Reverse varistor?

[Shadow_Storm56]

What component.. if it exsists would be able to block power when voltage passes a certain point.... So for example if you had an light that you wanted never to get more than 12v. It would open the circuit at 12.1v and allow current to flow again at 12 again.

[Enderman]

They're called voltage regulators.

It doesn't open the circuit past 12v, it would limit the output to 12v.

[Shadow_Storm56]

8 minutes ago, Enderman said:

They're called voltage regulators.

It doesn't open the circuit past 12v, it would limit the output to 12v.

What's their tolerance? Like if you had one rated for 48v and voltage spiked to 74 would it still hold 48v or would it pop? Maybe I would need another component?

[Unter Dog]

9 minutes ago, Shadow_Storm56 said:

What component.. if it exsists would be able to block power when voltage passes a certain point.... So for example if you had an light that you wanted never to get more than 12v. It would open the circuit at 12.1v and allow current to flow again at 12 again.

Add a comparator with small hysteresis controlling a FET.  Requires designing it though. 

 

 

[Enderman]

8 minutes ago, Shadow_Storm56 said:

What's their tolerance? Like if you had one rated for 48v and voltage spiked to 74 would it still hold 48v or would it pop? Maybe I would need another component?

Depends on the input range of the regulator.

There are regulators for all different types of voltages or ranges.

[Unter Dog]

You could also use a zener/bjt combo for a poor mans regulator.  An in-line LDO would work also.  

[Shadow_Storm56]

3 minutes ago, Enderman said:

Depends on the input range of the regulator.

There are regulators for all different types of voltages or ranges.

All this stuff is so complicated rawr

[Unter Dog]

5 minutes ago, Shadow_Storm56 said:

All this stuff is so complicated rawr

Is this for a 12 volt automotive system like a car?  

[mariushm]

A linear regulator takes in some voltage and outputs a lower voltage by throwing out the difference between input and output as heat.

 

So if you want to power a product with 12v and 100mA (like a relay for example) , your linear regulator takes in 48v and outputs 12v at 100mA , but throws out  (48v-12v)  x 0.1A = 3.6 watts as heat . This means the chip must be on a heatsink and at over 3-4 watts a fan blowing over the heatsink would be a very good idea - by itself, a linear regulator chip can usually only dissipate around 1 to 1.5 watts of heat without overheating.

.

Linear regulators are very simple to use, they have an input voltage and ground pins and an output pin ... the adjustable kind can be configured using two resistors between the output and ground pins

 

 

Do keep in mind that most linear regulators are only designed to work up to 30-35v .  There are high voltage versions of some common linear regulators (with HV in their name) which work up to 60v. 

You mentioned in another thread that you want a relay to trigger when you go over 48v and that the voltage may go up to 70-80v - in this case these common linear regulators won't help you.

 

There are special linear regulators which can work with even 450v but they generally can only output a very low amount of current, let's say up to 10 mA, and they have some other limitations.

 

For example, LR8 from Microchip can output minimum 1.2v but and can do it even with as much as 450v DC at the input, as long as the input voltage is around 12v higher than output voltage : 

Digikey : https://www.digikey.com/product-detail/en/microchip-technology/LR8K4-G/LR8K4-GCT-ND/4918746

Farnell : http://uk.farnell.com/microchip/lr8k4-g/linear-volt-reg-0-02a-440v-to/dp/2448524

 

(note that this LR8 regulator can be adjusted like in the picture above, but since it's designed for much higher input voltages, it uses different - bigger- value resistors...  it likes around 6 kohm for R1 so for 5v output you'd need around 18 kOhm for R2)

 

So for example, you could configure this chip to output 5v and the chip will automatically start working and produce 5v on the output as soon as the input voltage is above 17v DC, even at 100v DC the output will still be 5v.

You can use this 5v to power a chip (a microcontroller or voltage monitor like I suggested in another post) which can measure the real voltage and if that voltage goes above a certain value (like 48v), it can send a signal to another circuit ( if you want isolated, through an optocoupler or digital isolator).

The amount of power the linear regulator produces (mA) is not enough to energize relays or do anything very complex, but it would be enough to send a signal through an optocoupler / isolator.

An optocoupler is basically a led and a light sensor and some air between them inside a rectangular package.  When you send a signal, the led inside the optocoupler lights up and the light sensor (basically a super tiny solar panel) produces energy and that generates the signal on the other end of the optocoupler.

That signal can be powerful enough for example to turn on a transistor which lets energy flow through the relay windings and therefore engage the relay.

 

 

 

That other circuit receives the signal and turns on that relay which does

[Shadow_Storm56]

1 hour ago, mariushm said:

A linear regulator takes in some voltage and outputs a lower voltage by throwing out the difference between input and output as heat.

 

So if you want to power a product with 12v and 100mA (like a relay for example) , your linear regulator takes in 48v and outputs 12v at 100mA , but throws out  (48v-12v)  x 0.1A = 3.6 watts as heat . This means the chip must be on a heatsink and at over 3-4 watts a fan blowing over the heatsink would be a very good idea - by itself, a linear regulator chip can usually only dissipate around 1 to 1.5 watts of heat without overheating.

.

Linear regulators are very simple to use, they have an input voltage and ground pins and an output pin ... the adjustable kind can be configured using two resistors between the output and ground pins

 

 

Do keep in mind that most linear regulators are only designed to work up to 30-35v .  There are high voltage versions of some common linear regulators (with HV in their name) which work up to 60v. 

You mentioned in another thread that you want a relay to trigger when you go over 48v and that the voltage may go up to 70-80v - in this case these common linear regulators won't help you.

 

There are special linear regulators which can work with even 450v but they generally can only output a very low amount of current, let's say up to 10 mA, and they have some other limitations.

 

For example, LR8 from Microchip can output minimum 1.2v but and can do it even with as much as 450v DC at the input, as long as the input voltage is around 12v higher than output voltage : 

Digikey : https://www.digikey.com/product-detail/en/microchip-technology/LR8K4-G/LR8K4-GCT-ND/4918746

Farnell : http://uk.farnell.com/microchip/lr8k4-g/linear-volt-reg-0-02a-440v-to/dp/2448524

 

(note that this LR8 regulator can be adjusted like in the picture above, but since it's designed for much higher input voltages, it uses different - bigger- value resistors...  it likes around 6 kohm for R1 so for 5v output you'd need around 18 kOhm for R2)

 

So for example, you could configure this chip to output 5v and the chip will automatically start working and produce 5v on the output as soon as the input voltage is above 17v DC, even at 100v DC the output will still be 5v.

You can use this 5v to power a chip (a microcontroller or voltage monitor like I suggested in another post) which can measure the real voltage and if that voltage goes above a certain value (like 48v), it can send a signal to another circuit ( if you want isolated, through an optocoupler or digital isolator).

The amount of power the linear regulator produces (mA) is not enough to energize relays or do anything very complex, but it would be enough to send a signal through an optocoupler / isolator.

An optocoupler is basically a led and a light sensor and some air between them inside a rectangular package.  When you send a signal, the led inside the optocoupler lights up and the light sensor (basically a super tiny solar panel) produces energy and that generates the signal on the other end of the optocoupler.

That signal can be powerful enough for example to turn on a transistor which lets energy flow through the relay windings and therefore engage the relay.

 

 

 

That other circuit receives the signal and turns on that relay which does

 

Sounds really complicated. I like the idea of using the one that can do up to 450v. So if I set it to output to 48v it would need to hit 60v... So I would need to make the sensor do 36 or so volts so when this regulator hit 48 it would output 36v from there on out and then this signal would go to a voltage sensor with the led like you described.

 

Also can mosfetts be used to open and close circuits that are high current.. like 80 amps dc or something like  that.

[mariushm]

Well, the thing is it doesn't work quite like that.

 

The only rule is that the input voltage MUST be higher than the output voltage by some amount (called dropout voltage the datasheet) in order for the linear regulator to output a very steady output voltage at your desired level.

The regulator will constantly try to reach that output voltage even if the input voltage is below that minimum amount.

 

For example, in the case of the LR8 regulator above, as soon as the input voltage is above the reference voltage (1.2v) PLUS the dropout voltage (12v), then the linear regulator starts working and will work hard to output the desired voltage. 

So if you configure the LR8 regulator to output 5v, it will not output anything until the input voltage goes above around 13v.  Once the input voltage is above 13v and until the input voltage is at least 17v (5v + dropout voltage of 12v), it will output [ input voltage - 12v ], so something between 1.2v and 5v.  It's not off until 17v is reached, and instant on once 17v is reached. 

Once the voltage is above 17v, it will continue to output 5v even when the voltage goes up to 400v.

 

You can have some microcontroller or maybe an Arduino, which can run from 5v, and you can use a potential divider (2 resistors like I explained in another thread) to reduce the input range you say you're gonna have  0.. 80v down to 0.. 4v ( divide by 20 ) or something like that, and inside your Arduino program, you'll know that if you measure more than 2.4v  (2.4v x 20 = 48v) , then your Arduino knows the input voltage is more than 48v and can send a signal like I explained, through an optocoupler to your other circuit.

The other part of the circuit which runs from 5v or 12v or 24v (depending on what relay you use)  picks up that signal and turns on the relay which starts the circuit you want.

 

You can replace the relay with a mosfet or transistor but if you want isolation between them you still want an optocoupler, or something else (isolation transformer or digital isolator).  A mosfet alone does not create isolation between the two circuits. 

 

 

 

 

[Shadow_Storm56]

46 minutes ago, mariushm said:

Well, the thing is it doesn't work quite like that.

 

The only rule is that the input voltage MUST be higher than the output voltage by some amount (called dropout voltage the datasheet) in order for the linear regulator to output a very steady output voltage at your desired level.

The regulator will constantly try to reach that output voltage even if the input voltage is below that minimum amount.

 

For example, in the case of the LR8 regulator above, as soon as the input voltage is above the reference voltage (1.2v) PLUS the dropout voltage (12v), then the linear regulator starts working and will work hard to output the desired voltage. 

So if you configure the LR8 regulator to output 5v, it will not output anything until the input voltage goes above around 13v.  Once the input voltage is above 13v and until the input voltage is at least 17v (5v + dropout voltage of 12v), it will output [ input voltage - 12v ], so something between 1.2v and 5v.  It's not off until 17v is reached, and instant on once 17v is reached. 

Once the voltage is above 17v, it will continue to output 5v even when the voltage goes up to 400v.

 

You can have some microcontroller or maybe an Arduino, which can run from 5v, and you can use a potential divider (2 resistors like I explained in another thread) to reduce the input range you say you're gonna have  0.. 80v down to 0.. 4v ( divide by 20 ) or something like that, and inside your Arduino program, you'll know that if you measure more than 2.4v  (2.4v x 20 = 48v) , then your Arduino knows the input voltage is more than 48v and can send a signal like I explained, through an optocoupler to your other circuit.

The other part of the circuit which runs from 5v or 12v or 24v (depending on what relay you use)  picks up that signal and turns on the relay which starts the circuit you want.

 

You can replace the relay with a mosfet or transistor but if you want isolation between them you still want an optocoupler, or something else (isolation transformer or digital isolator).  A mosfet alone does not create isolation between the two circuits. 

 

 

 

 

Couldn't I use a varistor to keep resistance really high until near the required voltage. So even if the regulator started producing at a lower voltage it would still be blocked by the varistor until near 36 or whatever?

[Shadow_Storm56]

Plus the mosfet would need a certain voltage to kick in as well.

[Shadow_Storm56]

4 hours ago, mariushm said:

Well, the thing is it doesn't work quite like that.

 

The only rule is that the input voltage MUST be higher than the output voltage by some amount (called dropout voltage the datasheet) in order for the linear regulator to output a very steady output voltage at your desired level.

The regulator will constantly try to reach that output voltage even if the input voltage is below that minimum amount.

 

For example, in the case of the LR8 regulator above, as soon as the input voltage is above the reference voltage (1.2v) PLUS the dropout voltage (12v), then the linear regulator starts working and will work hard to output the desired voltage. 

So if you configure the LR8 regulator to output 5v, it will not output anything until the input voltage goes above around 13v.  Once the input voltage is above 13v and until the input voltage is at least 17v (5v + dropout voltage of 12v), it will output [ input voltage - 12v ], so something between 1.2v and 5v.  It's not off until 17v is reached, and instant on once 17v is reached. 

Once the voltage is above 17v, it will continue to output 5v even when the voltage goes up to 400v.

 

You can have some microcontroller or maybe an Arduino, which can run from 5v, and you can use a potential divider (2 resistors like I explained in another thread) to reduce the input range you say you're gonna have  0.. 80v down to 0.. 4v ( divide by 20 ) or something like that, and inside your Arduino program, you'll know that if you measure more than 2.4v  (2.4v x 20 = 48v) , then your Arduino knows the input voltage is more than 48v and can send a signal like I explained, through an optocoupler to your other circuit.

The other part of the circuit which runs from 5v or 12v or 24v (depending on what relay you use)  picks up that signal and turns on the relay which starts the circuit you want.

 

You can replace the relay with a mosfet or transistor but if you want isolation between them you still want an optocoupler, or something else (isolation transformer or digital isolator).  A mosfet alone does not create isolation between the two circuits. 

 

 

 

 

Also I assume the optocoupler outuputs the same voltage as it inputs? 5 volts on one side 5 volts on the other...or whatever 

[mariushm]

Please stop quoting my long posts, it's just 2-3 people answering so it's obvious  what you're replying to.

 

No, the voltage is not 5v. It's not voltage.  The "optical sensor" is a transistor... when light is detected, the transistor opens up and lets energy flow from one pin to the other.

 

Watch the video above, it explains things in detail.

 

That particular optocoupler in the video will open the transistor inside when signal is detected, and that transistor  can handle up to 50mA between its pins and up to 70v difference between the pins.

If your relay uses more than what the optocoupler's transistor can let through (which is often the case with 12v relays, they usually use 100-120mA), then you can use an additional transistor between the optocoupler and the relay - see the last example in the video where he uses an extra npn transistor to turn on something that used up to 70mA or so.

 

 

[Shadow_Storm56]

Ok. So then the regulator would produce a set voltage when it hit a certain point.... this voltage could be dropped to a much lower voltage using the dual resistor thing. Then a voltage detector could be used after that which would allow power flow when it hit a certain voltage... this could power the photocoupler which then would allow power to the relay or mosfet. Sounds like it may work. I would prefer to use a mosfett but I'm not sure if mosfetts are made that big.

[Shadow_Storm56]

23 minutes ago, mariushm said:

 

I thought I had to quote for you to know I replied?

[mariushm]

// I have "auto subscribe"  to any thread i post to, so i get notifications every time someone posts somewhere i also posted, unless i manually disable for that particular thread.

 

The regulator is only needed if you need a stable voltage to power a chip or a microcontroller or a circuit... a microcontroller can be something like an Arduino board (which is basically just the chip with easy to access pins) which you'll basically use instead of a voltage detector chip.

 

You can write a program into your Arduino which has only one job , keep measuring the voltage and if it's above a threshold, send a signal through a pin.  That's exactly what the voltage detector chip does, it monitors its own power supply and if the voltage is above some threshold (you can buy different versions of the chip, each with a fixed threshold preset inside them) then the output pin is activated.

 

The voltage detector chips I suggested to you in another thread don't need a separate power supply, they measure directly the voltage they're powered from. Go back to that other thread and look how that voltage detector chip works

 

I'm not 100% sure this is right but with just the voltage detector chip, it would work kinda like this :

 

 

 

You pick the two resistors on the left so that when you reach some voltage, in the middle of those two resistors you're gonna have the voltage which is the threshold for the voltage detector chip you pick.

The one I linked to had a threshold voltage of 4.4v, so as an example I'd want the voltage detector to trigger when voltage goes above 44v, so I'd pick the two resistors in such a way that the input voltage will be divided by 10.  This way, at 44v input voltage, the chip will see 4.4v and start doing it's job.

By doing it's job i mean the chip will internally connect the output pin to ground, like flipping a switch.

 

So we can use this to create a circuit with the input voltage to turn on the led inside the optocoupler.  The LED inside the optocoupler turns on at around 1.5v - this is called forward voltage and varies from optocoupler to optocoupler and depends on what kind of led they use inside. For this example, it's safe to use 1.5v and will probably be very bright inside with anything between 2mA and 10mA so we can sort of aim for 5mA and we'll be fine.

 

In my example, we'll have at least 44v and we may have up to 70v on the input, so if we connect the led directly it would burn up... we put a resistor there to throw away the extra voltage and let only the amount of current we want through the led.

It's a simple formula  Voltage equals current times resistance .. so in our case  44v - 1.5v  = resistance * 0.005 A  so resistor value is 42.5 / 0.005 = 8500 ohm   

What happens if you have 70v ? Well 70v - 1.5v / 8500 ohm = 0.008 mA ... that's still low enough that the led won't be damaged.

But, to be on the safe side, you could just go with a 10k resistor and have around 3mA of current going through the led.

 

The microcontroller  / arduino thing would look like this :

 

 

 

 

The regulator's only job is to give your arduino a stable 5v.

The program on your arduino starts up and constantly measures the voltage on the ADC pin (which can be between 0v and 5v depending on how you pick those two resistors). Unlike using a fixed threshold voltage detector, you can write in your program at which threshold should your arduino react.

When whatever threshold you desire is hit, your arduino sends power (5v) through a pin , the resistor limits the current to 5-10mA and the led inside the optocoupler turns on.

Arduino or some other microcontroller also gives you more flexibility. 

 

For example, you pick those two resistors so that when the voltage goes above 44v for more than 5v seconds, then turn on the signal... and turn off the signal only when the voltage dropped down to 42v or when it was below 48v for more than 10s. Basically, you don't want to turn on and off a relay or something ten times a minute, because that's how many times the voltage went above 48v. 

 

The voltage detector chips have something similar to this but more basic, they have a "window" of around 0.2v .. for the chip i linked to, once the threshold of 4.4v is reached, the chip will continue to work until the voltage drops below around 4.2v

[Shadow_Storm56]

Ohhh.... So I could just use a usb phone charger or anything that outputs 5v for the Arduino if I have access to it. Toss a diode in as extra precaution. No need for a regulator at all. Plus if I program the delay in the Arduino, by the time the relay goes to break the resistive load it will be very low and preserve contacts. Plus the Arduino outputs the correct voltage to the led. So it comes down to this... how much do they cost and how hard are they to program.. I guess also what one would I need. I would likley do the divide by 20. So select the resistance values so at 100v it's 5v, just to be extra safe and keep it easy. Then at 48v it would be 2.4v. also if the led is steady at 5v then I just need to calculate a resistor to limit current to the proper level at 5v.

The other side of the coupler all I would have to do is have a very small relay power the large one so the load is kept low on the PHC.

[mariushm]

They're not hard to program, you can find loads of tutorials on youtube for them.

Genuine arduino boards are maybe 20-30$ , but you can get clones from eBay for like 5-6$

 

Genuine

 

Newark:

DIP :  23.38$ : http://www.newark.com/arduino-org/a000066/dev-board-atmega328-arduino-uno/dp/78T1601

SMD: 22.00$ : http://www.newark.com/arduino-org/a000073/dev-board-atmega328-arduino-uno/dp/63W3545

 

Sparkfun :

 

DIP : 20$ : https://www.sparkfun.com/products/11021

SMD: 24$ : https://www.sparkfun.com/products/11224

 

Adafruit:

 

DIP : 25$ : https://www.adafruit.com/product/50

 

 

eBay clone example :

 

3.6$ (33582 units sold) : https://www.ebay.com/itm/NEW-UNO-R3-ATmega328P-CH340-Mini-USB-Board-for-Compatible-Arduino/311155383820

5.22$ : https://www.ebay.com/itm/ATmega328P-CH340G-UNO-R3-Board-Mini-USB-Board-for-Compatible-Arduino-USB-Cable/281988689542

 

There are example programs that you can download from the Arduino website, including one that makes a led blink (so you can basically replace led with optocoupler) and one which measures a voltage using an ADC pin and does something with it (i think it just prints it by default, but you can change the code to make the led turn on or off)

 

This here is a very good series of videos that teach arduino (it's a playlist, bookmark it) :

 

https://www.youtube.com/watch?v=JMMamSVy1Zs&list=PLE72E4CFE73BD1DE1

 

 

 

[Shadow_Storm56]

I would buy genuine, Building these things to replace the built in controllers that fail will save me a lot of money. I am going to draw a final diagram, you can look at it and tell me if I am getting all this ok. I feel like I should pay you for all the help but I guess were all here to help eachother

[Shadow_Storm56]

 

What was the program someone (I think you) used for calculating the resistor values? 

[mariushm]

You can't put the relay directly on the optocoupler. Watch again all the way to the end that video that explains optocouplers.  The relay uses too much current and the optocoupler can't move that much current through it.

 

You don't put a diode on the ground (negative) wire. If you want to, you can put one on the +5v wire.  It won't do much.

Note that a genuine arduino board has a built in linear regulator, so if you power it from the DC In jack, you'll need to use any wallwart adapter with 7.5v ..12v voltage (arduino won't care, it takes that down to 5v) , or you can power it through the usb port with 5v , or you can power it through two of those pins in the long headers again with 5v.

You could also power it with 4 rechargeable AA batteries ( 4 x 1.2v..1.35v = 4.8v .. 5.1v) .. the arduino would use less than 10mA so the batteries would last for weeks.

 

The picture above is wrong.

 

The ground from the big voltage must be connected together with the ground wire from the power supply of arduino.

If the ground wires are not connected together, Arduino can't measure the voltage.  It's like you're trying to measure the height of a guy by comparing it to yours, but the other guy sits on a bench that you don't know the height of. Connecting the grounds together basically means you make sure the person you're trying to measure is on the same flat piece of land as you.

 

The potential divider works like this :

 

+V --------- resistor 1 ------- to ADC pin --------resistor 2 ------- to ground

 

The ratio between the two resistor values determines what voltage the ADC pin "sees"

 

The picture was from a software called Electronics Assistant because i was too lazy to type, but the formula is simple : Output voltage = [ R2 / (R1+R2) ] * Input Voltage

 

 

[Shadow_Storm56]

Ah so the ground goes before the resistor....cool. Besides that we are good? Also I have 12v in all the places these would go so I should be able to power it easy. So even a really small relay just large enough to power a big one would still draw too much I take it? No biggy, transistors can't be any harder than this to add. Also I have not watched the video yet, I am going too though. The resistors I would use would be a 16kohm on R1 and 820ohm on the other to keep it at 5v max at 100v (being safe) What about the capicitors? do I need them if I am not using the regulator? 

[mariushm]

The arduino board has linear regulators along with capacitors and everything required on the board. You just need to power it with 5v or more and the board does everything else.