# A beginner's question /may be too easy

1. Apr 3, 2009

### electrostatic

a beginner's question /may be too easy :)

hi

i recently started learnning electronics from several books and tutorials online
i ran into this circuit on madlab.org website:

i have two things troubling me in this circuit that i havnt figured out
1. why does it say that the capacitor (C1) is charged through the resistor (R1)
when there is a diode there. how does the capacitor charge ?
doesnt current must flow through it in order for it to begin charging?

2. why dont the diodes turn on and off both at the same time,
and if individually, what decides which one turns on first?

i hope someone can make things clearer for me

thank you

here is the original circuit page with an explanation :

Last edited by a moderator: May 4, 2017
2. Apr 3, 2009

### Bob S

Re: a beginner's question /may be too easy :)

1) The diode in the collector ckt is conducting whenever the transistor is ON.

2) Note the two cross coupled capactors from one collector to the other transistor's base, plus the base-biasing resistors. Only one diode is conducting at any given time.

3. Apr 3, 2009

### electrostatic

Re: a beginner's question /may be too easy :)

so, the left side of the capacitor (C1) is being charged through the flow comming from emmiter-base of TR2 (o.6v ?)?

being more specific :

http://img141.imageshack.us/img141/7609/circuit.jpg [Broken]
will the capacitor in this circuit charge?
and if it does, will the diode Light while charging?

Last edited by a moderator: May 4, 2017
4. Apr 3, 2009

### Staff: Mentor

Re: a beginner's question /may be too easy :)

If the 47uF cap starts out discharged, and you connect the 9V battery to the circuit, there will be a current briefly (on the order of 2ms). The flash will likely be too short for you to see.

Quiz Question -- where did I get the 2ms number from?

Last edited by a moderator: May 4, 2017
5. Apr 3, 2009

### Bob S

Re: a beginner's question /may be too easy :)

No.. One of the two npn transistors is ON and the other is OFF. For the ON transistor, the collector voltage is about 0.2 volts above ground, so there is about 8.8 volts across the 470 ohm resistor plus the LED diode.

Last edited by a moderator: May 4, 2017
6. Apr 4, 2009

### bitrex

Re: a beginner's question /may be too easy :)

When transistor TR2 is on, TR1 is off and capacitor C1 will charge up through the LED and R1 because the LED is forward-biased. The capacitor will charge up to around 7 volts because there is a about a 1.2 volt drop across the LED (type dependent) when it's in conduction and about 0.6 volts from transistor TR2's base to emitter when it's turned on. When it's charging up the LED shouldn't glow, at least not much, because over most of the charging time the current is small (as the capacitor voltage increases exponentially, the current decreases in an exponential curve, and it's current that determines LED light output).

7. Apr 4, 2009

### bitrex

Re: a beginner's question /may be too easy :)

I'm not sure why Bob S above is saying that the capacitor in the second circuit diagram won't charge - if the capacitor starts out uncharged with positive at zero volts and you put an LED and resistor across it the cap it is going to charge until it hits the supply minus the LED drop, just like a power supply capacitor charging through a rectifier.

8. Apr 4, 2009

### Staff: Mentor

Re: a beginner's question /may be too easy :)

I believe Bob was referring to the first circuit, even though he quoted the 2nd circuit.

9. Apr 4, 2009

### Bob S

Re: a beginner's question /may be too easy :)

The multivibrator circuit as pictured in the first post is conditionally stable because it is AC coupled. Here is a LTSpice modified version that starts when the power supply is switched on, because of added capacitor C3. Also added is an emitter follower output.

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Last edited: Apr 5, 2009
10. Apr 5, 2009

### Bob S

Re: a beginner's question /may be too easy :)

Here is waveform for multivibrator in previous post.

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• ###### Multivibrator_wvfm.jpg
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11. Apr 6, 2009

### electrostatic

Re: a beginner's question /may be too easy :)

apparently i got confused by the definitions of a capacitor
i saw an animation which showed electrons flowing through a capacitor from - to +
and thus charging the capacitor.

but then I've read the definition on the forum here , that no current flows through a capacitor.

i understand from all of your replies (which i very much appreciate, thank you all
that in the second 2nd circuit , the capacitor will charge through the diode.
my question is,
how is it possible? ,
if there is only one direction allowed. from - to + how will the capacitor charge?

i may not fully understand the electrons physics.
but it seems to trouble me enough to prevent me from understanding
any referrals to reading material will be appreciated as well.

thank you all

12. Apr 6, 2009

### Bob S

Re: a beginner's question /may be too easy :)

The capacitors will charge through the 470 ohm ohm resistors (and the LEDs) when the transistor collectors are open, or discharge through the transistor collectors when they are saturated, but only when the capacitor is charging through the base pullup resistors, or being dicharged by the transistor base currents. This sounds more complicated than it really is.

Capacitors can charge only when there is a voltage across them. There are two sources of current: the 470 ohm (and LEDs) or 4.7 kohm resistors, and only two sinks of current: The saturated collectors or the base current. The two capacitors actually change polarity during each cycle, because the saturated collector voltage of one transistor is less than the base voltage of the other transistor, even when it is off.

[EDit] I have added the waveforms of the voltage on both ends of C1. See my previous post on the LTSpice circuit model. You will need to click on magnifier image to the the two traces well. The capacitor actually does change polarity during the cycle.

#### Attached Files:

• ###### Multivibrator C1.jpg
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Last edited: Apr 6, 2009
13. Apr 6, 2009

### bitrex

Re: a beginner's question /may be too easy :)

It's possible to get quite a good intuitive sense for the behavior of a capacitor by studying its basic characteristic equation, $$I = C\frac{dv}{dt}$$. This shows that current can flow through a capacitor when the rate of change of voltage with respect to time, $$\frac{dv}{dt}$$ is not zero. That's one reason capacitors come in so handy, they can block a DC current and still let AC pass, for example when one needs to couple signals between two stages in an amplifier that are operating at different bias voltages.

Again, from the characteristic equation of the capacitor, $$I = C\frac{dv}{dt}$$, we see that the voltage across a capacitor cannot change instantaneously. The smaller you make the rate of change of voltage across the capacitor, $$\frac{dv}{dt}$$, the larger the current becomes. The capacitor could only charge in zero time if the current were infinite, which is of course impossible. So in the second circuit, let's say you have the LED and resistor hooked up to the capacitor and apply the power. Boom, at that instant the positive end of the capacitor is at 9-LedVdrop = 7.8 volts. But because the voltage across the capacitor can't change instantaneously, that means the negative end of the capacitor at that instant is also at 7.8 volts! So if you stuck a voltmeter across the capacitor and read the voltage across the capacitor at that instant, 7.8 volts - 7.8 volts = 0 volts, the capacitor is uncharged. However, now we have the negative end of the capacitor at 7.8 volts connected to ground at zero volts. Things can't stay like that forever, two objects at different potentials interacting with each other for any length of time would violate the conservation of energy! Electrons are attracted to the positive electric field applied to the positive plate of the capacitor and move up from ground to the negative plate of the capacitor. The voltage across the capacitor begins to increase, until the capacitor is fully charged at 7.8 volts. In electronics we talk about current flowing from positive to negative, of course that's not really what's happening, but it makes it so you don't have to constantly take the square root of negative numbers when doing simple circuit analysis.

The fact that a capacitor cannot change the voltage across it instantaneously is used to good effect in the multivibrator circuit above. When the capacitor C1 is charged to 7 volts or so, and transistor TR1 turns on, its positive plate is immediately connected to 0 volts. Since the negative side was at 0 volts to begin with, that side has no choice but to drop down to -7 volts. That large negative voltage across TR2's base-emitter junction immediately shuts TR2 down, allowing the cycle to repeat.

Here are two videos from one of MIT's undergraduate Electromagnetics courses that might help your understanding. The first discusses the way in which capacitors store charge in relation to the work-energy theorem. The second goes into the concept of capacitor dielectrics and dielectric polarization and explains why for example capacitors with certain dielectrics (like electrolytic caps) can store much more energy than one would expect based on their size. The whole lecture series itself is excellent and definitely worth watching in its entirety if you're able.

Lecture 7:
Lecture 8:

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