Thanks for another excellent video. In the second question about coupling capacitors, it's not quite correct to say there's NO phase shift. The small-signal AC voltage on the transistor base will have a slight phase shift (and attenuation) relative to the voltage produced by the microphone. For AC signals, the two biasing resistors plus the emitter resistor are effectively in parallel between the transistor base and ground. Their equivalent resistance forms the "bottom half" of a voltage divider with the impedance of the coupling capacitor. Normally, the capacitance is chosen so that its reactance (at the lowest frequency of interest) is less than 10% of the equivalent resistance. Then, to a first approximation, the capacitor can be regarded as a short circuit for AC signals. But there will in fact be some attenuation and phase shift as the signal passes from the microphone to the transistor base, although certainly much less than 90°. Similarly for the amplified signal that passes from the transistor collector across the coupling capacitor to the load.
I have a question, based upon a circuit that was supposedly a phase shift circuit that I saw many years ago. If i had two AC sources running at the same frequency but at different amplitudes wirh respect to ground, across a capacitor, can you describe the electron flow, if there is any. I would expect that after the AC sources stabalized, there would be a constant voltage difference across the capacitor and no charge would flow. i hope that I am not correct about that.
Another homing in on a misunderstanding haunting me for YEARS stunting my electronics understanding and thus no real improvement in forever. Phase shift of a component vs. phase shift of a circuit. WOW, thank you once again professor Bob. Another one was current sources. I just COULD not envision a current source since I've always approached sources from a battery standpoint instead of an induction standpoint. I can now see how a current source (microphone) creates current which then "pushes" a voltage vs. a battery which creates a voltage and "pushes" a current.
Is a microphone a current source? An ideal current source would produce the same current no matter what the resistance connected across it. I don't know a lot about microphones, but I think they're closer to ideal voltage sources. An example of a current source would be the collector (or emitter) current of a transistor operating in its active region. If the base current is kept constant, the collector current will be almost constant over a wide range of load resistances. You can make an even better current source using an opamp instead of a single transistor. Perhaps current sources are difficult to envision because they're somewhat difficult to create, unlike voltage sources. It's fairly easy to make a battery. But imagine if you could just open your parts bin and grab a current source. It would be a dangerous device. You'd have to be very careful to store it with its leads connected together. Otherwise, an arc of current would flow through the air from one lead to the other. And I suppose all that constant current flow would rapidly deplete all its energy. Ultimately, of course, any linear circuit has both a Thevenin equivalent and a Norton equivalent, the first using an ideal voltage source and the second an ideal current source. So a simple battery can be viewed as either a voltage source with a small series resistance, or a current source with a very large parallel resistance.
I'm sorry to have to correct you but, in an AC circuit, current leads voltage by 90 degrees. Your diagram shows it lagging and this is quite a terrible representation of the truth. This why we use the formula I = C dv/dt and get it right.
I am sorry but your formula makes no sense. By definition C=Q/V for a capacitor. Then it follows up that V=1/C * Q. Assuming the charge on the capacitor changes over time it is the result of accumulation of charges brought by current traveling through the circuit and voltage on the capacitor is the sum of all charges brought by i(t) over dt time intervals.
@@dan-florinchereches4892 How does that formula makes no sense? I (capacitor current) = C (capacitance in F) * dV/dt (rate of change of voltage across the capacitor). If dV/dt is close to zero (voltage is not changing) then current (I) will be close to zero. If dV/dt is high, then current through capacitor (I) will be high. Current can also be thought of as a rate of change of charge I = dQ/dt.
@@oriole8789 I see the differential equations are pretty much equivalent. My problem would be more along the lines of chicken or egg problem. I see it as the voltage at the capacitor plates is changing as a result of current depositing charge. The formula you are using is from the perspective of the charge existent on capacitor producing a current.
@@dan-florinchereches4892 Your perspective of the voltage between the capacitor plates changing as a result of current depositing charge is perfectly correct, and consistent with all the formulas provided (V = Q/C, I = C*dV/dt, I = dQ/dt etc). In that sense, V = Q/C directly leads to dV/dt = 1/C * dQ/dt. The perspective depends on what you're measuring and controlling. A voltage (potential) difference causes current to flow. Current cannot flow if there is no potential difference. Current depositing charge is a process that happens only as long as a potential difference exists. Once the system is in equilibrium (a capacitor is fully charged), there is no current flow, because there is no potential difference (with respect to either side of the capacitor, and the circuit nodes those sides are connected to, respectively). Between the legs of the capacitor, a potential difference is naturally high since the capacitor is charged. If a load is then applied to the capacitor, a new potential difference now exists with respect to that load, and if we assume that the load is 0ohm short circuit let's say, charge will flow out of the capacitor in the form of current (limited by the ESR of that capacitor in the real world), until the potential difference between the capacitor and the load reaches 0. Current flowing in or out of the capacitor doesn't change the formulas as long as sign conventions are followed. Hope this helps!
Since 0v is not an absence of voltage I can see how you can have 0V and max amps, but how can you have 0I and max volts? According to Ohm's law you need amps to have voltage, right?
A capacitor is similar to a battery in that regard. It can be charged up so that there's a voltage difference between its two plates. But, like a battery, even with a large voltage, there may or may not be any current flowing into or out of it. It all depends on what else it's connected to. Conductors and resistors follow Ohm's Law. They're sometimes called "Ohmic devices". But a lot of devices aren't Ohmic. In fact, most of the interesting ones aren't, including capacitors, inductors, diodes, transistors, etc. For a capacitor, the current flowing through it isn't proportional to the voltage across it, but rather to the *rate of change* of the voltage. So when the voltage reaches its maximum, for a brief moment the voltage isn't changing, and thus the current is zero. Also, when the voltage is zero, it's changing at its maximum rate, and so the current is flowing at a maximum. So if the voltage takes the form of a sine wave, oscillating between positive and negative, the current will also be an oscillating sine wave, but the oscillations will occur with a 90° phase shift relative to the voltage. Hope that helps.
I think you misinterpreted Ohm's law. This law defines a relationship between the current value given a certain voltage applied to a closed circuit with a certain *impedance*. You don't need a current (whose unit is given in amps) to have a voltage. A voltage is created, for instance, when you bring together opposing electrical charges separated by a certain distance (aka insulator). The result of doing that is that you have created a disturbance in the electric potential. In our universe, everything tends to seek a more stable state. In the same way that a body will tend to fall to the ground if it is lifted, the opposed charges will be delighted to meet each other, and if you give them that chance by connecting a wire, they will take it for sure. The value of the current will depend on the *resistance* , in the case of DC power source, or on the *impedance* of the wire in the case of a AC power source. Hope this helps you to understand the matter better. The electric and electronic world is a very tricky one.
I don't understand how current could be delayed 90 degrees in the capacitor and be 0 degrees delayed in the series circuit. I am imagining it like it is a stream of water on a tube, and I cannot see how the flow is delayed in a part of the tube but not in the other.
Thanks for that explanation! I understand a lot more clearly!
Another perfect repetition of a movie done earlier! Thanks for that!
You have a natural-born talent as a teacher and educator! 👍
i think that this is the best explanation of phase shift of a capacitor in the whole internet, seriously
Thanks for another excellent video.
In the second question about coupling capacitors, it's not quite correct to say there's NO phase shift. The small-signal AC voltage on the transistor base will have a slight phase shift (and attenuation) relative to the voltage produced by the microphone.
For AC signals, the two biasing resistors plus the emitter resistor are effectively in parallel between the transistor base and ground. Their equivalent resistance forms the "bottom half" of a voltage divider with the impedance of the coupling capacitor. Normally, the capacitance is chosen so that its reactance (at the lowest frequency of interest) is less than 10% of the equivalent resistance. Then, to a first approximation, the capacitor can be regarded as a short circuit for AC signals. But there will in fact be some attenuation and phase shift as the signal passes from the microphone to the transistor base, although certainly much less than 90°. Similarly for the amplified signal that passes from the transistor collector across the coupling capacitor to the load.
Wonderful explanation 👏👏👏👏👍👍👍👍
Very informative knowledge, thanks for remembering me the old college concepts.
I have a question, based upon a circuit that was supposedly a phase shift circuit that I saw many years ago. If i had two AC sources running at the same frequency but at different amplitudes wirh respect to ground, across a capacitor, can you describe the electron flow, if there is any. I would expect that after the AC sources stabalized, there would be a constant voltage difference across the capacitor and no charge would flow. i hope that I am not correct about that.
if you was my teacher i can learn nothing
Another homing in on a misunderstanding haunting me for YEARS stunting my electronics understanding and thus no real improvement in forever.
Phase shift of a component vs. phase shift of a circuit. WOW, thank you once again professor Bob.
Another one was current sources. I just COULD not envision a current source since I've always approached sources from a battery standpoint instead of an induction standpoint. I can now see how a current source (microphone) creates current which then "pushes" a voltage vs. a battery which creates a voltage and "pushes" a current.
Is a microphone a current source? An ideal current source would produce the same current no matter what the resistance connected across it. I don't know a lot about microphones, but I think they're closer to ideal voltage sources.
An example of a current source would be the collector (or emitter) current of a transistor operating in its active region. If the base current is kept constant, the collector current will be almost constant over a wide range of load resistances. You can make an even better current source using an opamp instead of a single transistor.
Perhaps current sources are difficult to envision because they're somewhat difficult to create, unlike voltage sources. It's fairly easy to make a battery. But imagine if you could just open your parts bin and grab a current source. It would be a dangerous device. You'd have to be very careful to store it with its leads connected together. Otherwise, an arc of current would flow through the air from one lead to the other. And I suppose all that constant current flow would rapidly deplete all its energy.
Ultimately, of course, any linear circuit has both a Thevenin equivalent and a Norton equivalent, the first using an ideal voltage source and the second an ideal current source. So a simple battery can be viewed as either a voltage source with a small series resistance, or a current source with a very large parallel resistance.
I'm sorry to have to correct you but, in an AC circuit, current leads voltage by 90 degrees. Your diagram shows it lagging and this is quite a terrible representation of the truth. This why we use the formula I = C dv/dt and get it right.
I am sorry but your formula makes no sense. By definition C=Q/V for a capacitor. Then it follows up that V=1/C * Q. Assuming the charge on the capacitor changes over time it is the result of accumulation of charges brought by current traveling through the circuit and voltage on the capacitor is the sum of all charges brought by i(t) over dt time intervals.
@@dan-florinchereches4892 How does that formula makes no sense? I (capacitor current) = C (capacitance in F) * dV/dt (rate of change of voltage across the capacitor). If dV/dt is close to zero (voltage is not changing) then current (I) will be close to zero. If dV/dt is high, then current through capacitor (I) will be high. Current can also be thought of as a rate of change of charge I = dQ/dt.
@@oriole8789
I see the differential equations are pretty much equivalent. My problem would be more along the lines of chicken or egg problem.
I see it as the voltage at the capacitor plates is changing as a result of current depositing charge.
The formula you are using is from the perspective of the charge existent on capacitor producing a current.
@@dan-florinchereches4892 Your perspective of the voltage between the capacitor plates changing as a result of current depositing charge is perfectly correct, and consistent with all the formulas provided (V = Q/C, I = C*dV/dt, I = dQ/dt etc). In that sense, V = Q/C directly leads to dV/dt = 1/C * dQ/dt. The perspective depends on what you're measuring and controlling. A voltage (potential) difference causes current to flow. Current cannot flow if there is no potential difference. Current depositing charge is a process that happens only as long as a potential difference exists. Once the system is in equilibrium (a capacitor is fully charged), there is no current flow, because there is no potential difference (with respect to either side of the capacitor, and the circuit nodes those sides are connected to, respectively). Between the legs of the capacitor, a potential difference is naturally high since the capacitor is charged. If a load is then applied to the capacitor, a new potential difference now exists with respect to that load, and if we assume that the load is 0ohm short circuit let's say, charge will flow out of the capacitor in the form of current (limited by the ESR of that capacitor in the real world), until the potential difference between the capacitor and the load reaches 0. Current flowing in or out of the capacitor doesn't change the formulas as long as sign conventions are followed. Hope this helps!
Since 0v is not an absence of voltage I can see how you can have 0V and max amps, but how can you have 0I and max volts? According to Ohm's law you need amps to have voltage, right?
A capacitor is similar to a battery in that regard. It can be charged up so that there's a voltage difference between its two plates. But, like a battery, even with a large voltage, there may or may not be any current flowing into or out of it. It all depends on what else it's connected to.
Conductors and resistors follow Ohm's Law. They're sometimes called "Ohmic devices". But a lot of devices aren't Ohmic. In fact, most of the interesting ones aren't, including capacitors, inductors, diodes, transistors, etc.
For a capacitor, the current flowing through it isn't proportional to the voltage across it, but rather to the *rate of change* of the voltage. So when the voltage reaches its maximum, for a brief moment the voltage isn't changing, and thus the current is zero. Also, when the voltage is zero, it's changing at its maximum rate, and so the current is flowing at a maximum.
So if the voltage takes the form of a sine wave, oscillating between positive and negative, the current will also be an oscillating sine wave, but the oscillations will occur with a 90° phase shift relative to the voltage.
Hope that helps.
@@zenbum2654 I think I understand. Thank you.
because the circuit looks like it is open.
I think you misinterpreted Ohm's law. This law defines a relationship between the current value given a certain voltage applied to a closed circuit with a certain *impedance*. You don't need a current (whose unit is given in amps) to have a voltage. A voltage is created, for instance, when you bring together opposing electrical charges separated by a certain distance (aka insulator). The result of doing that is that you have created a disturbance in the electric potential. In our universe, everything tends to seek a more stable state. In the same way that a body will tend to fall to the ground if it is lifted, the opposed charges will be delighted to meet each other, and if you give them that chance by connecting a wire, they will take it for sure. The value of the current will depend on the *resistance* , in the case of DC power source, or on the *impedance* of the wire in the case of a AC power source.
Hope this helps you to understand the matter better. The electric and electronic world is a very tricky one.
I don't understand how current could be delayed 90 degrees in the capacitor and be 0 degrees delayed in the series circuit. I am imagining it like it is a stream of water on a tube, and I cannot see how the flow is delayed in a part of the tube but not in the other.