Newbie project, need clarification before proceeding.

[krappleby]

Hi guys,

im working on a simple circuit to drain a capacitor of its charge. and wondered if this schematic looks right.. Im a newbie to electronics and want to do this right lol.. here is the schematic.



The plan is simple.. providing a 5v power supply the system will light the Green LED to inform that the system is active.. when a cap with power is connected to the cap connections, the relay will switch from pos 5 to pos 4 providing a current through the transistor allowing the RED led to glow. (this is the warning led) when the charge in the cap is too low the red LED will go out as the relay will switch back to 5, meaning the cap is safe.

i have some 12v 10A 5 pin relays at hand.. is this good enough. or would i need better. or possibly a second resistor before the relay.

is this feesable and will it work..

thanks in advance..
keith

[tpowell1830]

I don't understand your schematic. Perhaps if you showed the location/orientation of the cap under test and also the relay symbols internal to your relay I could get a better understanding of your circuit. I could then give you an opinion.

[CheapRussianBeer]

   As mentioned, your schematic is hard to understand because you did not portray what's inside a relay, but what if you try some circuit like

[Marinated]

What voltages do you expect to be across the caps you plan to discharge?

[Caio Negri]

Why are pins 1 and 3 shorted? If those are the coil terminals, no current will flow through the coil and the relay will never switch positions.
Please explain how the relay is connected internally with a shcematic like this:

I'm sure it can me made to work with relays and discrete transistors, but I think a simpler way to do this will be using a comparator op-amp to drive the red LED while the voltage across the cap is higher than the minimum safe voltage.

[krappleby]

hi,

ok thanks for the comments.. let me explain.

a general capacitor discharge is two wires to a resistor, that will slowly discharge a capacitor before working on a circuit, to make sure you dont get electrocuted..

in the example i posted, pins 1&3 are the coil wires.. pin 2, is input and pins 4*5 are the switch.. so if there is no power running through pins 1&2 the switch is set to 5, when power is applied (ie through a discharging cap) the switch activates and switches to pin 4.

i wasn't sure about the short between pin 1&3 and can remove it. i was just thinking about the possible power running through the caps, if its a 40v cap for instance. although the relay runs up to 48v i did it this way so that not all the power was routed through the relay.

as explained earlier the idea is to create a visual aid when discharging caps, as caps dont instantly discharge. so whilst the cap is at a high voltage the red led will light.. when its low enough, (6-8)v the red light will go off. (this is the minimum for the relay to switch)

[krappleby]

here is the relay layout

[Marinated]

There are a couple of problems with the schematic, but first let's step back. It's important to start with a specification.

What range of voltage across the discharging cap do you want to be able to handle? If your relay coil is specified, as you say, up to 48V, then caps with seriously dangerous voltages across them will be seriously dangerous for your relay.

You have 10R in series with the discharging cap, but the relay coil is also in series, with a higher resistance and much lower power rating.  Did you allow for that?  At any rate, do you understand how to calculate the resulting RC time constant, and discharge time for the cap?  If your red LED never stays on for more than a small fraction of a second then the device might not feel as responsive as you intend?

Comments about the switching part of the design could turn out to be a waste of time due to xy problem. Let's nail down some numbers for what the device should do before going further on how it should do it. 

[Caio Negri]

hi,

ok thanks for the comments.. let me explain.

a general capacitor discharge is two wires to a resistor, that will slowly discharge a capacitor before working on a circuit, to make sure you dont get electrocuted..

in the example i posted, pins 1&3 are the coil wires.. pin 2, is input and pins 4*5 are the switch.. so if there is no power running through pins 1&2 the switch is set to 5, when power is applied (ie through a discharging cap) the switch activates and switches to pin 4.

i wasn't sure about the short between pin 1&3 and can remove it. i was just thinking about the possible power running through the caps, if its a 40v cap for instance. although the relay runs up to 48v i did it this way so that not all the power was routed through the relay.

as explained earlier the idea is to create a visual aid when discharging caps, as caps dont instantly discharge. so whilst the cap is at a high voltage the red led will light.. when its low enough, (6-8)v the red light will go off. (this is the minimum for the relay to switch)

If you remove the short between pins 1 and 3 and adjust the resistors (10k for the LED current limiting resistors seems too much) the circuit should work but, as Marinated said, you might exceed the power rating of your relay or not even notice the lights turning on depending of the voltage and capacitance in your cap.

In summary:

• Specify the range of voltages and capacitances you expect to find in your caps
• Calculate the resistance in series based on the average time constant you want https://en.wikipedia.org/wiki/RC_time_constant
• Remove the short between pins 1 and 3. If your coil can't handle the full current, consider adding a current divider
• Adjust the LED current limiting resistors based on the nominal current of your LEDs

[krappleby]

hi and thanks again..

ok...

i am working as a beginner on a small station that will have several functions..

1. variable power supply (using an ATX power supply) (using pentometer)
2. small simple oscilloscope (using a pre purchased self build of amazon)
3. cap discharge circuit
4. 5v usb power supplies

those four things are going to be the main parts of the unit that i am going to build. the reason is funds.. i am just starting out, and don't have a great deal so by doing this in this way, should enable me to have some functionality to the small all in one system. basically a simple modular system.

the design i put above, is for a cap discharge system (i could just use the 10R 50 Watt) resistor on its own to discharge the caps. but that would give no visible details on whether the cap was safe. my idea was to have a system that whilst the cap was dangerous (say above 12v) then there would be a visible warning (red led)

now.. as once this system is built i wont be doing much in the way of working on large power units. however it may come in handy at some point, so i just wanted the option there if needed.

as i said i am really new to this, having only some basic design when i was a kid, my first design was a simple house alarm consisting of 1 9v batter, about 100m of wire, 4 reed switches and a relay and buzzer.. Very simple.

as for cap power, yes there may be some very large ones (power supply). how would i protect the relay, or is there a better way?
would a mosfet be a better solution instead of the relay as i know they can take some pounding with heatsink?

with regard to the 2 10k resistors for the led, i was working on the actual maths to get a figure, and came across a website that stated it was normal to use 10k resistors, thats why they are there.

All in all, im not really fused about having the unit, i just thought it may be better than having three/four seperate things clutering up my desk

 If your red LED never stays on for more than a small fraction of a second then the device might not feel as responsive as you intend?
Why would the red led only stay on for a fraction of a second. if there is power left in the cap, the switch should always be on 4 providing a through route for the red LED>. the red LED should only go out when the power from the cap is reduced to less than what the relay needs to switch?


cheers
keith

[alexanderbrevig]

Have you looked at something like this:

[krappleby]

electronics is confusing.. lol..

thanks for the post alexander, no, i have not..

ok.. im assuming that probe and ground are the two points that the cap will connect to. so if a large cap (lets say for argument sake 100v) is connected fully charged.. that 100v will run through this system.

now it will run through the first resistor, (not sure the strength) and will then have two paths to travel, one is down diodes 567 and 8 and the other is through resistor 2 to the led..

in this plan, will all power run through the diodes except that which is blocked.. ie if its a 5v diode, then everything over that will run through, and only the 5v will be blocked. resulting in that running through the resistor to the led.?

now. why is there also diodes the opposite way.? the current would just run to ground wouldnt it?

sorry as i said new.. lol. lots to learn.

[alexanderbrevig]

On my phone so sorry if it's short and misspelled.

The diodes will allow you to discharge positive and negative sides of a supply. They also provide a voltage differential for the LEDs to light up.
That resistor must handle the rest of the voltage after the for diode drops.
Just ask if you want some maths

[tpowell1830]

hi and thanks again..

--------------------------- SNIP -----------------------------------------
 If your red LED never stays on for more than a small fraction of a second then the device might not feel as responsive as you intend?
Why would the red led only stay on for a fraction of a second. if there is power left in the cap, the switch should always be on 4 providing a through route for the red LED>. the red LED should only go out when the power from the cap is reduced to less than what the relay needs to switch?


cheers
keith

MOST capacitors will discharge through a 10 ohm resistor in a fraction of a second.  They are not like  batteries, holding many units of power over a long period. Your circuit will not work as you describe, since the coil is shorted in your schematic. Plus, even if the coil was not shorted and you are using the capacitor charge to the relay, the time of operation is so short, you would only see a very brief pulse from the LED, if any. Remember, the coil is a wire that is wound up and its' resistance is very low as well, further shorting your capacitor. The 10k ohm resistors in series with the LEDs looks a bit large for most LEDS, I would expect something in the range of a few hundred ohms to a couple thousand ohms at best.

The drawing with all of the diodes stacked in both directions allows you to place a capacitor connected to the probes in any orientation and the diodes will short the cap regardless of direction of charge. The 2 LEDS are also oriented in both directions so as to theoretically turn on regardless of direction of charge as well. This will not drain your cap under test completely due to the forward voltage drop of the diodes stacked up. There could potentially still be ~2 to 4 volts left on the cap.

You mentioned some of the tools that you are trying to cobble together on a budget, however, one of the most essential "tools" that is critical for a hobbyest is knowledge of electrical circuits and devices and how they work. You have a lot of homework to do. Study is free and there are hundreds of tutorials online and thousands of specification sheets available in order to gather the required information to configure devices.

I wish you happy hunting for knowledge, but I also urge you to do this before attempting any endeavors to actually build any devices. This hobby can be very dangerous, if mistakes are made.

[krappleby]

thanks for the replies guys,

tpowel, yes i know about all the training, i am constantly watching youtube and have been for the last 4-5 months (that is how i ended up here, with daves blog), several videos a day, and know a lot more than i did a few months ago, however some things still occasionally confuse me, i assumed that the circuit could run both ways. however, what i also see with that circuit is that the (say 100v) power from the cap can flow straight to the LED's causing them to blow.

now, some of the things that i would ask is..

1. does the .5 watt resistor after the LED's only allow .5 watt through it at any one time? (ie, no matter the watt into it. it will only allow .5 watt through?) this has never been clarified for me. ie is that the maximum it will pass. (from the videos, all you are told points to a .5 watt resistor should be used in a .5 watt circuit etc.. )

All training videos are not detailed enough they don't tell you what you need to know. i know what a resistor is meant to do, i know how to work out the values needed, but no where does it say, you can use a .5 watt resistor in a 20 watt circuit and it will only allow .5 watt through. for that reason i assumed, that i would need to use a 20 watt resistor for that 20 watt circuit)

its hard to find some stuff on the net.. For example. it took me a few months to find out that electricity in a circuit will go via the easiest route possible.

now another question, the diodes, i know they have a overflow feature (cant remember the name ), ie the diode is say 5v, anything above that is let through, however im sure they have a cap too.. if it goes above that it completely decimated the diode, so would the 100v. not destroy them too.


based on question one.. the easiest way to say it is.. if i took power from the wall, and connected it to a .5w resistor, and sent it back.. would the resistor and LEDS not be destroyed. ? in other words, would a large cap not potentialy destroy this circuit ?

as i said before, i coudl just put the big resistor to two wires and use them directly to discharge the cap, however, i just wanted some sort of warning for safety reasons.

cheers
hope my questions are worded correctly

[krappleby]

I wish you happy hunting for knowledge, but I also urge you to do this before attempting any endeavors to actually build any devices. This hobby can be very dangerous, if mistakes are made.

Just so you are aware, i dont plan on creating major power circuits, most of my projects will run around 6 volts max, batteries .. the cap discharger is just for INCASE of need. there may be the odd time when i look into for example changing caps in a digital TV, or other units that use power however that wont be till later anyway.

[tpowell1830]

thanks for the replies guys,

tpowel, yes i know about all the training, i am constantly watching youtube and have been for the last 4-5 months (that is how i ended up here, with daves blog), several videos a day, and know a lot more than i did a few months ago, however some things still occasionally confuse me, i assumed that the circuit could run both ways. however, what i also see with that circuit is that the (say 100v) power from the cap can flow straight to the LED's causing them to blow.

Yes, electronics is very confusing at first, and even much much later when you are well versed.  You said 100v "power", voltage is not a measure of power, watts are a measure of power. Voltage is a measure of electromotive force or potential. The resistors are the limiting factor in the current that will potentially prevent blowing the LED.

Look at it from Ohm's law V=RI, where V is voltage, R is resistance and I is current. So to calculate the limiting current in a series circuit, the voltage drop across the resistor divided by the resistor value will give you the current flowing through the series circuit.

The power in the circuit is VI and watts is the unit. So, W=VI.

now, some of the things that i would ask is..

1. does the .5 watt resistor after the LED's only allow .5 watt through it at any one time? (ie, no matter the watt into it. it will only allow .5 watt through?) this has never been clarified for me. ie is that the maximum it will pass.

No, the wattage rating of the components in a circuit is only the maximum allowable watts before destruction of the component.  This has nothing to do with the amount of wattage  going into the device. Refer back to Ohm's law.

its hard to find some stuff on the net.. For example. it took me a few months to find out that electricity in a circuit will go via the easiest route possible.

now another question, the diodes, i know they have a overflow feature (cant remember the name ), ie the diode is say 5v, anything above that is let through, however im sure they have a cap too.. if it goes above that it completely decimated the diode, so would the 100v. not destroy them too.

The diodes have a forward bias voltage, typically between .4 volts to .9 volts. The specification sheet for a specific diode will tell you what the typical forward bias voltage is for a particular diode. Also, a reverse bias voltage is also specified for each diode. This is sometimes called the zener voltage. This can vary greatly and is specified on the specification sheet for each diode.
Take a look at this video.

based on question one.. the easiest way to say it is.. if i took power from the wall, and connected it to a .5w resistor, and sent it back.. would the resistor and LEDS not be destroyed. ? in other words, would a large cap not potentialy destroy this circuit ?

as i said before, i coudl just put the big resistor to two wires and use them directly to discharge the cap, however, i just wanted some sort of warning for safety reasons.

cheers
hope my questions are worded correctly

Refer back to Ohm's law.
Also, find and study Thevenin, and Kirchoff circuit analysis.

[krappleby]

thank you for your respsones, however it does not answer the questions i had, your comments, are exactly as i have already learned from programs, i know what zener diodes do, i know what normal diodes do, i know what resistors do. you did answer one question which was regarding the wat limit for the resistor, and thank you..

after looking for over an hour on the net, i found this

http://www.digikey.com/Web%20Export/Supplier%20Content/tt-electronics-welwyn-985/docs/tt-electronics-capacitor-discharge-calculator.xls?redirected=1

from this i am surmising, that the power of the entire circuit is based on the bleed resistor itself, so for example if i used a 10,000ohm resistor as in my first example  the circuit would be almost 1 watt throughout the circtuit and would take around 4 and a half seconds to distribute 100v of charge. The circuit would run at 1 watt.

if i used only a 10ohm resistor then the unit would run at 1000watts based on that excell spread sheet and a 195uf cap.

am i right?

based on question one.. the easiest way to say it is.. if i took power from the wall, and connected it to a .5w resistor, and sent it back.. would the resistor and LEDS not be destroyed. ? in other words, would a large cap not potentialy destroy this circuit ?

Or better still, how do i ensure that the circuit is only .5 watts. is that the bleed resistor, because if its 5 watts that resistor will blow. etc.

from this, i surmise that the bleed risistor would need to be large wattage to handle the charge, but using a 10,000 ohm resistor (50 Watt) would reduce the circuit to a 1 Watt circuit.