Capacitor circuit and current flow

[DerStrom8]

This is not a "normal circuit." It is a very clever circuit.
You will notice no-one is even close to describing its operation.
It does not matter what load the circuit is supplying, the operation of the circuit can be described when supplying a LOW, MEDIUM and HIGH impedance or resistance load and the features of the circuit can be explained.
This has not been done correctly and no-one has even the slightest idea of how it operates.

Well, to be honest, you're not exactly giving crystal clear answers yourself....

I don't mean to be rude, but I'm getting the feeling that YOU don't even know how it operates! Your "answers" are very dodgy and broad, and do not explain the operation at all. You're talking everyone down for not knowing how it works, and when someone asks you to explain to us "dump people", you simply keep going with your insults. IMHO you're being nothing but a hindrance until you start answering with clarity.

With all due respect

 

[colin55]

I know exactly how it works. I designed it and have written it up on my website. I am going to leave it for 3 days to see if anyone can come up with any sort of discussion on its operation.

 

[DerStrom8]

I know exactly how it works. I designed it and have written it up on my website. I am going to leave it for 3 days to see if anyone can come up with any sort of discussion on its operation.

Colin, this thread wasn't started to test people to see if they can figure out how it works. It was started so that people could help the OP figure out why it works. You're really not helping much. Even a link to your site would be appreciated....

 

[colin55]

Three days are not up.

 

[DerStrom8]

Three days are not up.

Okay, I've just about had it. Why comment on the thread when you're not going to help? This is not a forum for quizzing the other members! If you feel you have to do that, start your OWN thread in the members' lounge. Don't hijack someone ELSE'S thread and start teasing him with useless replies! If you know the answer, you can either help the OP now, or stop posting. I'm not going to participate in your "quiz". I don't care if you think you know better than anyone else here. Believe what you like--it doesn't bother me. But don't deprive the OP from an answer he requested and deserves! This forum is here to HELP PEOPLE, not to show them that "you're smarter than everyone" by quizzing them!

 

[duffy]

Yeah, this isn't "clever" so much as it's "nonsense". By the way - these model railroad "point motors" are just solenoids.

And like I said before, this one's easy enough to prove:
Remove the transistor, remove the diode with the "?", and replace that last 1N4004 on the right with a piece of wire, and it work just the same.

 

[JimB]

OK, as the apparent cause of all this invective comment, I will hold my hand up and say that my judgement that is was a "truely" awfull circuit was based on my initial mis-reading of the circuit by missing the series connected 1N4004.

Maybe at the time I should have said that it was a bizarre circuit with no obvious application.

Now that I see how and why it is used (from the link provided by edeca), I will concede that the circuit does have merit.

JimB

 

[colin55]

JimB Thank you.

The thousands of model railroad hobbysits would not have bought the kit if it did not have a purpose and did not work.

duffy You just do not know what you are talking about.

It's not about the two circuits working the same. It's about WHY the circuit was designed.

 

[duffy]

When the points switch is open, node 8 in the circuit floats high, allowing the base of the darlington to go high, turning on the transistor and charging the caps. When the switch connects to the solenoid, it pulls the same node lower than the emitter, so the transistor is off, the caps discharge through the coil, and some holding current is maintained through the parallel resistors. The 1N4004 with the "?" is probably there for backspike suppression (but please don't tell me the reason you put it ahead of the final 1N4004 was to recover energy and recharge the caps).

It it DOES have some merit. It should have a faster charge time on the caps than the simpler half-wave rectified RC circuit I'm suggesting it can be replaced with. But still - a "very clever circuit" that "no-one has even the slightest idea of how it operates" is a bit much.

 

[colin55]

Why didn't you come up with this explanation in post #3?

Now, from all the other "smart" posters, why don't you come up with a better and simpler circuit that protects the point motor from being burnt-out.

 

[duffy]

Didn't know what it was for, none of us did. Looked like some fool screwed up a current regulator. My apologies.

 

[edeca]

I already linked to the original site in post #9.

 

[ljcox]

I have just noticed this thread.

Yes Colin55, it is a clever circuit.

Let me see if I can explain it. I know little about model railways, so I don't know what the solenoid does, but that does not matter.

When the power is first applied, the 1000 uF caps are discharged, D3 (the horizontal one at the output) is off, there is no resistance across the output (presumably no solenoid at this point in time) & so the transistor is on. Note that D3 prevents a short between the base & emitter at this point since the base is more positive than the emitter.

Thus the Caps charge via the transistor until the base/emitter voltage falls to the point where the transistor goes off.

Because of the LED current, the cap voltage will decay so, the transistor will remain on (slightly) in order to maintain the voltage

Now, when the soleniod (for whatever reason) is applied across the output, the caps discharge into it and presuambly cause it to operate briefly (ie. until the current through its coil falls below its hold level)

The solenoid (because of its low impedance) keeps the transistor off & so the only current flowing through it is that through the parallel resistors (I estimated about 37 mA peak).

Diodes D2 & D3 provide a path for the solenoid's back EMF, which according to Lenz's Law, will be a negative voltage.

If D2 was not provided, the back EMF would be applied to the caps & emitter thus possibly doing damage.

 

[MrAl]

The circuit is designed to connect to a 4 to 6 ohm load.

Hello,

If you want someone to know how your circuit works you at least have to tell them what kind load it is to drive. Again you throw a curve in declaring it to be a "4 to 6 ohm load". From the other posts i can see now it is supposed to be a solenoid. That's not a 4 to 6 ohm load, but has lots of inductance too.
To fully understand this however, it would be nice to know the typical operating current of the solenoid, the current to pull in and the current to hold, etc. Knowing the coil resistance is 4 to 6 ohms helps too.

Also in a cloud of obscurity is the chosen title for the circuit, which immediately makes the circuit sound like something totally different. Note:
1. It does not discharge a capacitor, it drives a solenoid, so "Capacitor Discharge Circuit" is not very appropriate for the name of the circuit.
2. Capacitor discharge circuits are used to discharge external capacitors, usually either high voltage or of large capacity or both.

Since it was made for and is used for driving solenoids, it should have been called "Solenoid Driver" at the least, or "Safe Solenoid Driver", or "Protected Solenoid Driver" or something like that. People would have a MUCH better idea what it was that way without going drudging through all the hedges first.

It does not matter what load the circuit is supplying, the operation of the circuit can be described when supplying a LOW, MEDIUM or HIGH impedance (or resistance) load and the features of the circuit can be explained.

Not really. A solenoid is not a resistor. It contains inductance which greatly impedes the flow of current during the first instants of time, and generates back emf that must be accounted for in the design.

Now you see why i asked what the load really is.

 

[colin55]

"the back EMF would be applied to the caps"
Where do you get back EMF from ???????????

"It contains inductance which greatly impedes the flow of current during the first instants of time, and generates back emf that must be accounted for in the design."

Both of you have no idea what you are talking about.

The coil simply produces a back voltage that slows down the inrush current.

This is not technically "back EMF that destoys a capacitor diode or other semitconductor device."
That's why I knew you didn't know how the circuit works.

 

[DerStrom8]

"the back EMF would be applied to the caps"
Where do you get back EMF from ???????????

"It contains inductance which greatly impedes the flow of current during the first instants of time, and generates back emf that must be accounted for in the design."

Both of you have no idea what you are talking about.

The coil simply produces a back voltage that slows down the inrush current.

This is not technically "back EMF that destoys a capacitor diode or other semitconductor device."
That's why I knew you didn't know how the circuit works.

The back EMF appears when the voltage to the coil shuts off. It is the same idea behind needing a snubber circuit or freewheel diode on a motor or coil.

 

[colin55]

The voltage gradually dies to 15mA
It NEVER SHUTS OFF.

Look at the circuit and try to work out how it works.

 

[ljcox]

"the back EMF would be applied to the caps"
Where do you get back EMF from ???????????

"It contains inductance which greatly impedes the flow of current during the first instants of time, and generates back emf that must be accounted for in the design."

Both of you have no idea what you are talking about.

The coil simply produces a back voltage that slows down the inrush current. Yes

This is not technically "back EMF that destoys a capacitor diode or other semitconductor device." Wrong!
That's why I knew you didn't know how the circuit works.

When you switch off the current flowing through an inductor, there is a back EMF due to the decaying flux. Have you heard of Lenz's Law?

V = L di/dt.

If there is no protection against (eg. a diode) the back EMF, then there will be a large voltage spike which can damage components.

In the case of your circuit, the capacitors will slow the decay of the flux & thus the back EMF will be relatively small provided that the caps have low ESR.

 

[ljcox]

The voltage gradually dies to 15mA
It NEVER SHUTS OFF.

Look at the circuit and try to work out how it works.

You have made a typo in this. Please tell us whether you're talking about voltage or current.

 

[DerStrom8]

You have made a typo in this. Please tell us whether you're talking about voltage or current.

Hahaha!