Author Topic: Basic circuit questions  (Read 248 times)

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garethashenden

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Basic circuit questions
« on: October 29, 2019, 09:49:53 AM »
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Electricity isn’t my strength and I have some questions that are probably pretty basic to some of you.
I’m currently installing a  Loksound 5 Micro in an Arnold SW1. I got it installed and it works,  it definitely need a capacitor. The diagram on ESU’s website says to include a resistor in the capacitor circuit, is that actually needed?

Now for the really basic questions. Does it matter which pole of an LED has the resistor attached or is its presence somewhere in the circuit good enough? Can I attach one resistor to the common lighting wire or do I need one for each headlight?

C855B

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Re: Basic circuit questions
« Reply #1 on: October 29, 2019, 10:00:01 AM »
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Yes, the resistor is absolutely necessary with the capacitor, otherwise it will want to "dump" its stored energy all at once, defeating the keep-alive function.

Resistor on the LED is fine on either pole (a/k/a "lead", pronounced "leed"). If the headlights or other lighting are never on at the same time you can use a common resistor, but if you set the CVs for dim-in-reverse, you won't get the lighting levels you expect. Use one resistor per LED to be safe.
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ednadolski

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Steveruger45

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Re: Basic circuit questions
« Reply #3 on: October 29, 2019, 12:56:05 PM »
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Electricity isn’t my strength and I have some questions that are probably pretty basic to some of you.
I’m currently installing a  Loksound 5 Micro in an Arnold SW1. I got it installed and it works,  it definitely need a capacitor. The diagram on ESU’s website says to include a resistor in the capacitor circuit, is that actually needed?

Now for the really basic questions. Does it matter which pole of an LED has the resistor attached or is its presence somewhere in the circuit good enough? Can I attach one resistor to the common lighting wire or do I need one for each headlight?

Page 43 of the loksound 5 manual shows a 2200uF cap with not only a resistor but also a diode.
The resistor and diode would be needed for such a large capacitor.
It has been common practice to install stay alive capacitors without the resistor and diode Iro between about 400 - 600 uF  in the v4 micro and select.  These are not true keep alive or power packs and will not power the whole locomotive for any noticeable time.  They do help with keeping the sound going at a momentary power interruption.
LED’s.... in a series circuit it doesn’t matter on the sequence of the components, you can put your resistor on either positive or negative end of the LED.  The Positive and negative orientation of the LED is important in the circuit though, the LED is polarized, if you get it backwards it won’t work.
« Last Edit: October 29, 2019, 02:02:57 PM by Steveruger45 »
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peteski

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Re: Basic circuit questions
« Reply #4 on: October 29, 2019, 06:25:43 PM »
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Yes, the resistor is absolutely necessary with the capacitor, otherwise it will want to "dump" its stored energy all at once, defeating the keep-alive function.

I wouldn't say "absolutely necessary", but it is good idea to limit the current.  Actually it is the initial capacitor charge current that needs limiting, not its discharge current.  The available discharge current of a keep-alive circuit should be as high as possible.

A capacitor that is not charged will appear a a dead short when the power is first applied to it.  The ideal capacitor that is, but even real-world capacitors will be close to a dead short. Then as the capacitor charges up, the current drops until it charges up to the same voltage as the DC power source. At that time no current will flow either way.  Because the initial current is so high, if we were to hook a large-value cap directly across the output of the decoder's bridge rectifier, that would place a strain on the rectifier (because the initial charge current would be so high).  That is why it makes sense to limit the charging current by installing a resistor in series with the keep-alive cap.

But when the track power is interrupted, and the capacitor begins to supply the power to the loads (decoder, motor and headlights), it has to provide all the current it can for those devices to stay running. The loads all have internal resistances which will only us as much current as they need.  If you limited the cap's discharge current too much (with a series connected resistor), that would result in voltage drop across that resistor. If the voltage drop was too high, then the load devices would start stop working.

Many  keep-alive circuits use both, a resistor and a reverse-connected diode in parallel with the resistor.  This way when the cap is charging, the diode is not conducting and the resistor is limiting the charging current. When the capacitor is called to supply power, the current flowing in the opposite direction will bypass the resistor, so a higher current can be supplied to the loads on the decoder.  This is routinely done on keep-alives with very high capacitance (thousands or hundreds of thousands of micro Farads).

Some more basic keep-alives which have lower value caps can work with just a properly chosen resistor in series with the cap.  Its value will be high enough to prevent excessive initial charging current, yet low enough not to drop too much voltage to supply some useful power to the decoder/motor/headlights when track power is lost.

Then sometimes a coil is used in series with the cap instead of a resistor. Coil's electric properties are opposite of a capacitor's (its resistance is high when the current is changing - like when a capacitor is first getting charged).  There are also some hybrid designs using coils, diodes, and resistors to control the current in keep-alives.

Quote
Resistor on the LED is fine on either pole (a/k/a "lead", pronounced "leed"). If the headlights or other lighting are never on at the same time you can use a common resistor, but if you set the CVs for dim-in-reverse, you won't get the lighting levels you expect. Use one resistor per LED to be safe.

Easy thing to remember is the rule that the location of each component (their order) is irrelevant in a series-connected (or daisy-chained) circuit. 
« Last Edit: October 29, 2019, 09:42:31 PM by peteski »
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Point353

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Re: Basic circuit questions
« Reply #5 on: October 29, 2019, 09:14:52 PM »
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If the voltage drop was too high, then the load devices would start working.
Care to elaborate?

peteski

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Re: Basic circuit questions
« Reply #6 on: October 29, 2019, 09:48:12 PM »
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Care to elaborate?

LOL, I meant "stop". I fixed my post.  I hope that makes more sense now.

If the charged capacitor is the source of power, and it powers the rest of the locomotive (the electrical load) through a resistor connected in series, if that resistor's resistance it too high, most of the voltage will be across the resistor, and the decoder (the electric load which needs the power) will only see a small part of the voltage, and the current will also be limited, based on the combined resistance of the series resistor plus whatever the load's resistance is.
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Point353

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Re: Basic circuit questions
« Reply #7 on: October 29, 2019, 10:41:19 PM »
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LOL, I meant "stop". I fixed my post.  I hope that makes more sense now.
Infinitely.