Load between battery and capacitor

[jacau]

I don't have a specific example for this question but I have read decoupling caps have to be right near the VCC pins on an IC so their placement is easy to understand.

But with the larger capacitors, should power from the battery also be behind them from the perspective of the load?

If a large cap was at the end of the power line with the load in the middle, would that setup be less that ideal?

Thanks

[Zero999]

Decoupling capacitor should ideally be as close to the load, as possible. The idea is the voltage drop doesn't drop, as the load draw large current pulses. It's true, larger capacitors are normally a little further away from the load, but they should still be as close as possible.

[JustMeHere]

[Capernicus]

The issue with bypass caps,   is they discharge backwards as well as forwards so watch out for that one.

[David Hess]

The best location for the bulk decoupling capacitors depends on how power distribution is laid out because it may serve as an AC termination.  Usually bulk decoupling capacitors are found at the output of the regulator, where they are also part of the regulator's frequency compensation, and at the power input of each printed circuit board or independent power plane, where they serve as an AC termination.  They are sometimes also found as local decoupling between the power distribution on a circuit board and a specific circuit which is susceptible to or or generates noise.

But with the larger capacitors, should power from the battery also be behind them from the perspective of the load?

Yes.

If a large cap was at the end of the power line with the load in the middle, would that setup be less that ideal?

Yes.

In both cases, the result is extra lead length which adds to the ESR and ESL of the capacitor.  Sometimes you can find circuit layouts where a 4-wire connection is made to the bulk capacitors, as shown below, for exactly this reason.  The original much longer capacitor was wired the same way, and the circuit board has two attachment points for each capacitor lead.

[ogden]

The issue with bypass caps,   is they discharge backwards as well as forwards so watch out for that one.

Sorry, but you better explain what you are talking about here.

But with the larger capacitors, should power from the battery also be behind them from the perspective of the load?

It usually depends on impedance of the battery versus it's capacity. For example there are applications which require infrequent high current bursts of load - then bulk storage capacitor (or even supercap) is needed, to avoid requirement for hi-current battery. One example would be - low power sensor which is RF- transmitting it's state few times per day.

[jacau]

Great, thanks everyone for the information!

[Capernicus]

The issue with bypass caps,   is they discharge backwards as well as forwards so watch out for that one.

Sorry, but you better explain what you are talking about here.

In the case of a ac->dc converter,  the smoothing cap will discharge back into the diode bridge.     And more,  if you want to try out more topologies.

[Zero999]

The issue with bypass caps,   is they discharge backwards as well as forwards so watch out for that one.

Sorry, but you better explain what you are talking about here.

In the case of a ac->dc converter,  the smoothing cap will discharge back into the diode bridge.     And more,  if you want to try out more topologies.

The diode bridge doesn't conduct in the reverse direction, so the smoothing capacitor can't discharge back into it.

[Capernicus]

If a diode is open, it allows current in both directions as long as the net flow isn't reverse biased.   

[tooki]

If a diode is open, it allows current in both directions as long as the net flow isn't reverse biased.   

Bruh… 

[LaryPant]

If a diode is open, it allows current in both directions as long as the net flow isn't reverse biased.   

For any beginner reading this in the future: The quoted statement is completely wrong!

If something is open in electronics then there is by definition no electrical current flow.

It is, however, true that there is current flowing across the p-n junction of an open diode, namely the drift and diffusion currents, which are equal but in opposite direction and the net current is therefore zero. This does NOT mean that electrical current flows through an open diode, but just internal currents which can be explained with semiconductor physics.

In circuit theory an open diode means NO current flow!

[Zero999]

If a diode is open, it allows current in both directions as long as the net flow isn't reverse biased.   

No. Look at the diodes. They all point upwards and will be reverse biased by any voltage applied to the output.

[Capernicus]

I meant the diode being "closed",  when its passing current.

Here is a Walter Lewin drawing->   (but I added a diode to it.)  - its like a H-bridge, except just with resistors.



The diode has more current going through it downward, because of the resistance, so then it conducts,   but when it conducts itll also allow the reverse current at the same time, so it goes back through the diode as well as forward through.

If the diode was upwards -   then all current will be stopped in the cross-line,  but it will still be blasting through the side channels.

[Zero999]

I meant the diode being "closed",  when its passing current.

Here is a Walter Lewin drawing->   (but I added a diode to it.)  - its like a H-bridge, except just with resistors.



The diode has more current going through it downward, because of the resistance, so then it conducts,   but when it conducts itll also allow the reverse current at the same time, so it goes back through the diode as well as forward through.

If the diode was upwards -   then all current will be stopped in the cross-line,  but it will still be blasting through the side channels.

That's not a bridge rectifier and does not represent one.

If you charge up a large capacitor and connect it to the output of a bridge rectifier, + to anode and - to cathode, no current will flow, except a tiny leakage.


Look at the schematic I posted above. The cathodes go to the + output and anodes, to the - output, so current can't flow in to the bridge, from the output, using conventional current flow.

[Capernicus]

Its not supposed to be a bridge rectifier,   but its demonstrating a point that if a diode is forward biased,  then it is open to conduct in both directions, not just forwards.   if its reverse biased, it wont conduct in either direction.

So what I'm meaning is to do with ac->dc conversion -> The "smoothing" capacitor might conduct backwards, if the ac power source has opened for conduction the diode for it.

[Zero999]

Its not supposed to be a bridge rectifier,   but its demonstrating a point that if a diode is forward biased,  then it is open to conduct in both directions, not just forwards.   if its reverse biased, it wont conduct in either direction.

You're contradicting yourself. A diode can't be forward biased and conduct in the opposite direction. If a diode conducts in both directions, then it's broken.

So what I'm meaning is to do with ac->dc conversion -> The "smoothing" capacitor might conduct backwards, if the ac power source has opened for conduction the diode for it.

No, because the diodes will be reverse biased. The smoothing capacitor will not conduct backwards, as you say.  And the last bit "the ac power source has opened for conduction the diode for it." is meaningless.

[Capernicus]

Take it or leave it,   that's what I believe up till now.

[fourfathom]

Take it or leave it,   that's what I believe up till now.

Are you perhaps thinking about a diode biased to pass or block an AC signal?  Such as a PIN diode?  In this case, yes, a signal can flow "backwards" (source and load can be on either side of the diode), but still, the net current flow is forward when the diode is "on", and when the diode is reverse-biased ("off") there is no current or signal flow (ignoring leakage, parasitic capacitance, etc.)

[Zero999]

Take it or leave it,   that's what I believe up till now.

Well you're belief was mistaken. If there's no load on the rectifier, current only flows into the capacitor, as it charges up, then falls to zero.

[Capernicus]

My belief isnt mistaken,   but yes, it looks like you are correct that it doesnt happen through a diode bridge from the smoothing cap, on my own re-inspection.   I hadn't looked at it that hard before, but it looks like one of the diodes conducts, when the other doesnt,  and that one that isnt conducting stops the cap from self-discharging behind.

But it can happen in different machines than this,  there's always the possibility that a bypass cap is going to charge back down the machine as well as in front.

It comes up when you have a "multi-rail"  when your powering more than one unit off the same battery,  it can happen there too, and can just go straight backwards through the diode, given the specific situation.

[fourfathom]

it can happen there too, and can just go straight backwards through the diode, given the specific situation.

Can you please describe this specific situation?  Something to do with minority carrier lifetime?  Yes, this can allow a small amount of reverse current to *briefly* flow.

[Capernicus]

it can happen there too, and can just go straight backwards through the diode, given the specific situation.

Can you please describe this specific situation?  Something to do with minority carrier lifetime?  Yes, this can allow a small amount of reverse current to *briefly* flow.

It can happen on your rail,  negative and positive.    Maybe there is a way to correctly guard a rail with diodes (I havent thought about it enough yet.),  but theres also incorrect ways, which is what I'm talking about.

[Doctorandus_P]

but its demonstrating a point that if a diode is forward biased,  then it is open to conduct in both directions, not just forwards.

Both directions but not forward? How many directions does that leave open?

...  if its reverse biased, it wont conduct in either direction.

The only difference between "forward" and "reverse" -biased is the sign of the applied voltage. So you can simplify the whole mumble jumble by stating that a diode only conducts current in one direction.

[fourfathom]

it can happen there too, and can just go straight backwards through the diode, given the specific situation.

Can you please describe this specific situation?  Something to do with minority carrier lifetime?  Yes, this can allow a small amount of reverse current to *briefly* flow.

It can happen on your rail,  negative and positive.    Maybe there is a way to correctly guard a rail with diodes (I havent thought about it enough yet.),  but theres also incorrect ways, which is what I'm talking about.

Then please describe an incorrect way, where the diode is conducting in the reverse direction.  A "rail" is just another node in a circuit.  Can you be more specific?