how to do circuit analysis ?

Discussion in 'General Electronics Chat' started by abhimanyu143, Jan 29, 2015.

  1. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
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    1
    hello ,
    I have breadboard , multimeter and some components. I made some connection on breadboard. I want to learn basics. I want to learn with practical readings.
    breadboard connection

    upload_2015-1-30_7-10-41.png

    I am posting this image. I just want to show that I am really doing something for learning electronics, this picture is not good quality . I am attaching image that I have been design on software for batter understanding
    upload_2015-1-30_7-11-12.png

    question : how to do circuit analysis ?

    my first step:
    component - diode
    Q1 how to calculate different parameter for diode ?
    datasheet
    diode 1N4007 rating
    Dc bloking voltage Vr= 1000 volts
    Peak reverse voltage VRSM = 1200 volts
    RMS reverse voltage VR (RSM)= 700 volts
    Rectified forward current IO = 1.0 Amp
    Peak Surge current I FSM = 30 for 1 cycle amp
    Operating and junction temprature Tj = -60 to 175 *C
    Maximum forward voltage drop VF = 1.1 volts
    Average forward voltage drop (AV VF) = 0.8 volts
    Maximum reverse current IR = 10 to 50 uAmp
    Averge reverse current IR (AV) = 30 uAmp

    I can measure voltage , current and resistance with multimeter. but how to do with theory ?
    according to me ,
    formula V=IR
    If we know the two value than we can find unknown value
    I read theory, how to use some formula for practical value, how to work with some practical reading ?
     
  2. wayneh

    Expert

    Sep 9, 2010
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    I cannot tell what you want to do. I suggest that you determine, for each component, the voltage on every pin, the current through the device, and the power being dissipated by each device. This exercise will reveal a lot of details and give you a good understanding of the circuit.
    The diode is not necessary, but protects the circuit from opposite polarity hookup to the power supply.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    wayneh,
    I have a different idea. I think the diode is there so that the circuit can powered with a wall-wart plug-in transformer, like a one that puts out 9Vac...

    Amhim,
    Sounds like you know all about the diode specs, but are asking about what the diode is doing in this circuit?

    First. Here is a schematic electrical diagram of your circuit. It gives the components names, (like your diodeD1), and it gives names to the electrical connection points (nodes). Engineers think of a circuit like yours in terms of a schematic. They only think about a layout of the circuit, like your Fritzing picture, much later in the design process. You need to learn to think about circuits at the schematic level...

    VrLed1.gif
     
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  4. wayneh

    Expert

    Sep 9, 2010
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    Excellent point. There's a battery in his picture and I was thinking that drawing was a USB port but that was an assumption. Protection against reverse polarity AND protection against AC power supply.
     
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The schematic shows the parts of your circuit required to understand how it works, and what each component is doing. From left to right, the transformer V1 is connected through a half-wave rectifier D1 to a filter capacitor C1. The rectified, filtered dc voltage at node rec goes into a 5V Integrated Circuit Regulator U1, and comes out as regulated 5Vdc at node dc. The 5Vdc is used to light an LED D2, but as with all LEDs, there must be a current-limiting resistor R1 in-series with the LED.

    The schematic shows the input of your circuit being a low-voltage transformer, represented as a 9Vrms AC voltage source V1 which is plugged in to the jack on your picture. Note that the jack has no symbol on the schematic. It is there for mechanical convenience, and doesn't really serve an electrical function...

    A couple of things you will learn about transformers. Unless they labelled as being a DC output power supply, they put out an AC wave, at the line frequency (likely 50Hz in your country). Many output voltages are possible, but here I arbitrarily chose one that puts out 9Vac. You will learn that 9Vac in this context is also equiv to 9Vrms, which you can think of as being 9Vac average. An AC sinusoid wave has a peak value ~1.4 times higher than its average, and the polarity reverses 50 times per second...

    Let me show you some expected voltages and current flows in your circuit. First, we have to agree about how we talk about them. For this discussion, it is necessary to understand what point of reference voltages are measured with respect to. Look at the schematic. Note I have shown a small diamond symbol at the bottom. It is convenient (and convention) to use this to show the "common" node in the circuit. Think of it as being the 0V point where you would connect the negative lead of your voltmeter while measuring voltage at all other nodes.

    Now look at the attached plot pane. It shows three different voltages in the circuit vs time (the x-axis) and with respect to com. The first voltage V(ac) green trace is just the transformer voltage. The bottom end of the simulated transformer is tied to common (follow the jack wiring in your Fritzing), so the top end connected to node ac shows the familiar sine wave. Note that the peak value of a nominal 9V sine is about 12V. The time between successive zero-crossings of the wave is directly related to the 50Hz line frequency (1/50 = 0.02 = 20ms).

    The diode D1 is used as a half-wave rectifier. It is forward biased (conducts) only when the voltage V(ac) is more positive than the previous voltage V(rec) red trace on the filter capacitor C1. You will learn that a Silicon diode will conduct current from anode to cathode only when the anode voltage is more positive than the cathode voltage by ~0.65V. Look at the green and red traces. Can you see where green is above red?

    Now look at V(rec) red trace again. Notice how the capacitor filters the current pulses delivered to it by D1. Think of it this way: The diode passes current into the filter for the time while the transformer voltage is higher than the capacitor voltage (minus the forward drop of the diode itself). This charges the capacitor to its peak value. After that, the capacitor must deliver the current required by the IC regulator (and the ultimately the LED) until it gets its next infusion of current one AC cycle later... During this interval, the capacitor is losing charge (its voltage is dropping). The capacitor must be large enough so as to keep the voltage from falling too far before it gets recharged again. Note that if this condition is met, the IC regulator U1 can make pure DC at V(dc) blue trace.

    VrLed2.gif
     
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  6. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
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    Ok , I can make schematic diagram.
    as I said before If we know the two value than we can find unknown value
    formula V=IR

    I measure voltage resistance and current for each component with multimeter. so I have reading on paper for voltage , current and resistance .
    than I write any two value on paper and I start to solve for unknown value. I can calculate value using ohms law. so I have two result one from multimeter (practically) and another from formula (theoretically). I just want to learn by doing some real work after that I will match both value If both value are same than I will sure that my theory is correct
     
    Last edited: Jan 29, 2015
  7. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Ok, this gets to what the diode D1 is actually doing in your circuit. I have made a new plot which shows two traces, the voltage from anode to cathode of the diode D1, which is shown as V(ac,rec) yellow trace, and I(D1) violet trace which is the current from anode to cathode of diode D1. Note that on the plot voltage is shown on the left y-axis, but current is shown on the right y-axis, both wave still plotted vs time.
    Note the special way of showing the differential voltage across the diode. Read V(ac,rec) as "the voltage from node ac to node rec"

    Note that the yellow trace clearly shows that when the diode is conducting, its forward voltage is clamped to ~0.65V by the internal behavior of the diode itself. When it is not conducting, the voltage actually reverses to a negative value of ~-21V. This is the peak-inverse voltage (piv) that the diode D1 ever sees in this circuit. Note that your 1N4007 that you selected has a piv rating of 1000V. Do you think that this circuit even comes close to stressing the diode? You could substitute one that has a much lower piv rating, like a 1N4001 and still have plenty of reserve.

    Now to show you the current through the diode D1 violet trace. Note that the current only flows for a small part of the 20ms period of the 50Hz power line frequency. We have already discussed that the diode conducts only when the yellow trace shows a forward voltage of ~0.65V. This show that no current flows the rest of the time...

    Note that the first current pulse is bigger than the others because the capacitor starts out uncharged, so more current is required to fill it during the first current pulse. The subsequent current pulses are all the same and smaller. Note that the peak current at each subsequent pulse is almost 0.25A, even though the average diode current is only a few tens of mA. Do you think that this current level stresses the 1N4007 diode you selected?

    VrLed3.gif
     
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  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Make a schematic, and post it here.
    Please use the same names for the components and the nodes as I have already showed you.

    What are you powering the real circuit with? Do you have a small transformer as I guessed?
    Power the circuit, an use your multimeter to make measurements as I will direct you.
    Please post the values you actually measure.
    Does the LED light up as expected?
    I will have you make various measurements, and we will compare them to what I have already showed you.
     
  9. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    I don't have transformer
    upload_2015-1-30_10-59-28.png

    I have taken some reading for above picture

    Power supply
    Voltage (7.7V – 7.9 V Dc)
    Current (12.4 mA – 10.8 m A)

    Diode
    Voltage (0.6 v Dc)
    Current (5.4 mA – 4.8 m A)

    Capacitor
    Voltage (2.98V – 2.79 V Dc)
    Current (12.34 mA – 12.38 m A)

    Regulator IC
    Voltage (3.54V – 2.74 V Dc)

    Resistor
    Voltage (0.42 V Dc)
    Current (5.30 mA – 4.60 m A)

    LED
    Voltage (1.98V Dc)
    Current (9.5 mA – 6.8 m A)

    I am not sure that readings are correct ?
     
  10. ScottWang

    Moderator

    Aug 23, 2012
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    To buy a 90~240Vac to 12Vdc/2A adaptor, and using a LM317(and heatsink) and some resistor and diodes and capacitors, you also can be adding a LM7805, and then you can get a 0~9V adjustable power and a +5V fixed voltage power.

    Or you want to do some more as 2 sets and 4 kinds ±5~15V fixed dual power supplies..
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi there,

    It's nice to see you are interested in this stuff. You will find that it is an interesting field.

    If you want to start with the basics and get a well rounded idea how to analyze circuits, you should start with resistors and voltage supplies. That will teach you about Ohm's Law and parallel and series resistors and this information is widely applicable.

    But to help you best, it would be a good idea if you would state what your math background has been up to this point. The more math you already have had the faster you'll learn how to deal with more complicated circuits. If you have limited math background then you'll do well to learn more and more math as you go.
     
  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Before I comment on the actual readings, it looks to me that your 9V battery is dead. Can you get a fresh one, and repeat the measurements. For now, just measure the open-circuit voltage of the fresh battery. Then connect it to the circuit, and measure only the voltage at nodes ac, rec, dc, and led with the negative lead of your mulimeter at com.
     
  13. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is your circuit repeated with a 9V battery in place of the transformer. Since the battery is a DC source, then the diode D1, nor the capacitor C1 is needed for rectification/filtering. However, I will leave them in the circuit for now just to point out some things...

    I have let the simulator predict the voltages based on the assumption that the battery voltage is exactly 9.000V, which it will not be exactly.
    Note that V(rec) is about 0.6V less than V(ac), which is now 9.000V. The current through the diode D1 is ~19mA, which is equal to the current that the 9V battery is supplying.

    On the output side of the regulator V(dc), the voltage is very close to 5.00V, which is what the voltage regulator is supposed to do: make 5.00V out of the ~8.4V at its input. Note the current through C1. Why is it (very close to) zero?

    Now look at the current through R1 I(R1) and the Led I(D2). Kirchhoff states that they are same. Look at the forward voltage across the LED V(led). That is typical for a Red 20mA LED.

    Finally, notice that the Led current is about 14mA, while the current into the regulator (same as the current through D1) is about 19mA.
    Where did the unaccounted for 4mA go?

    VrLed4.gif
     
  14. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    I will buy transformer very soon

    I have taken another reading
    battery with no connection
    voltage= (6.55 - 5.32 volt DC)
    current= (20.2 - 19.2 m Amp )

    battery with connection
    voltage= (4.13 - 3.3 volt DC)
    current= (18.7 - 17.2 m Amp )

    Diode
    voltage= ( 3.45 volt DC)
    current= (14.3- 12.2 m Amp )

    capacitor
    voltage= ( 3.11 volt DC)
    current= (12.6 m Amp )

    regulator IC
    Input voltage= (3.9 volt DC)
    input current= (12.0 m Amp )
    output voltage= (2.5 volt DC)
    output current= (9.1-9.2 m Amp )

    resistor
    voltage= ( 1.95 volt DC)
    current= (0.75 m Amp )

    LED
    voltage= (1.94 volt DC)
    current= (0.74 m Amp )
     
  15. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    Normal light 3mm LED rated current 2V/10mA, you can provide it about 80% of 10mA as 8mA.
    Normal light 5mm LED rated current 3V/20mA, you can provide it about 80% of 20mA as 16mA.
    High light 3mm LED rated current 3V/10mA, you can provide it about 80% of 10mA as 8mA.
    High light 5mm LED rated current 3V/20mA, you can provide it about 80% of 20mA as 16mA.

    The voltage range of 2V led about 1.8V~2.2V
    The voltage range of 2V led about 3V~3.4V
    Some leds maybe has higher voltages.
     
  16. abhimanyu143

    Thread Starter Member

    Aug 25, 2014
    211
    1
    that's ok , I will buy new battery but I want to know about reading. thats reading are correct ?
     
    Last edited: Jan 30, 2015
  17. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    LESSON 1:
    Learn how to read all the posts that were posted when you were away from the forum before you reply another time.
    You missed several posts before yours that could have helped you a lot. The last post you see here wont always be the only one that was posted while you were away, there could be several more.
     
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  18. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Lesson for today...

    Battery is so dead as to be useless... Get a new battery!
    It is ok to connect a voltmeter directly across the battery terminals, with or with-out a load connected. A voltmeter has a high input resistance, and will not harm the battery.

    You never, never connect a ammeter directly across battery terminals. That short-circuits the battery, and instantly ruins it. An ammeter has a very low (fraction of Ω) input resistance, so it draws huge currents from the battery. An ammeter is always inserted into a branch of a circuit where the other components in that branch limit the current.

    To measure how much current a battery is delivering, you leave whatever load the battery is connected to, and break into one of the two conductors going to the load (doesn't matter if it is the positive or negative lead), put the ammeter in-series with the load, and then read the current. Note that the current is limited by the load resistance..
    Battery is so dead as to be useless... Get a new battery! Dont ruin the new one by connecting the ammeter directly across it. Voltmeter is ok.
     
    Last edited: Jan 30, 2015
  19. ScottWang

    Moderator

    Aug 23, 2012
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    The new 9V battery has about 9.6V, you have to get the power right to about 9V, otherwise all other measuring values are useless.
     
  20. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    No, let him make the measurements with the new battery. I will rerun the sim after we learn what the new battery voltage actually is...
     
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