Tech Forum Questions

All but one of the answers here didn't answer the OP question, which was about a diode across the CONTACTS, not the coil. As the one answer that addressed the issue indicated, the reason is to suppress arcing across the relay CONTACTS when the coil is de-energized (particularly if the contacts are connected to anotrher inductive load).

It is used to protect the components of the circuit from burning out from a release of the induced voltage when the relay is turned off. The energy stored in the magnetic field of the relay coil is now released and the voltage can get quite high and case a spark. If high enough, this can cause damage, but even if it’s a low voltage, it can cause RFI and cause a circuit to misbehave. This diode has a number of names, but I like calling it the flyback diode.

Example: I once built a 8080 based microcomputer and programed it to turn lights on and off in the house when we were on vacation. But, it was always glitching, the program would get lost and hang up. Until I realised I was not using any flyback diode on all the relays. Once these were installed the RFI noise was removed, or at least reduced enough, so that the computer would not lose its place in the program and all worked fine from then on.

Aside: A flyback transformer where this flyback voltage is put to good use, see: https://en.wikipedia.org/wiki/Flyback_transformer

I looked this up to find more for you, check out these web pages: This is a good explanation: https://resources.altium.com/p/using-flyback-diodes-relays-prevents-electrical-noise-your-circuits

Here’s a good article on Wiki: https://en.wikipedia.org/wiki/Flyback_diode

The diode is called a flyback diode. The purpose is to dump the energy from the relay coil when it is disconnected from power. The coil acts as an inductor and when the power is disconnected, as when a transistor turns off, the current can not change instantly. This results in the voltage across the coil spiking to try and keep the same current flowing.

The diode lets the current flow thru it when the power is turned off keeping the inductive kick from destroying the transistor. No current flows thru the diode when the coil is energized because it is reverse biased then.

Some relays have the diode built in and do not require an external one. If you replace the relay and there was no diode across it the old one may have had an internal diode. If your replacement does not have a built in diode you should add one externally.

When a voltage is removed from the relay coil, the magnetic field collapses, inducing an inverse pulse to flow. This inverse pulse can damage transistors and - more easily - ICs. The diode sorts this pulse so that it doesn’t damage any sensitive components.

If you are controlling a relay with any silicon device - e.g. transistor, IC, etc - you probably need to use a diode (called a snubber diode, by the way). If you are controlling the relay directly, say with a switch, then you don’t really need it.

When the energizing contacts of a relay (or any inductor) open, the collapsing magnetic field forces the current to keep flowing. If the applied voltage was large and the coil is large enough, a substantial voltage will build up across the ends of the coil. If the voltage is very large, it will cause damage to the circuit. The diode is to short the voltage and prevents damage.

Not across the contacts, but across the coil. It is called freewheeling or flyback diode. It provides a path for the high voltage created by the collapsing magnetic field when current is cut off. Otherwise the high voltage can destroy the driving circuit. Also used across solenoids and sometimes DC motors.

Mr. Abend asks about diodes placed backwards across relay contacts.  NOTE that this technique is applicable only in DC circuits. 

The diode is called a “snubber” and it serves to protect the relay contacts from overvoltage in inductive circuits. Relay contacts in DC circuits are subject to sparking as they open when they control an inductive load. This opening can occur intentionally, or can occur during contact closure due to contact bounce. Examples of inductive loads include motors, solenoids, or other relay coils.

Current and voltage in an inductive circuit are related by Faraday’s Law, one form of which is V = -L di/dt. It states that the rate-of-change of current (amperes per second) in an inductor is proportional to the voltage applied across the inductor and inversely proportional to its inductance (henries). The minus sign indicates that the inductor resists a change in current by opposing the applied voltage.

Consider a voltage applied across an inductor through a set of relay contacts: At the instant of contact closure, load current is zero. As time advances, and dependent upon the load inductance, the current through the inductor will increase at a constant rate determined by the ratio V/L. At some point, the inductor will saturate — i.e., it cannot be magnetized further — whence the limiting current will then be governed by the winding resistance.

Now consider what happens when the relay contacts open: The magnetic field in the inductor immediately begins to collapse, and as it does so, it will attempt to drive current back into the voltage source. But since the voltage source no longer exists — we have an open circuit — there is nothing (other than the inductor’s winding resistance) to limit the magnitude of the voltage induced, and this kickback voltage will therefore appear across the relay contacts that are in the process of opening — contacts that have, in fact, barely moved from their closed position. The polarity of the kickback voltage is opposite to that of the original voltage that was used to energize the inductor. The kickback voltage creates a spark that degrades the contact surface, and over time, this sparking can cause contact failure.

A backwards diode connected across the relay contacts will switch into conduction when kickback voltage is applied. In this case, the decreasing rate-of-change of inductor current will be governed by the inductor winding resistance (ignoring the forward voltage drop in the diode).

Practical examples: An inductor having an inductance of 1 henry and a winding resistance of 1000 ohms, energized by a 25-volt source. The rate-of-change of increasing current in the inductor will be 25 amperes per second (25 milliamperes per millisecond) and the limiting steady-state current will be 25 milliamperes — which will be attained in about one millisecond.

When the voltage source is removed, as by opening a relay contact protected by a snubber diode, the initial value of the inductor current will be the same as its limiting value earlier — 25 milliamperes. The decrease in inductor current over time (t) is an exponential function of winding resistance (R) and inductance (L) proportional to e (exp -tR/L).

When the switching device is a transistor, the snubber diode serves to protect the transistor from destruction due to the imposition of kickback overvoltage.

More often the diode is placed across the coil such that it is reverse biased. Its purpose is to absorb the counter EMF when the coil is de-energized . When the relay coil is energized it stores energy in its magnetic field. When the current is turned off that magnetic field reverses and tries to keep current flowing in the circuit. However, since the path is open that energy must dissipate. The energy can produce a rather high-voltage spike. This is how the Kettering ignition system works in early automobiles spark generation.

Since in most small electronics we do not want that high voltage to damage components the diode across the coil shorts out that reverse Electro-Magnetic-Field (EMF) and thus saves delicate components. You may find a diode or other suppressing component across contacts for the same reason. i.e. To prevent a spark which would shorten the contact life or damage other components.

In the automobile ignition system there is usually a capacitor placed across the points in the distributor. The ignition coil produces the spark when the contacts open because the magnetic field collapses sending that energy to the spark plug(s). I hope this helps you understand a little bit more about inductance in coils.

When a relay (or other) coil is de-energised, the stored charge can create a substantial back e.m.f. amounting to several hundred volts. This voltage can damage delicate components (FETs etc.) in the ciruit.

The reverse connected diode will clamp this voltage to 0.6 or 0.7 volts for a silicone diode. This limits any fast going high voltage spike in the circuitry.

The problem the diode is solving occurs when the current flowing through the relay coil is switched off. This on-to-off transition is relatively fast and the collapsing magnetic field (the field generated to actuate and hold the relay in its switched condition) generates a very high reverse voltage across the relay coil. The relay coil current is usually switched by a switching-transistor.

This reverse voltage can easily exceed the capacity of the switching-transistor’s reverse-breakdown voltage and, thus, destroy the transistor. To solve the problem, the backwards diode provides a forwards path for the collapsing current to cycle back through the relay, and thus the backwards current flow dissipates within the relay.

A diode is used on relays to keep back EMF from taking out the device driving the relay when power to said relay is turned off. When power to a coil goes off, the magnetic field collapses. This causes very high voltages and currents on the relay coil. They can go backwards into the drive circuit, damaging the active device. They are used where damage is very likely. Hope this helps.

The coil of a relay is a fairly large inductor. Inductors store energy in a magnetic field that builds up around the coil when current if flowing through the coil. If the current is turned off suddenly the magnetic field collapses and "induces" voltage in the coil, just like a generator. The stored energy has to go somewhere and the induced voltage tries to keep the current flowing in the same direction.

The voltage will continue to rise with a polarity opposite of what had been applied. That means the end of the coil that had been connected will generate a voltage much higher than, and added to, the supply voltage at the other end of the coil. That voltage can easily rise to a level high enough to damage the circuitry. The diode placed "backward" across the coil provides a path for that current. It effectively shorts out the voltage induced in the coil and allows the stored energy to dissipate in the diode and the wiring without reaching dangerous levels. The diode should be rated somewhat higher than the supply voltage and should be able to handle at least the same amount of current as the relay coil.

When is it necessary? The full answer requires quite a bit of math and physics but the short, simple answer is anytime an inductive device is switched off and on. It applies not only to relay coils, but also to motors, solenoids, and any other type of devices with coils/inductors. In some cases the switching circuit will have protective diodes built in. For instance, the ULN2803 darlington array IC is intended for driving such loads and has the diodes built in. The L293D (D for Diode) stepper motor driver / dual H bridge IC also has protective diodes built in.

The diode is there to protect any solid state device driving the relay. The current flow through the energized coil creates a magnetic field around the coil. When switched off the collapsing field results in an instantaneous polarity change in voltage across the coil. Without the diode this voltage can become quite large exceeding the voltage rating of the switching device. The changed polarity of the voltage induced into the coil forward biases the diode limiting or clamping the voltage rise to around 0.7 volts.

Sometimes referred to as a snubber diode or snubber circuit. Sometimes an RC circuit is used in this fashion but diodes have become more common. Some 12 V automotive relays include this diode internally, others do not. I would not replace one with a diode with one without. The diode is "backwards" because we do not want it to conduct until the voltage reversal across the coil happens. Automotive relays with the diode need to be connected so the internal diode is reverse biased. Typically shown as a rectangle with no polarity indications.

This is one of those $1,000,000 questions and one which is difficult for those of us who are only somewhat into electronics, like me.

I think the basic answer is yes, they can be used to do the same job but other things will determine which is best to use. By best, I mean many things, for instance, there may be no electronic reason for using one over the other, then again there could be. It might be easier to use one over the other because of what is being done, see Below. There could also be a cost reason, one way might require more parts than the other to make say a PNP work in the same job as using an NPN, or vice versa.

I looked this up because it’s a question that has always bothered me that I don’t really know, I just think I might know.

Below:
Here’s a site that helps: How to decide between PNP and NPN https://www.controldesign.com/articles/2016/how-to-decide-between-pnp-and-npn/

While you may not understand everything said in the above article, I don’t either, it does hit upon my two reasons, one is easier to use than the other, one uses more parts than the other.

For a home circuit this may not matter but in industry where extra parts means much more overhead cost to make something, it does matter.

Also, as in software, sometimes it’s easier to understand a circuit of one type rather than the other. I had this experience in software development. I had two developers who know some really interesting ways of doing things, but no one else could understand the method as it was written. My feeling was that if that person ever left the company no one else would be able to debug such a section of code, so I kept lobbying for code that was easier to understand by a majority of software developers.

Such is also the case with electronics, while it might be possible to do it that way, the other way, - NPN vs PNP - might be much easier for a majority of electronics engineers to understand, so that’s the one you should use.

I hope that helps, it did help clarify things for me.

Mostly, yes. There are not that many general types of transistors. NPN and PNP bipolar transistors are by far the most common. They are used for small signal amplification and switching.

If high frequencies are involved, you must use a type with useful gain at the frequency of interest. If high power is involved, then that must be considered. Some circuits use higher than common voltages, and high current, and that must be observed.

Darlington types are used where extremely high gain is needed, but the small differences between the thousands of types are simply not that important. I have only a very few that I use for everything.

Then there are MOSFETs which are mostly used for high power control of motors and such. One big consideration here is the maximum voltage and current and how you provide heat sinking. Specialty types such as enhancement mode FETs and Unijunction are not that common, and there are not that many different types available.

Mr. Bodine raised a question about transistor choice and interchangeability for a given circuit. In short, choice depends upon purpose, circuit polarity, operating characteristics such as signal input voltage or current, power-handling, operating temperature, power supply, etc. etc. Some circuit designs are more amenable to using MosFETs rather than bipolar junction transistors, and vice-versa.

The book that launched me into transistor circuit design is by Albert Malvino, entitled Transistor Circuit Approximations. While it is currently out of print, a number of used copies are available through Amazon beginning at under twelve dollars.  See https://www.amazon.com/Transistor-circuit-approximations-Albert-Malvino/dp/007039878X.

There are generally two types of bi-polar transistors. PNP and NPN. Most small signal (i.e. not hi power) transistors can sometimes be substituted for a like transistor. That is a PNP can be replaced by a PNP but NOT a NPN. Likewise a NPN can be replaced by another NPN but NOT a PNP. There are lots of cross references to look up equivalent replacements. My favorate is https://www.nteinc.com/ I may be mistaken, but I think the most popular type of transistor in use today are of the NPN type.

The type of transistor used depends on what your doing.(amp, buffer, volt. regulator, ECT.) Also circuit specs. PNP and NPN can be swapped if the emitter and collector connections are reversed. This is true so long as device/circuit specs are similar. Hope this helps.

This circuit consists of two main parts. The first part is the oscillator which is a standard circuit consisting of everything on the left hand side up to and including R4 and Q2. On the right hand side Q3 and R6 form a current source that drives the LED. It is controlled by the voltage on the base of Q3. R5 controls how fast C3 charges and discharges, controlling the voltage on the base of Q3. If the oscillator is to slow, decreasing C1, C2 and/or R2, R3 will speed it up. If the fading effect is to slow, decreasing C3 will speed that up.

Ah, this is a difficult thing to do simply. First an LED is not like an old tech incandescent bulb which gets dimmer when the voltage goes down and then brighter as the voltage goes back up to its correct operating voltage.

An LED turns on at a specific voltage and turns off at a lower voltage, it will continue to give light for some voltage range as long as the current is constrained to not exceed too high a current when the LED will burn out.

So,

1. An LED turns on or off depending on its driving voltage
2. We can not exceed its operating current or we burn out the LED
3. LEDs do not have different brightness depending on voltage.

The only way I know of making this happen is with a circuit that is called a pulse width modulator that changes the pulse width that drives the LED ON vs. OFF time. I have used an Arduino and a program sketch to do this, but I’m sure there’s someone out there who may have done this with something like a 555 timer IC which might be a “simpler” circuit. In fact I found this: Generate Pulse Width Modulation (PWM) Signal using 555 Timer IC https://circuitdigest.com/electronic-circuits/555-timer-pwmgenerator-circuit/

Notice here that the author uses a pot, variable resistor, to drive the 555 Timer IC to change it’s pulse width. This means you’ll have to hire someone to continually adjust the resistor up and down and up and down, etc, for two minute cycles to get what you want, definitely NOT realistic.

OK, so what does that resistor do? It probably changes the voltage going into the 555 timer that causes the 555 to increase or decrease the pulse width. This sounds like a sine wave of varying voltage wich could be made with another transistor circuit.

You might want to also watch: https://www.youtube.com/watch?v=QmB1Ev-h3y4 If anyone knows of a simple circuit that does this I’d like to know as well. BUT, I think the best you will be able  to do is to find a Pulse Width Modulating IC to do the job and then a driving circuit to produce whatever is needed to tell the PWM IC what pulse width is needed now.This might be the IC needed, I know you said you didn’t want an IC, but even the simplest circuit used a 555 Timer IC: How to use TL494 pulse width modulation control IC 94 https://microcontrollerslab.com/tl494-pulse-width-modulationcontrol- ic/ Good-luck and I hope this helps at least a little.

Ms. Carrillo raised a question about a means of driving an LED from bright to dim, then reverse, and repeat.  She further stipulated that it must be a non-IC-based circuit.

Since LED brightness is a nearly-linear function of its forward current, this suggests a triangle-waveform driver circuit. An examination of the information available on the internet for a transistor-based circuit uncovered a suitable circuit¹ for which I have modified timing component values and added an output buffer, all of which is shown as Figure 1 below.  

Figure 1:  Schematic diagram

The circuit operates from a 12-volt source. It is comprised of two current-mirror elements Q1-Q2 and Q3-Q4, and a three-way differential comparator circuit Q5, Q6, and Q7. Current mirror Q3-Q4 functions such that the current drawn by the collector of output transistor Q4 will always be identical to the current supplied to the collector of input transistor Q3; a similar explanation can be made for the operation of current mirror Q1-Q2. 

Figure 2:  Charge C1

In the initial state, capacitor C1 is fully discharged. Upon application of power, R1 provides about 3.5 mA into current mirror Q3-Q4. Since the voltage at the upper end of C1 [V(C1)] is zero, Q6 is off. Q5, however, is forward-biased by the voltage divider R3-R4, thereby completing the circuit Q1-Q5-Q4, such that 3.5 mA is drawn by Q4 out of Q1. Because Q1-Q2 is also a current mirror, a current of 3.5 mA will be applied to C1 by Q2, whence V(C1) increases from zero. C1’s voltage increase is linear because the charging current is a constant value. See interval t₁ in Figure 2 above.

Figure 3: Switch

Transistor Q6 turns on at the point where C1 has risen to be equal to Q5’s base voltage. In its conducting state, Q6 steals current from R3, removing base drive to Q5, turning it off, thereby turning off current mirror Q1-Q2 whence charging current is removed from C1. See interval t₂ in  Figure 3; “V(n004)” is the base voltage at Q5.

Figure 4:  Discharge C1...

C1 discharges into current mirror Q3-Q4 via the base-emitter junction of Q6. This action provides base current to Q6, keeping it turned on until such time as the terminal voltage of C1 [V(C1)] falls below the base voltage at Q7 (set by voltage divider R1, D1 and Q3). At this point, Q7 begins to turn on, rapidly shutting off Q6. See interval t₃ in  Figure 4 above.

Figure 5: ... and repeat

Once Q6 is turned off, R3 can once again provide base current to Q5, turning it on; the cycle repeats. See interval t₄ in  Figure 5 above, being the same as interval t₁ in Figure 2.

Figure 6: ... and repeat

Figures 2 through 5 show V(C1), the terminal voltage of C1, having a triangular waveform with a period of two seconds. This voltage is impressed upon the gate of MOSFET M1; diode D3 provides a DC voltage shift to better match the MOSFET’s characteristics. M1 drives LED D2 through current-limiting resistors R5 and R6, whose values have been chosen to enhance current linearity. 

Figure 6 above shows the excellent match between drive voltage V(C1) and LED current I(D2). A linear drive current will produce a very nearly linear variation in light output; see Figure 7.

Figure 7:  LED Luminosity vs Forward Current

Some words about the components used: M1 is shown as a 2N7002, which is a surface-mounted device, for which leaded part 2N7000 is a functional substitute. BS170 is another leaded version of this part.  LED QTLP690C was a random choice from the LTSpice component library; choose your own with particular emphasis on current-vs-luminosity linearity. All resistors are 5% tolerance except R1, which is a 1% value. Given that the tolerance of C1 is probably not better than 10%, it would be wise to replace R1 with a 5K-ohm variable resistance to allow for adjustment if the timing is at all critical.

Current mirrors: Proper operation of a current mirror circuit requires that the silicon bodies of both transistors of the pair be at the same temperature. This is nearly impossible to achieve with discrete components, especially as in the present case when the power dissipated by the two transistors differs significantly, one from the other. Quad-transistor through-hole and surface-mount package equivalents of the 2N3904 and 2N3904 are available from Mouser and others; look for MPQ3904 and MPQ3906, respectively.

Ms. Carrillo did not specify the power of the LED to be driven. I assumed a small indicator, so that the maximum drive current provided by this circuit is just shy of 20 milliamperes. For larger, more-powerful devices, a larger MOSFET would be needed (with attendant modifications of the values of R5 and R6). A preferred approach in this case would be to use the drive current available at the drain of M1 as a means of modulating a multivibrator circuit, driving the large LED with a pulsed current whose pulse width varies according to the linear triangle-wave modulation. While such a circuit is beyond the scope of the present discussion, several, ^2.3 have been found in the literature.

¹ https://www.electronicspoint.com/forums/threads/7-transistors-triangle-wave-generator-650khz.264497/
² See US Patent 3445788A, Camenzind, Pulse-Width Modulation Circuits (https://patents.google.com/patent/US3445788A/en)
³ See US Patent 3587002, Brown, Voltage-Controlled Transistor Multivibrator  (https://patents.google.com/patent/US3587002)

There are several simple circuits around that will do it, just google "breathing LED circuit," but there are also LEDs with the circuit built in. The only supplier I know of is Lighthouse LEDs in the US (I forget where, I buy online).

Mr. Lambdin inquired about a circuit to monitor the AC ripple on the output of a linear power supply.  The inference is that the frequency response of the circuit need not be excessive. He also wanted a digital display.

I found what ought to be a suitable device, in kit form, from Amazon, for $29.95. See https://www.amazon.com/Oscilloscope-Handheld-Pocket-size-Electronic-Learning/dp/B01LWK49W3/ref=asc_df_B01LWK49W3/?tag=hyprod-20&linkCode=df0&hvadid=309778489815&hvpos=&hvnetw=g&hvrand=18068906155916055584&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=1018200&hvtargid=pla-465542683176&psc=1

In a linear supply the vast majority of the ripple will be at the input frequency which will normally be the power line frequency (50 or 60 hz) or two times that frequency (100 or 120 hz.) A digital multimeter on the AC volt ranges is intended primarily for that frequency range. By blocking out the DC from reaching the meter any DMM set to the AC volt ranges should measure it fairly well. A "True RMS" type meter is best since the ripple probably will not be a sine wave.

To block out the DC part simply put a capacitor in series between the positive side of the supply and the positive input of the DMM. A value of 1 microFarad or more should work. If you use any type of polarized capacitor such as an electrolytic make sure to connect the positive end to the positive output of the supply. Of course, make sure the voltage rating of the cap is higher (16 or 25 Volts should be fine) than the supply. You may need to put a fairly high value resistor (say 10 K Ohms) across the meter leads (from the negative end of the capacitor to the negative side of the power supply.

Searching for "spdif input" under "Computers, Tablets, and Networking" seems to show a number of things like what you want. Here's the list: https://www.ebay.com/sch/i.html?_from=R40&_trksid=p2047675.m570.l1313&_nkw=spdif+input&_sacat=0 And here's one that looks useful: https://www.ebay.com/itm/184216799887?hash=item2ae42d4a8f:g:6jMAAOSwJ81gZjs1 They're in the $20-$40 range, which doesn't seem too bad.

I did a quick Google and found this in my second hit: www.amazon.com/Ebetter-External-Surround-Recording-Compatible/dp/B07BT6CM6Q Says it does SPDIF to USB when set for digital input. And it's under $20. I'm lucky, I still have a few SPDIF compatible sound cards from the late 90's/early 2000's that are PCI based, and PCs with PCI cards.

I would differentiate between the situation where the volatile parts of the lubricant have evaporated (e.g. dried out) versus a scenario where the lubricant has become stiff due to adherence of dust to it. In the second case flushing it out would be warranted. Not sure what I would use in that case. It depends on if there are plastics in the immediate vicinity that could be damaged. I would probably try some CFC if you have any, otherwise a generic contact cleaner spray. I would keep cotton balls and swabs to blot up the excess as quickly as possible. WD40 is another option, but again, capture the excess or it will leave a residue. In the dried up case I have had very good luck using additional lubricant to loosen up the viscosity of the original without flushing it out. Unless the plunger is plastic it is highly likely the original lubricant is petroleum based. A very small amount of general purpose lubricant such as 3 in 1 sewing machine oil can be applied and worked in. Mineral oil (from the drug store) will also work. Even very small amounts of WD40 will work, but since it will fairly quickly evaporate (over a few months) you will need to combine it with the oil. If you believe that the original is not pertroleum-based, then a small amount of dishwashing liquid such a Dawn may help. I hope this helps.

I really hate to be a wet towel, but this is really not a good idea. Arbitrarily adjusting things like fuel and timing is most likely to result in a smoking, noisy engine. If run to long the engine is more likely to self-destruct than increase power. Not to mention tailpipe pollution. Even "professionally" reprogrammed engines emit excess pollution, and shorter lifespan.

A Passive Infrared (PIR) detector is much less expensive and easier to use. There are many available for outdoor use, with battery operation and remote monitoring by Wifi or other means. I don’t know of any DIY road loop circuits, and they are being phased out due to high installation and maintenance costs in favor of cameras, which are another choice.

I once bought a little board from a guy in Australia which monitored standard NTSC video and detected movement. Worked pretty well.

I would suggest an ebay search for guitar tune”. There are several listed in the ten dollar range with free shipping. The tuners don’t generate a tone but receive a tone and show the note on the lcd. I think parts cost for self build would be higher than purchase.