Reply
 
Thread Tools Display Modes
  #1  
Old 05-13-2019, 12:23 AM
dstarfire is offline
Guest
 
Join Date: Oct 2009
Location: Tacoma, WA; USA
Posts: 1,536

Why do so many circuits have resistors and capacitors going straight from + to - side?


Been watching some electronics videos lately that include circuit diagrams of real-world devices (mainly UK devices). In almost every circuit there's a resistor or capacitor creating additional paths from positive to negative (sides of the power supply) that don't seem to really do anything.

What purpose/function do these paths serve?
Don't these alternate paths consume power?
Why don't they function as a short-circuit?

Just to clarify, I DO understand that electricity flows down parallel paths in proportion to their resistance (there's no "only" in the adage about electricity following the path of least resistance).
__________________
Dion Starfire, grammar atheist.
  #2  
Old 05-13-2019, 01:03 AM
snfaulkner's Avatar
snfaulkner is online now
Guest
 
Join Date: May 2015
Location: 123 Fake Street
Posts: 7,443
Are you talking about a pull up(or down) resistor?

https://en.wikipedia.org/wiki/Pull-u...or?wprov=sfla1
__________________
It may be because I'm a drooling simpleton with the attention span of a demented gnat, but would you mind explaining everything in words of one syllable. 140 chars max.
  #3  
Old 05-13-2019, 01:03 AM
Saffer is offline
Guest
 
Join Date: Feb 2009
Posts: 380

Why do so many circuits have resistors and capacitors going straight from + to - side?


Im not sure about a resister but this is exactly what you want to do with a capacitor. The capacitor doesn’t allow current to flow right through. Instead it builds up charge and releases it during dips, to smooth the flow of current. Think of it like a high speed battery.


Sent from my iPhone using Tapatalk

Last edited by Saffer; 05-13-2019 at 01:04 AM.
  #4  
Old 05-13-2019, 01:38 AM
Leaffan's Avatar
Leaffan is offline
Member
 
Join Date: Aug 2005
Location: Ottawa
Posts: 24,265
Yes. Sounds like smoothing capacitors to me.
  #5  
Old 05-13-2019, 01:57 AM
engineer_comp_geek's Avatar
engineer_comp_geek is offline
Robot Mod in Beta Testing
Moderator
 
Join Date: Mar 2001
Location: Pennsylvania
Posts: 23,847
If we are talking about DC circuits, then the capacitor is almost always for filtering and voltage stability. Think of all of the little wires and connections as resistors. If a component draws an excessive amount of current suddenly, the voltage at that component will drop, possibly enough to cause a malfunction. Or, worse, the voltage change will cause an oscillation inside the part that feeds back into the current draw making the oscillation worse and suddenly your circuit is howling like a banshee.

Capacitors are small energy storage devices that will help to smooth out those fluctuations.

There are two basic types of capacitors. You have large electrolytic capacitors, which are used for heavy duty bulk filtering. These have large capacitance values, and they look like little tin cans, because you can basically think of them as kind of an electrical fruit roll-up. They have two metal foil sheets with a bunch of goo (the electrolyte) between them, and the whole thing is rolled up and shoved into a metal case. Electrolytic capacitors are great for high capacitance value, but at high frequencies they suck. So for higher frequencies, we use chip capacitors or small ceramic capacitors (mostly chip capacitors these days). Often in power supplies you will have big electrolytic capacitors paralleled with small chip or ceramic capacitors. The electrolytics do the bulk filtering, and the smaller capacitors do the high frequency stability.

In order to keep the power supply stable at each integrated circuit, I personally (as well as many other electronics designers) will just out of habit put a small capacitor between the positive and negative voltages right at the chip, keeping the circuit traces very short.

Sometimes you'll put a resistor across the power supply to give capacitors (and maybe inductors or relays if you have any) a place to dump their stored energy once the power is shut off. Otherwise that stored energy can go elsewhere in the circuit, possibly somewhere bad. Often instead of just a simple resistor you'll have a reverse biased diode in series with a resistor. The advantage of this is that no current is being wasted heating up the resistor while the power is on.

Similarly, high voltage circuits will often have bleeder resistors across energy storage devices, for similar reasons. Old cathode ray type televisions usually had bleeder resistors on the tube, and microwave ovens usually have a bleeder resistor across the big capacitors. Once the power is shut off, the energy in the capacitors gets turned into heat in the resistors. Without the resistors, those capacitors can store energy for quite some time, and could give you a nasty shock if you were to open up the microwave oven and touch them.

For AC circuits, you also have what is called a Metal Oxide Varistor, or MOV. Sometimes you'll have these in DC circuits for similar reasons. An MOV has a fairly high resistance until you exceed its turn-on voltage, at which point its resistance drops significantly. The point of using these on power supplies is to clamp down on any over-voltages, thus suppressing electrical spikes by shorting them out quickly. The MOV is often destroyed when this happens, but the idea is you kill the MOV so that the circuit it is protecting survives.

If you have a question about anything specific, give us a link to the video or web page you are looking at.

Last edited by engineer_comp_geek; 05-13-2019 at 02:00 AM.
  #6  
Old 05-13-2019, 04:55 AM
septimus's Avatar
septimus is offline
Guest
 
Join Date: Dec 2009
Location: The Land of Smiles
Posts: 19,119
Note that electrolytic capacitors are not symmetric — both the circuit symbol and the physical device will indicate which connection is for +, which for -.

This was brought home to me painfully when I worked as a contractor for a low-budget manufacturer of electronic subsystems. After installing many of his circuit boards into a very expensive system and powering on we started hearing "Pop ... Pop Pop" — sounds like popcorn popping loudly. His assemblers had installed all the electrolytic capacitors backwards. It was particularly humiliating since it showed that Mr. Fly-by-Night hadn't even bothered to "burn in" those boards.
  #7  
Old 05-13-2019, 10:05 AM
casdave is offline
Member
 
Join Date: Mar 2000
Posts: 8,268
Also worth noting that in R.F circuits and t.v sets you often have capacitors of low values placed in this way to provide r.f phase coupling.

It is not unusual to have larger electrolytic capacitors used to assist in maintaining volt rail stability, however in switching circuits - such as found in t.v sets there is usually a high frequency component associated with the effects of high speed switching (relating to edge frequency) which needs to be grounded to prevent interference.

It might seem odd that a 50uF capacitor is in parallel with a 10nF cap but its because at higher frequencies electrolytic capacitors tend to have quite a high level of inductance which would effectively prevent them from allowing those high frequency signals through them.

In some circuits, local oscillators rely on a number of capacitor/resistor bridges to achieve a phase change in order to ensure positive feedback - typically such a network will enable a phase change of around 60 degrees - so you need a network that will give you 180 degrees to get that full positive feedback. Its common practice for one of those phase changes to take place across the voltage rails.

One of those unwritten rules of home electronics -and a variation of Murphys law is Amplifiers will tend to oscillate, and oscillators won't.
  #8  
Old 05-13-2019, 02:32 PM
dstarfire is offline
Guest
 
Join Date: Oct 2009
Location: Tacoma, WA; USA
Posts: 1,536
Quote:
Originally Posted by snfaulkner View Post
Are you talking about a pull up(or down) resistor?

https://en.wikipedia.org/wiki/Pull-u...or?wprov=sfla1
No, but that's something else that has puzzled me in circuit diagrams. Those are usually in what I think of as the 'working' part of a circuit.

What I'm talking about are the capacitors and resistors that sort of off to the side and run directly from the positive end to the negative end. It looks like taking a resistor or capacitor and connecting it's leads directly to either side of a battery.

When that capcitor discharges, isn't it creating a temporary short circuit?
__________________
Dion Starfire, grammar atheist.
  #9  
Old 05-13-2019, 02:47 PM
Dobbs is offline
Charter Member
 
Join Date: Apr 2004
Posts: 401
Quote:
Originally Posted by dstarfire View Post
No, but that's something else that has puzzled me in circuit diagrams. Those are usually in what I think of as the 'working' part of a circuit.

What I'm talking about are the capacitors and resistors that sort of off to the side and run directly from the positive end to the negative end. It looks like taking a resistor or capacitor and connecting it's leads directly to either side of a battery.

When that capcitor discharges, isn't it creating a temporary short circuit?
Might I trouble you for a picture?
  #10  
Old 05-13-2019, 03:00 PM
beowulff's Avatar
beowulff is offline
Member
 
Join Date: May 2001
Location: Scottsdale, more-or-less
Posts: 16,437
Quote:
Originally Posted by dstarfire View Post
No, but that's something else that has puzzled me in circuit diagrams. Those are usually in what I think of as the 'working' part of a circuit.

What I'm talking about are the capacitors and resistors that sort of off to the side and run directly from the positive end to the negative end. It looks like taking a resistor or capacitor and connecting it's leads directly to either side of a battery.

When that capcitor discharges, isn't it creating a temporary short circuit?
I’ve never seen a resistor across a battery, although high-value “bleeder" resistors are put across filter capacitors (which may be directly across the power supply), for safety.

Capacitors can be put directly across the battery for two main reasons: bulk filtering and current reservoirs. They will draw a lot of current when they are charging, but that is not usually a problem. If the designer thinks that the inrush current is too high, they will use a current limiter.
  #11  
Old 05-13-2019, 03:27 PM
dstarfire is offline
Guest
 
Join Date: Oct 2009
Location: Tacoma, WA; USA
Posts: 1,536
Quote:
Originally Posted by Dobbs View Post
Might I trouble you for a picture?
Here you go. I've circled the portion in red.
http://i.imgur.com/vfQ2fNp.png

This is a screencap from this youtube video: https://www.youtube.com/watch?v=UerFD3AgJHE

Quote:
Originally Posted by engineer_comp_geek View Post
In order to keep the power supply stable at each integrated circuit, I personally (as well as many other electronics designers) will just out of habit put a small capacitor between the positive and negative voltages right at the chip.
That sounds a lot like what I've seen. A small capacitor running between the positive and negative rails, presented as so common and expected you'd be mildly surprised if it weren't there.
__________________
Dion Starfire, grammar atheist.
  #12  
Old 05-13-2019, 03:48 PM
beowulff's Avatar
beowulff is offline
Member
 
Join Date: May 2001
Location: Scottsdale, more-or-less
Posts: 16,437
Quote:
Originally Posted by dstarfire View Post
Here you go. I've circled the portion in red.
http://i.imgur.com/vfQ2fNp.png

This is a screencap from this youtube video: https://www.youtube.com/watch?v=UerFD3AgJHE



That sounds a lot like what I've seen. A small capacitor running between the positive and negative rails, presented as so common and expected you'd be mildly surprised if it weren't there.
That’s the main filter cap, right after the bridge rectifier.
  #13  
Old 05-13-2019, 05:22 PM
engineer_comp_geek's Avatar
engineer_comp_geek is offline
Robot Mod in Beta Testing
Moderator
 
Join Date: Mar 2001
Location: Pennsylvania
Posts: 23,847
Quote:
Originally Posted by dstarfire View Post
Here you go. I've circled the portion in red.
http://i.imgur.com/vfQ2fNp.png
Keep in mind that the little box he drew on the left isn't a battery. It's a full wave rectifier. He probably drew it as a box because it's a single component, but it's really 4 diodes inside a single part.

The power is an AC sine wave coming in from the left.

This seems like a decent description of a full wave rectifier.
https://www.elprocus.com/full-wave-r...orking-theory/

The thing labeled "bridge rectifier" in this picture (the 4 diodes) is the full wave rectifier that he drew as a box in the video.
https://www.elprocus.com/wp-content/...2015/01/21.jpg
  #14  
Old 05-13-2019, 05:44 PM
The Tooth is offline
Guest
 
Join Date: Dec 2001
Location: Calgary, Alberta
Posts: 4,673
Quote:
Originally Posted by dstarfire View Post
Here you go. I've circled the portion in red.
http://i.imgur.com/vfQ2fNp.png

This is a screencap from this youtube video: https://www.youtube.com/watch?v=UerFD3AgJHE



That sounds a lot like what I've seen. A small capacitor running between the positive and negative rails, presented as so common and expected you'd be mildly surprised if it weren't there.
I would be. They're there to keep the power supply clean by providing a path to ground for high-frequency noise; their reactance is inversely proportional to frequency.
__________________
"It would never occur to me to wear pink, just as it would never occur to Michael Douglas to play a poor person." - Sarah Vowell
  #15  
Old 05-13-2019, 08:50 PM
Melbourne is offline
Guest
 
Join Date: Nov 2009
Posts: 4,851
Quote:
Originally Posted by dstarfire View Post
Here you go. I've circled the portion in red.
http://i.imgur.com/vfQ2fNp.png
So, capacitors yes, and resistors less common. I can think of several reasons to put a resistor across rails, but only in relay circuits would I would be surprised by their abscence.
  #16  
Old 05-13-2019, 10:08 PM
Francis Vaughan is offline
Guest
 
Join Date: Sep 2009
Location: Adelaide, Australia
Posts: 5,000
I had a feeling it was BigClive being cited. He has a penchant for pulling apart evil little mains powered devices. Things with direct rectified mains. Any of these is going to feature a bleeder resistor to bleed down the charge in the smoothing capacitor. At least they should. Low voltage circuits don’t tend to have or need them. This means you will see an over representation of circuits with a rail to rail resistors in his examples.
  #17  
Old 05-14-2019, 05:29 PM
dstarfire is offline
Guest
 
Join Date: Oct 2009
Location: Tacoma, WA; USA
Posts: 1,536
And here's one with resistors going from positive to negative rails. The varistor was explained, so I understand what the one before the rectifier is for. This is for a set of LED grow lights that connect directly to household AC (or "mains" as Brits call it).

http://i.imgur.com/I761svJ.png
__________________
Dion Starfire, grammar atheist.
  #18  
Old 05-14-2019, 08:19 PM
Crafter_Man's Avatar
Crafter_Man is offline
Guest
 
Join Date: Apr 1999
Location: Ohio
Posts: 11,258
Quote:
Originally Posted by dstarfire View Post
And here's one with resistors going from positive to negative rails. The varistor was explained, so I understand what the one before the rectifier is for. This is for a set of LED grow lights that connect directly to household AC (or "mains" as Brits call it).

http://i.imgur.com/I761svJ.png
Here's the issue: if you unplug a device from the wall receptacle, the DC voltage on the right side of the bridge rectifier will decrease to around 0 V. (At least you hope this happens.) Why? Because capacitor(s) located throughout the circuit will be discharging into things like resistors, diodes, transistors, light bulbs, motors, etc., resulting in the voltage on the right side of the bridge rectifier going to 0 V. How quickly this voltage decreases depends on a number of factors (number of caps, size of caps, resistances of loads, etc.). Suffice to say, the voltage may decrease in a couple milliseconds, or it may take minutes; it all depends on the circuit.

So... let's say it takes 10 seconds for the DC voltage on the right side of the bridge rectifier to decrease to around 0 V after you unplug the device. If you touch the two prongs on the plug immediately after you unplug it, will you receive a shock?

Well, if each diode in the bridge rectifier has the exact same leakage current characteristics when reverse biased, then - at least theoretically - there won't be a voltage between the two prongs. But in real life this is never the case, and thus there will be a voltage between the two prongs when there is still a voltage on the right side of the bridge rectifier. (The voltage on the prongs will have a fairly high source impedance, which helps matters. But we shouldn't count on it.) The magnitude of the voltage on the prongs will depend on the whatever the voltage is on the right side of the bridge rectifier at the instant you touch the prongs, and it will depend on the mismatch in the leakage current characteristics of the diodes.

You can't do a whole lot about the latter. But, you can control how quickly the voltage "dies down" on the right side of the bridge rectifier by simply sticking a resistor across it. The resistance value of the resistor isn't super critical, but it shouldn't be too high (else it won't be effective at reducing the voltage bleed-down time) and it shouldn't be too low (else you will have to use a high wattage resistor that gets hot and it will make your device inefficient during normal use).
  #19  
Old 05-15-2019, 02:17 PM
dstarfire is offline
Guest
 
Join Date: Oct 2009
Location: Tacoma, WA; USA
Posts: 1,536
Quote:
Originally Posted by Crafter_Man View Post
You can't do a whole lot about the latter. But, you can control how quickly the voltage "dies down" on the right side of the bridge rectifier by simply sticking a resistor across it. The resistance value of the resistor isn't super critical, but it shouldn't be too high (else it won't be effective at reducing the voltage bleed-down time) and it shouldn't be too low (else you will have to use a high wattage resistor that gets hot and it will make your device inefficient during normal use).
Okay, that makes sense. Do you compare it to the rest of your circuit?

If your bleed-down resistor is an easier path than the working part of your circuit (i.e. the part that does the thing you want done) than most of your current is going to be wasted heating up that resistor.

Also, do you have to account for losses from this resistor when determining your power requirements
? While most of the devices I've seen torn down appear to function okay with reduced current I imagine there could be situations or devices where power isn't as abundant or the other parts require a certain level of power, or it's essential that the device doesn't drop below a certain level of operation.
__________________
Dion Starfire, grammar atheist.
  #20  
Old 05-15-2019, 02:36 PM
engineer_comp_geek's Avatar
engineer_comp_geek is offline
Robot Mod in Beta Testing
Moderator
 
Join Date: Mar 2001
Location: Pennsylvania
Posts: 23,847
Quote:
Originally Posted by dstarfire View Post
If your bleed-down resistor is an easier path than the working part of your circuit (i.e. the part that does the thing you want done) than most of your current is going to be wasted heating up that resistor.
Keep in mind that most real-world power sources are voltage sources, not current sources. In other words, the power source will maintain a constant voltage and will let the current vary depending on the load. Adding a resistor increases the load, but since the resistor is in parallel with the rest of the load (the rest of the circuitry), it does not change the current going through the rest of the circuitry.

When power is removed, the resistor gives any stored up energy a place to go where it can be easily dissipated. This prevents that stored up energy from going somewhere you don't want it to, i.e. someone getting shocked or another component being stressed as the stored up energy goes through it.

Quote:
Originally Posted by dstarfire View Post
Also, do you have to account for losses from this resistor when determining your power requirements?
Yes.

Note that in the circuit you are looking at, the resistance is pretty high (1 meg). There isn't going to be a lot of current going through it, but it is going to generate a small amount of waste heat. So yes, you have to add that to the power requirements and also to the cooling requirements for your circuit. Neither of those is probably significant in this case, but in other cases it could be.
Reply

Bookmarks

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is Off
HTML code is Off

Forum Jump


All times are GMT -5. The time now is 04:49 AM.

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2019, vBulletin Solutions, Inc.

Send questions for Cecil Adams to: cecil@straightdope.com

Send comments about this website to: webmaster@straightdope.com

Terms of Use / Privacy Policy

Advertise on the Straight Dope!
(Your direct line to thousands of the smartest, hippest people on the planet, plus a few total dipsticks.)

Copyright © 2018 STM Reader, LLC.

 
Copyright © 2017