news is that I was unable to make both relays latch on, even after many attempts at pushing both buttons simultaneously.

This is one of the simplest quiz circuits I've come across. I've built several versions. Some as simple as the first one, and a very complex one, but still using the same concept of parallel SCRs. www.techlib.com/electronics/games.html

There is a very simple old school tie breaker indicator circuit which I have used several times, It's inexpensive and works well.

Use a small neon lamp such as an NE2, and NE51 or any of the other types available. Use as many lamps as necessary for the number of stations desired. A latching type pushbutton or toggle switch in series with each lamp located at each players position. The trick here is that all of the series lamp/switch combinations are paralleled from one power source that being the 120 VAC line with a 68k-100k 1/2 watt resistor in series with one side of the line.

The principle is simple, with all of the switches off, there is no drop across the resistor. The first lamp that is energized, fires and pulls the voltage on the bus down to the point where no other lamp can fire. Neon lamps need a high voltage (generally around 65 volts) to fire but once fired will stay illuminated on much lower voltage, lower than any other lamp can fire.

The attached circuits (Figures 1 and 2) should do the trick. Each student station is equipped with a normally-open pushbutton switch connected to the instructor’s console via a cable of suitable length, terminating at a connector.

The instructor’s console consists of a bank of LEDs, one per student; a Master Reset pushbutton switch; and a suitable number of two-pin connectors into which the student pushbutton cables are plugged.

Each student readout consists of a “D” flipflop, an AND gate, an inverter, a diode, and an LED, plus associated wires, resistors, and capacitors as shown. The LED can be red or green, of any physical shape, having a maximum continuous current rating of 20 mA. The circuit is set for about 10 mA through the LED, which is very conservative and will provide quite adequate illumination.

The CMOS 4013B contains two “D” flipflops as shown, in a 14-pin package; power pins are 14(+) and 7(-). The CMOS 4081B contains four two-input AND gates as shown, in a 14-pin package; power pins are 14(+) and 7(-). The CMOS 4049B contains six inverters as shown, in a 16-pin package; power pins are 1(+) and 8(-). Be careful of this — applying “+” power to pin 1 is unconventional, but this is the way that the package is designed.

The circuit works in the following manner:  

• Assume that the Master Reset button has been pushed on the teacher’s console, asserting the +RESET bus. This forces all “D” flops to be reset, and all LEDs to be extinguished. The voltage level on the (diode-OR’d) -INHIBIT bus applies a logic-high signal to the D inputs of all flipflops, as well as to one of the inputs of each of the AND gates.    
• The first student to press a button enables the AND gate for her/his receiver, which sets his/her “D” flipflop to the “on” state. Immediately its Q-bar output goes low, pulling down (asserting) the -INHIBIT bus, which presents a logic-low signal to the D inputs of all flipflops and disables all of the AND gates, thereby inhibiting further clocking pulses to all “D” flipflops.   

This ensures two things:

1. that subsequent button depressions by other students cannot set the “D” flipflop in their receiver circuits to the “on” state; and
2. that a second depression of the winning student’s pushbutton will not set the “D” flipflop in her/his receiver to the “off” state, thereby negating her/his vote and inadvertently opening the process to a new round of voting.

The two resistors shown relative to the output and one input of each AND gate provides hysteresis and reduces noise sensitivity. As implemented, the input signal delivered to the 12K ohm resistor must rise above eight volts or fall below 4.5 volts before a high or low output level (respectively) will be realized at the gate output (assuming that the remaining input to the AND gate is held high).

The contacts in each pushbutton are debounced with a paralleled R-C network.

The 12 VDC power can be provided by a small wallwart regulated power supply. Anything that will provide 100 mA or more will suffice.

Construction Suggestions:

1. Note that the integrated circuit (IC) package schematics show the two 4013B elements, the six 4049B elements, and the four 4081B elements, and that each IC is bypassed by a 0.01 µF disk ceramic capacitor, to be connected with the shortest possible leads across the power input terminals of the IC.
2. Pin assignments in the circuit schematic are illustrative; consult the IC package schematics for actual pin-outs for each circuit element in any particular IC.
3. The tie-breaker circuit shows two 4049B elements connected in parallel. Both elements must reside in the same IC package.
4. Diodes are 1N914 or 1N4148 signal diodes. All resistors are 1/4 watt carbon composition. With one exception, the capacitors are all disk ceramic, of any type (Z5U or better), 25 volts rating. The polarized capacitor is tantalum, and is to be connected directly across the 12V power input to the circuit board.  
5. While the circuit schematic shows only three “student receiver” circuits, there does not seem to be any reasonable limit to the number that might be included, so long as +12V, circuit ground, and the +RESET and -INHIBIT busses are extended to all.
6. The input pins of all unused circuit elements must be connected to ground.
7. The student switches should be connected to the main console using twisted-pair cable to ensure minimum noise sensitivity. Mount the R-C debounce circuit directly across the pushbutton pins.
8. Dual-in-line package (DIP)-type ICs and IC sockets are recommended.

Datasheets for the three ICs may be found at:
www.nxp.com/documents/data_sheet/HEF4013B.pdf
www.nxp.com/documents/data_sheet/HEF4049B.pdf
www.nxp.com/documents/data_sheet/HEF4081B.pdf
A 1N914 diode datasheet is located at: www.vishay.com/docs/85622/1n914.pdf
Parts are all readily accessible from Digi-Key, Mouser, Jones Associates, Newark, etc.

I have modified a circuit that I used 27 years ago, which consists of a two input NAND (cd4011n) driving a set-reset flip-flop (cd4013n). The push button switch is connected to one 4011 input and goes high when pressed. The other input of the 4011 is held high by a resistor to VCC.

When the button is pressed, Q of the 4013 goes high, lighting the LED. At the same time, the NOT-Q output goes low and locks out all 4011 inputs including the one that pressed the button. The circuit remains in this state until the reset button is pressed.

There are several reasons an Arduino hobby type PIR is not a good substitute for a commercial PIR motion sensor.

1. You will need to step down the 12V to 5V with a voltage regulator. Coincidentally page 78 of the July 2015 N&V Tech Forum has answers for a 5V regulator question .
2. You will need to add a 5V relay from the TTL output of your Arduino PIR so that you can provide floating form C (COM and NC) contacts for Alarm COM and ALARM NC. Normally the ALARM contacts are closed unless an alarm condition is sensed. All sensors and switches in the ALARM loop are wired in series in a burglar alarm. Color code may vary, but in my experience a red and a black wire were used. You should pay attention to the colors used in your system.
3. Unlike a commercial PIR, the Arduino PIR does not have a optical shield and/or shutters to adjust the beam width. As such, it will have a broader sensitivity and be susceptible to false alarms from moving persons or objects outside windows or pets inside. Additionally, it may be blinded from ambient light sources.
4. The terminals marked TAMP 1 and TAMP 2 are tamper switch connections. Normally the tamper circuit is closed unless the sensor case is opened. In my experience a white and green wire were used for tamper circuits. Yours may differ. If the tamper switch is opened in a commercial alarm system, a supervisory signal is sent to the alarm office and a technician is sent to inspect the system. This is to prevent a would be burglar from opening the sensor and tampering with wires or placing tape over the PIR sensor. In a residential alarm system, a trouble light will appear locally at the control panel. With the above information, you can make the Arduino PIR function, however the possibility for false alarms will be much greater unless you are willing to spend the time constructing a proper light shield.

The $89 quoted by your alarm company sounds fair, and is an even better deal if they install it for that price. If you are under contract for alarm reporting to the central office, modification of the system may create problems with the agreement. You may be able to find an equivalent commercial sensor on E-Bay for a good price if you insist on replacing it. Personally I would stick with a commercial PIR for this application.

You only need to connect four wires for a burglar alarm motion detector to work properly. “GND” and “12V” are voltage for the motion detector - 12V is the positive, GND is the negative. “ALARM COM” and “ALARM NC” connect to the zone input on your burglar alarm panel. Typically, there is no need for polarity because this is a simple switch - with no motion in the area the switch is closed, motion is detected the switch opens.

“TAMP1” and “TAMP2” are connected to a mircoswitch within the motion detector housing that activates your alarm if someone were to remove the cover - called a “tamper.” If anything is connected to these terminals, you can simply twist the wires together, complete the circuit, and cap them off. It is not necessary to utilize the tamper feature and is often not recommended.

But, I would avoid any type of “inexpensive” replacements for your alarm system motion detector. One of the major sources of false alarms in burglar alarm systems tends to be the motion detector. Installing a motion detector not specifically designed for burglar alarm use may end up causing false alarms with your system and depending on your jurisdiction, fines from the local police department from the dispatching of those false alarms. Most burglar alarm motion detectors are designed to limit false triggers - so I would suggest going that route.

I did a quick search and could find a reliable motion detector listed for burglar alarm use for $25. Easy enough to replace yourself, just make sure you seal up any penetrations in the housing where the wire and screws go through. Spiders love to make homes in these and that will also cause false alarms.

Alternately, $89 may not be a bad price for your alarm company to replace the detector if that includes the labor. Nobody wants to work for free after all! Plus you can get them to quickly check your system to make sure it is communicating properly to your monitoring company and everything is working correctly.

TAMP1 and TAMP2 are the tamper contacts. They are normally closed as long as the device is not being tampered with. They are just ‘dry’ contacts and could be connected to the Arduino.

Mosfets come in many flavors: there are N-type and P-type of course but there are enhancement mode and depletion mode. Enhancement mode are off at zero gate voltage; you have to apply a positive gate voltage for N-type or a negative gate voltage for P-type to turn it on. N-type mosfets are available with drain voltage ratings from 30 V to 800 V or more. Most mosfets are designed for switching; an 800 V, 10 amp device would quickly burn up unless it could turn on and saturate even quicker.

Mosfets are characterized by their saturation resistance which can be very low (like .01 ohms). Bipolar transistors on the other hand, are characterized by their saturation voltage which can’t get as low power as mosfets. The threshold voltage (the point where the transistor just turns on) is not well controlled so you can’t really know what the drain current will be at a particular voltage. That makes it difficult to design a linear circuit. I avoid that problem by using bipolar transistors in linear circuits, or using pulse width modulation in a switching circuit which can be filtered to produce an analog signal.
 

Depletion mode mosfets are on at zero gate voltage and you have to apply a negative gate voltage to N-type to turn it off. Junction fets are also depletion mode devices and the zero gate voltage drain current is not well controlled so they are usually binned and labeled so you can have some idea of what you are designing with.

First I cannot help you with the information you are looking for on the Allied Space Spanner receiver. However, the other part of your question is about calculating the information for coil L1. The formula for calculating inductance needed for a resonate frequency as the value of the capacitor is shown.

To calculate the inductance of L1 for the standard broadcast band (535KHz to 1700KHz), L= the inductance in µH, f= the frequency in MHz, and capacitance is in pF. Since the capacitance is given as 10 to 365pF, and we know the minimum frequency is 535KHz (or .535MHz), L=25330/(.535 x .535 x 365). L=242µH. In the above formula, we replace the C an L can be interchanged. So if we know the value of the capacitance we can calculate the value of the inductance and visa versa.

With the calculated inductance (242µH) and the maximum frequency of the broadcast band (1700KHz or 1.7MHz), C=25330/(1.7 x1.7x242). C=32pF. So, with a coil wound very close to 242µH, the receiver would cover the standard broadcast band, and a little higher.

Let’s now calculate the number of turns of #30 and diameter of the coil needed to get 242µH. N=√(L(9r+10l) )/r is the formula for calculating the number of turns on a coil, but we have to know the values of radius, and the length of the coil. With a coil, the larger the diameter and the closer the turns are together increases the inductance. r = the mean radius, l = the length of the coil.

I recently did a project very similar to this so I have first hand knowledge of the approximate dimensions of the coil. Looking in a wire table, I find that #30 wire will make approximately 90 turns for a linear inch. I then choose the l of the coil of 1 inch long and the diameter to be 2 inches, so r=1. After plugging the values into the formula, 67 turns is the result. So close wind 67 turns of #30 wire on a 2 inch diameter form and then loosen the turns slightly, and spread them out so the length of the coil is 1 inch. With this information you can calculate the other resonate circuits.

L1 value us very close to 250µH. You could wind one on a form, but buying on would be much easier as the magnet wire may be much more expensive.

It’s pretty simple... you just need another switch that’s connected to the output of the first switch. These switches are commonly called Inverters, meaning that the output is inverted from the input. This means that for a high level input, the output is a low level, and vice-versa. So, all we need to create a bi-level output from a single input is a second inverter connected to the output of the first inverter. Bipolar transistors and FETs act pretty much the same way, the difference being that bipolar transistors are current-driven, whereas FETs are voltage-driven.

If you need more theory on transistor circuits, I refer you to a series of articles by Ray Marston in Nuts&Volts titled “Bipolar Transistor Cookbook”. This is an 8-part series that N&V has graciously made available online. www.nutsvolts.com/magazine/article/bipolar_transistor_cookbook_part_1. Be sure to have plenty of reading time available, because it contains a train-load of information. I’ve attached a circuit image depicting how to connect two transistors as an SPDT switch.

One thing that was not mentioned was a start run capacitor. Does the motor have a "hump" on it?  If it does, there is a capacitor under it and it is probably bad. You can check it with an ohm meter when it is disconnected. See if the capacitor has not charged and causes the ohm meter to give some funny reading, like a  negative reading. If it is shorted or open it is defective.

Mr. Olivo is experiencing nuisance fuse blowing in his drill press. The problem may well be that the fuses that he has been using cannot tolerate the inrush current demand of the motor. At the instant of power application, the magnetic state of the motor core may be such that the first half-cycle of utility power drives it into saturation.

Assuming that the fuse is a 1/4” x 1-1/4” cartridge fuse, a Bussmann MDL-5 fuse will open at five amperes, but will tolerate more than one hundred amperes during a one-half cycle of utility power. See the time-current curves for this family on data sheet #2004, which is downloadable from web page www.cooperindustries.com/content/public/en/bussmann/electrical/products/electronic_smalldimension/elx_1_4_x_1-1_4_/mdl-v_mdl.catalog_numbers_(amps).brands.cooper_bussmann.html

If the fuse is another size, check with your local electrical jobber for an appropriate fuse having similar inrush tolerance. Good luck.

Does the dill press sound any different? Maybe the bearing in the motor is wearing out causing the increased load.

What about the cord? Has it been pinched/damaged in any way? Or even the outlet that it is plugged into, are there visible signs of scorching?

Lastly, how old is it? Could the motor be electrically failing? Things to consider beyond stresses due to operator use.

If this is just an on/off, switched drill press, then you definitely have an excessive friction issue. If a good lubrication doesn’t help, then the brushes or commutator in the motor are worn and binding.

Worn bearings could be allowing the armature to rub on the stator. Remove the belt and try to move the motor shaft; if it moves at all, it is bad.

:You don't say what type of fuse you are using but if it is a buss type glass fuse, you might want to make sure you are using a slow-blow type fuse for your drill press.