Low Cost PCB's Low Cost Components

Author Topic: Creating a negative pulse using a capacitor  (Read 716 times)

0 Members and 1 Guest are viewing this topic.

Offline rodrigo_vda

  • Newbie
  • Posts: 4
  • Country: es
Creating a negative pulse using a capacitor
« on: May 20, 2017, 07:37:24 AM »
Hi! I'm new to this forum, so I hope this is in the right section of the forum.
Oh, and I'm Spanish, so I'm sorry if my English is a little bit fuzzy.

So I've been doing electronics for a few years, so I know all the basic stuff, but eventually I find myself thinking about something that looks simple but I don't completely get it. This particular example I've never fully understood how it worked, even when I understood it partially (Ignore the diode, it's a feedback element whose importance is not relevant to my doubt):
I saw this circuit somewhere used to trigger a 555 timer, and as we all know, it needs a negative input as a signal. The 555 is meant to trigger when the LDR(a potentiometer or a digital signal could be used) raises over certain level. I fully understand how the first part works:

  • Before the LDR raises its value, the voltage divider (adjustable by the potentiometer) blocks the transistor, so it doesn't let any current flow from the capacitor to ground, so the capacitor is fully charged, so V(TRG) is 12V.
  • When the potentiometer raises its value, the transistor allows current to flow from the capacitor, so it drops its value to (almost) 0, and it needs some time to charge the TRG side to 5, effectively getting a negative pulse there.
But I still have some doubts, like: Does the capacitor discharge through the NEGATIVE terminal? Whenever I see electrolitic capacitors, the negative terminal is usually directed to ground and it's the positive terminal the one that varies it's state, so the positive terminal charges the capacitor, and then it gets discharged, but in this case I can't see any logical current line going from the positive terminal of the capacitor to GND.

So that's why (paradoxically enough) I understand how it works, but still don't understand how it works. I've tried to google this but I never came into any good explanation of this method of creating a negative pulse. (How would this be called? inverse latch? Because it makes the opposite of a latch, it makes a pulse out of a "constant" change in value).

Thanks a lot for your help :)
« Last Edit: May 20, 2017, 07:40:04 AM by rodrigo_vda »
 

Offline james_s

  • Super Contributor
  • ***
  • Posts: 2600
  • Country: us
Re: Creating a negative pulse using a capacitor
« Reply #1 on: May 20, 2017, 07:51:01 AM »
You always charge and discharge a capacitor through *both* terminals, you have to have a complete circuit. In this case the transistor is pulling the negative terminal of the capacitor down relative to the potential where the positive terminal is connected. You have to consider the potential difference between the two points the capacitor connects, don't think of the charge going in and out through a specific pin, that's not what is happening.
« Last Edit: May 20, 2017, 09:02:22 AM by james_s »
 

Online suicidaleggroll

  • Super Contributor
  • ***
  • Posts: 1401
  • Country: us
Re: Creating a negative pulse using a capacitor
« Reply #2 on: May 20, 2017, 07:55:57 AM »
Before the LDR raises its value, the voltage divider (adjustable by the potentiometer) blocks the transistor, so it doesn't let any current flow from the capacitor to ground, so the capacitor is fully charged, so V(TRG) is 12V.

You sure about that?  If the transistor is off, BOTH sides of the capacitor are at 12V.  Therefore the voltage across the capacitor is 0V, it's completely discharged.  When the transistor turns on, it pulls the negative pin of the cap to ground.  You can't change the voltage across a capacitor instantaneously, so by pulling the neg pin to ground, you're also in effect pulling the pos pin to ground as well.  At that point you have a completely discharged cap, neg connected to ground, pos at 0V, and the 47k resistor feeds current into the pos terminal to start charging up the cap.  When the transistor is shut back off, the other 47k resistor discharges the cap and brings the neg terminal up to 12v to match the pos terminal.

In other words, when the transistor turns on, current flows from the 12V rail, through the 47k resistor on the right, through the cap, through the transistor, to ground.  When the transistor turns off, current flows from the 12v rail, through the left 47k resistor, through the cap, through the right 47k resistor, and back to the 12v rail.  You're essentially discharging the cap through two 47k resistors in series.
« Last Edit: May 20, 2017, 08:00:53 AM by suicidaleggroll »
 
The following users thanked this post: rodrigo_vda

Offline rodrigo_vda

  • Newbie
  • Posts: 4
  • Country: es
Re: Creating a negative pulse using a capacitor
« Reply #3 on: May 20, 2017, 08:34:08 AM »
You sure about that?  If the transistor is off, BOTH sides of the capacitor are at 12V.  Therefore the voltage across the capacitor is 0V, it's completely discharged.

When the transistor turns on, it pulls the negative pin of the cap to ground.  You can't change the voltage across a capacitor instantaneously, so by pulling the neg pin to ground, you're also in effect pulling the pos pin to ground as well.
You're right I understood that but I SAID it wrong, that's what I meant. I understand that both are at 12V and then suddenly they drop to 0V, I understand this, but from this exact moment, is where I got confused.

At that point you have a completely discharged cap, neg connected to ground, pos at 0V, and the 47k resistor feeds current into the pos terminal to start charging up the cap.  When the transistor is shut back off, the other 47k resistor discharges the cap and brings the neg terminal up to 12v to match the pos terminal.

In other words, when the transistor turns on, current flows from the 12V rail, through the 47k resistor on the right, through the cap, through the transistor, to ground.  When the transistor turns off, current flows from the 12v rail, through the left 47k resistor, through the cap, through the right 47k resistor, and back to the 12v rail.  You're essentially discharging the cap through two 47k resistors in series.
Here is where I get confused, because current can't flow through the capacitor itself. I mean, I get how it gets charged, 12V creates current through BOTH resistors, but the left one goes to ground so its voltage at that point stays in 0, but the right one gets charged to 12V. The direction of the current is "downward". But then, when the transistors shuts off, that charge at the right of the capacitor has not left yet, so even when I understand how the left part gets charged until it's voltage equals the right side, I don't understand how can the charges of the right side go. In my head's scheme, the right side of the capacitor only got current in, but never got it out. When does the current get OUT from the right sie of the capacitor? and where does it go? why?

P.S.: Thank you, I appreciate the effort to explain this.
 

Offline rodrigo_vda

  • Newbie
  • Posts: 4
  • Country: es
Re: Creating a negative pulse using a capacitor
« Reply #4 on: May 20, 2017, 08:42:01 AM »
I mean, I know that if you treat a capacitor as a "black box" you can get current through it, but what you're doing there is charging one of the sides of the capacitor positively and the other one negatively, so that current is limited until the maximum capacity of the capacitor is achieved. Once the capacitor is fully charged, it can't let current flow in the same direction, it must discharge. So for every "current input" there must be a "current output" later. I get how the current flows into the right side, but I don't get how/when it flows out of it. As far as I know, that point is never going to be above 12V so it can't have any current flowing TO the 12V line. And it doesn't have any clear path to any lower voltage so... where does it discharge to?
 

Online suicidaleggroll

  • Super Contributor
  • ***
  • Posts: 1401
  • Country: us
Re: Creating a negative pulse using a capacitor
« Reply #5 on: May 20, 2017, 08:53:05 AM »
I get how the current flows into the right side, but I don't get how/when it flows out of it. As far as I know, that point is never going to be above 12V so it can't have any current flowing TO the 12V line. And it doesn't have any clear path to any lower voltage so... where does it discharge to?
Why won't it ever be above 12V?  When the transistor shuts off, you have a complete circuit from the cap +, through one 47k, through another 47k, and finally to the cap -.  Current will flow in this loop until the cap is discharged.  Since the two resistors straddle the 12V rail, the cap pos should shift up to 18V and the cap neg to 6V temporarily, then as it discharges both terminals will approach 12V.
 
The following users thanked this post: rodrigo_vda

Offline rodrigo_vda

  • Newbie
  • Posts: 4
  • Country: es
Re: Creating a negative pulse using a capacitor
« Reply #6 on: May 20, 2017, 09:00:59 AM »
Why won't it ever be above 12V?  When the transistor shuts off, you have a complete circuit from the cap +, through one 47k, through another 47k, and finally to the cap -.  Current will flow in this loop until the cap is discharged.  Since the two resistors straddle the 12V rail, the cap pos should shift up to 18V and the cap neg to 6V temporarily, then as it discharges both terminals will approach 12V.
Oh I get it now! Yeah, that makes a whole lot of sense!!

Finally I understand it!! this has bugged me for about two years!! haha thank you so much!  :clap:
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf