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Circuit Design

Chr1s

Dec 20, 2016
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What would be a good circuit to build that would help me to understand more about the role of each component In the circuit? I would like to begin with a theoretical approach and then build it on a breadboard. Any ideas ? Any help would be much appreciated.

Thanks guys
 

Bluejets

Oct 5, 2014
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Most start with an LED a resistor a switch and a battery.
How long is a piece of string comes to mind.
Depends on a million things which way you go.
This is why it pays to be a bit more specific.
 

Chr1s

Dec 20, 2016
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I understand these basics circuits but I am having difficulty in understanding more complicated circuits. I need a circuit that consists of a few transistors , capacitors, inductors and resistors to try to really understand what is going on.
 

(*steve*)

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How good is your understanding of the components you mention?
 

Chr1s

Dec 20, 2016
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Capacitors I can understand but the inductor is proving troublesome.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Back in the day... well, about the middle of the previous century... the standard "introduction" to electronics was to build a crystal radio set with a one-transistor audio amplifier. This involved winding a coil (inductor) on a paper core (perhaps after applying a coat or two of shellac to improve the stiffness), adding a variable tuning capacitor of a few hundred picofarads, a 1N34 germanium crystal diode, a general-purpose small-signal transistor, a crystal ear-bud or very small loudspeaker, a fifty-foot or so long-wire antenna, and a cold-water pipe ground. Easy peasy and you could usually pick up at least one AM radio station with this rig. Check out this website for more information.

You don't learn much about electronics doing "monkey see, monkey do" projects, but it was (and still is) a way to get people interested in the hobby. For any real progress you need to learn the theory behind the components. That means math and a lot of study. You said you wanted begin with a theoretical approach and then build it on a breadboard. There is a LOT of theory involved with AM radio communications, but a crystal set on a breadboard is pretty simple and offers instant gratification because you can actually HEAR radio stations!. I can still remember the thrill of that sixty-something years later.
 

Chr1s

Dec 20, 2016
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why ?
what is it about them that confuses you ?

Ok. I'll try explain what's going on in my head. Say we have a coil of so many henries connected to a supply through a series resistor and a switch. When the switch is closed the inductor reaches 5tc , the inductor will have the full supply across it minus the drop through the resistor? Now what happens when the switch is opened again? How does the inductor behave?
 

Chr1s

Dec 20, 2016
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Back in the day... well, about the middle of the previous century... the standard "introduction" to electronics was to build a crystal radio set with a one-transistor audio amplifier. This involved winding a coil (inductor) on a paper core (perhaps after applying a coat or two of shellac to improve the stiffness), adding a variable tuning capacitor of a few hundred picofarads, a 1N34 germanium crystal diode, a general-purpose small-signal transistor, a crystal ear-bud or very small loudspeaker, a fifty-foot or so long-wire antenna, and a cold-water pipe ground. Easy peasy and you could usually pick up at least one AM radio station with this rig. Check out this website for more information.

You don't learn much about electronics doing "monkey see, monkey do" projects, but it was (and still is) a way to get people interested in the hobby. For any real progress you need to learn the theory behind the components. That means math and a lot of study. You said you wanted begin with a theoretical approach and then build it on a breadboard. There is a LOT of theory involved with AM radio communications, but a crystal set on a breadboard is pretty simple and offers instant gratification because you can actually HEAR radio stations!. I can still remember the thrill of that sixty-something years later.

Thanks hevans1944
 

(*steve*)

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How does the inductor behave?

That is not a good question. It is a bit like a question where you close a switch and short a capacitor.

If there is a resistor in parallel with the inductor in this case, the current will continue to flow through the inductor, reducing exponentially as the energy stored in its magnetic field is dissipated.

In your actual example several things could happen:
  1. An arc forms in the switch and current continues to flow for a short time before it is extinguished.
  2. An arc forms in the switch and current flows indefinitely.
  3. An arc from somewhere in the circuit to somewhere else for a short period.
  4. Nothing appears to happen
 

(*steve*)

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Capacitors I can understand but the inductor is proving troublesome.

A capacitor stores energy in an electric field.
An inductor stores energy in a magnetic field.

Inductors and capacitors are pretty much the reciprocal of each other. Whatever you know about voltage and current in capacitors you should understand as being about current and voltage in inductors.

When a capacitor has energy stored in it, a voltage appears across it
When an inductor has energy stored in it, a current flows through it.

A capacitor resists a change in voltage across it (i.e. you cannot instantly change the voltage, current must flow through the capacitor and energy is either added or removed from the capacitor).
An inductor resists a change in current through it (i.e. you cannot instantly change the current, voltage must be placed across the inductor and energy is either added or removed from the inductor).

A constant current flowing into a capacitor will cause a constant rate of change of the voltage across it.
A constant voltage across an inductor will cause a constant rate of change of the current through it.

When a source of voltage is disconnected from a capacitor the voltage across the capacitor is retained
When a source of current is disconnected from an inductor the current through the inductor is retained***


(***) This is weird and causes problems in the real world. To make this an exact reciprocal, the "disconnection" of the current is a shorting out of the inductor! Shorting a charged capacitor is potentially dangerous. Leaving open an energised inductor is also potentially dangerous. Open inductors kill far more electronics than shorted capacitors!
 

hevans1944

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Ok. I'll try explain what's going on in my head. Say we have a coil of so many henries connected to a supply through a series resistor and a switch. When the switch is closed the inductor reaches 5tc , the inductor will have the full supply across it minus the drop through the resistor? Now what happens when the switch is opened again? How does the inductor behave?
You have this all wrong. The ideal inductor has zero resistance, so in series with a current-limiting resistor, R, after five time-constants equal to L/R seconds, the current in the inductor will have increased exponentially and asymptotically to approximately I = V/R, where V is the supply voltage and R is the value of the current-limiting resistor. At this point there is zero voltage across the inductor. A pretty good introductory explanation with just a little math can be found here.

What happens when the switch opens depends on how fast the switch opens. If it could open instantaneously, and stay open, the current in the inductor, I = V/R, would instantly go to zero. The voltage across the inductor would immediately rise to Infinity because the time rate-of-change of current from some finite value to zero value occurred in zero time. In reality this cannot happen, but the voltage can rise to a high enough value to either (1) cause an arc across the switch contacts or (2) destroy a semiconductor switch, such as a bipolar junction transistor or a MOSFET.

In reality, stray capacitance in the circuit always limits how fast the voltage across the inductor can rise when the switch opens. In high-power pulsed-power applications every effort is made to minimize the stray capacitance to deliberately allow thousands of volts to appear across an inductor when it's current is interrupted by a fast-opening switch. For the rest of us, some protective circuits are necessary. For mechanical switches, you can place a capacitor, in series with a small-valued resistor, across either the switch contacts or the inductor. Either way, the capacitor slows down the rate at which the inductor current decreases when the switch opens, and the resistor dissipates the electromagnetic energy stored in the coil. You can also use a diode, with or without an energy-dissipating resistor, placed in parallel with the inductor to do the same thing. The diode is wired so it does not conduct when power is applied to the inductor. It only conducts (briefly) when power is removed from the inductor.
 
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Chr1s

Dec 20, 2016
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You have this all wrong. The ideal inductor has zero resistance, so in series with a current-limiting resistor, R, after five time-constants equal to L/R seconds, the current in the inductor will have increased exponentially and asymptotically to approximately I = V/R, where V is the supply voltage and R is the value of the current-limiting resistor. At this point there is zero voltage across the inductor. A pretty good introductory explanation with just a little math can be found here.

What happens when the switch opens depends on how fast the switch opens. If it could open instantaneously, and stay open, the current in the inductor, I = V/R, would instantly go to zero. The voltage across the inductor would immediately rise to Infinity because the time rate-of-change of current from some finite value to zero value occurred in zero time. In reality this cannot happen, but the voltage can rise to a high enough value to either (1) cause an arc across the switch contacts or (2) destroy a semiconductor switch, such as a bipolar junction transistor or a MOSFET.

In reality, stray capacitance in the circuit always limits how fast the voltage across the inductor can rise when the switch opens. In high-power pulsed-power applications every effort is made to minimize the stray capacitance to deliberately allow thousands of volts to appear across an inductor when it's current is interrupted by a fast-opening switch. For the rest of us, some protective circuits are necessary. For mechanical switches, you can place a capacitor, in series with a small-valued resistor, across either the switch contacts or the inductor. Either way, the capacitor slows down the rate at which the inductor current decreases when the switch opens, and the resistor dissipates the electromagnetic energy stored in the coil. You can also use a diode, with or without an energy-dissipating resistor, placed in parallel with the inductor to do the same thing. The diode is wired so it does not conduct when power is applied to the inductor. It only conducts (briefly) when power is removed from the inductor.

Very well explained. Thanks so much. I really appreciate it.
 

Chr1s

Dec 20, 2016
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A capacitor stores energy in an electric field.
An inductor stores energy in a magnetic field.

Inductors and capacitors are pretty much the reciprocal of each other. Whatever you know about voltage and current in capacitors you should understand as being about current and voltage in inductors.

When a capacitor has energy stored in it, a voltage appears across it
When an inductor has energy stored in it, a current flows through it.

A capacitor resists a change in voltage across it (i.e. you cannot instantly change the voltage, current must flow through the capacitor and energy is either added or removed from the capacitor).
An inductor resists a change in current through it (i.e. you cannot instantly change the current, voltage must be placed across the inductor and energy is either added or removed from the inductor).

A constant current flowing into a capacitor will cause a constant rate of change of the voltage across it.
A constant voltage across an inductor will cause a constant rate of change of the current through it.

When a source of voltage is disconnected from a capacitor the voltage across the capacitor is retained
When a source of current is disconnected from an inductor the current through the inductor is retained***


(***) This is weird and causes problems in the real world. To make this an exact reciprocal, the "disconnection" of the current is a shorting out of the inductor! Shorting a charged capacitor is potentially dangerous. Leaving open an energised inductor is also potentially dangerous. Open inductors kill far more electronics than shorted capacitors!

Thanks Steve. This is very helpful to me, however I just have a couple of questions based on your explanation?

What is the difference between an electric field and magnetic field?
 

Herschel Peeler

Feb 21, 2016
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What would be a good circuit to build that would help me to understand more about the role of each component In the circuit? I would like to begin with a theoretical approach and then build it on a breadboard. Any ideas ? Any help would be much appreciated.

Thanks guys

Power supplies are a good place to start. Then add a voltage regulator. Then current limiting.
 
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