# Vcc and GND in AC operation?

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#### xiongyw

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Hi,

In the attached picture, I saw a statement which I do not understand:

"The collector is directly connected to the DC power supply, so as far as AC operation is concerned, the collector is connected to the ground."

What does it actually mean? or there is a kind of assumptions (or background) I am not aware of (as a newbie)?

Thanks,
/bruin

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All circuits have DC and AC grounds. Sometimes they are the same, sometimes not. In your simple circuit, both are the same.

DC power supplies normally have a large decoupling/filter capacitor connected to ground. Therefore for AC signals the power supply impedance to ground is very low and it is thus considered to be at "AC ground". Normally AC and DC ground are the same point.

In this circuit you have your AC signals operating on top of a DC-bias (which makes the transistor influence the AC the way you want it to). The statement has to do with the difference between AC and DC analysis. You know what those are right?

When you are biasing the transistor you ignore the AC signals and only pay attention to the DC. You redraw the circuit the way that DC sees the circuit (inductors become short-circuits and capacitors become open-circuits, all AC sources are turned off). This sets the way the transistor behaves. THen you go to the AC analysis. You redraw the circuit the way AC sees the circuit (inductors and capacitors remain, all DC sources are turned off). You ignore the DC by "turning off the DC supply" so all the AC voltages are centered around 0V ...BUT you pretend the transistor still behaves as if it was still biased. You then add up the DC and AC voltages to get the actual voltage afterwards.

What happens when you turn off a DC current source? It goes to 0A flowing through both terminals- just like an open-circuit.
What happens when you turn off a DC voltage source? It goes to 0V across both terminals- just like a short-circuit.

SO in the AC analysis, it looks like +V is connected to ground via that short-circuit formed by turning off the DC voltage. In circuits where there is a current source, this effectively acts as an AC open-circuit.

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In this circuit you have your AC signals operating on top of a DC-bias (which makes the transistor influence the AC the way you want it to). The statement has to do with the difference between AC and DC analysis. You know what those are right?

When you are biasing the transistor you ignore the AC signals and only pay attention to the DC. You redraw the circuit the way that DC sees the circuit (inductors become short-circuits and capacitors become open-circuits, all AC sources are turned off). This sets the way the transistor behaves. THen you go to the AC analysis. You redraw the circuit the way AC sees the circuit (inductors and capacitors remain, all DC sources are turned off). You ignore the DC by "turning off the DC supply" so all the AC voltages are centered around 0V ...BUT you pretend the transistor still behaves as if it was still biased. You then add up the DC and AC voltages to get the actual voltage afterwards.

What happens when you turn off a DC current source? It goes to 0A flowing through both terminals- just like an open-circuit.
What happens when you turn off a DC voltage source? It goes to 0V across both terminals- just like a short-circuit.

SO in the AC analysis, it looks like +V is connected to ground via that short-circuit formed by turning off the DC voltage. In circuits where there is a current source, this effectively acts as an AC open-circuit.

Thanks, this is exactly the background information (DC/AC analysis and their respective treatment of power supply) I was not clear, though I saw similar phrases several times in books/webpages. I guess I need to grasp a text book to read related sections (I am sorry that I do not have the opportunity to do a systematical study on this field, so the information I obtained so far is piece by piece and not fit together yet).

I have been reading articles talking about linearity of circuit/elements, and/or linear region of circuit elements (e.g., it seems diode/transistors are not linear elements but have linear regions). So I guess why we can divide circuit analysis into DC/AC parts and "add" them together, is because the circuit in analysis is either linear, or it's assumed to be working in its linear region.

In terms of treatment of DC power supply in AC analysis, as you pointed out, they are treated as "short-circuit". My guessing is that, by "turn off" the DC power, it only means to remove the DC voltage from the ciruit, but not necessary means short-circuit the DC power terminals, especially considering the internal impedance of the power supply (which I believe is not going to disappear in AC analysis). Is this correct? If so, the following question is, why we can still consider the DC power terminals are short-circuited? Is this because the internal impedance of the DC power supply can be ignored (similar to what crutschow mentioned), or anything else?

Actually I am also a bit confused by the phrase like "power supply impedance to ground" as in crutschow's reply (as well as in other contexts talking about impedance "from power to ground"). My understanding of the DC power supply, take battery as an example, is that the battery is the "power supply", and it has two terminals, cathode and anode, and the cathode terminal is usually regarded as GND. So the impedance of the power supply (the battery) is the impedance from the cathode to the anode (or v.v.). Can I interprete the phrase "from power to ground" as a simplified form of "from cathode to anode", in this case?

Thanks again!

I have been reading articles talking about linearity of circuit/elements, and/or linear region of circuit elements (e.g., it seems diode/transistors are not linear elements but have linear regions). So I guess why we can divide circuit analysis into DC/AC parts and "add" them together, is because the circuit in analysis is either linear, or it's assumed to be working in its linear region.
Yes, that's the definition of linear. It's origin is from mathematics like linear equations and systems can be figured out separately and added up/superimposed.

In terms of treatment of DC power supply in AC analysis, as you pointed out, they are treated as "short-circuit". My guessing is that, by "turn off" the DC power, it only means to remove the DC voltage from the ciruit, but not necessary means short-circuit the DC power terminals, especially considering the internal impedance of the power supply (which I believe is not going to disappear in AC analysis). Is this correct? If so, the following question is, why we can still consider the DC power terminals are short-circuited? Is this because the internal impedance of the DC power supply can be ignored (similar to what crutschow mentioned), or anything else?
Only voltage sources are treated as short-circuits. Current source are treated as open-circuits. In real power supply we do this by sticking an open-circuit in between the power supply and everything else using a switch, but this is now how it is mathematically.

THe impedance of the power supply is not going to dissapear in reality when you turn off a source. In a real voltage source it's like a resistor in series with an ideal voltage source. In a real current source it is like a resistor in parallel with an ideal current source. But we ignore things like this in many theoretical analysis since usually an ideal source is being used (where the impedance is zero for a voltage source and infinite for a current source). However, if you were including these in your analysis, nothing changes in how you would turn off the supply since you would draw the supply as a resistor and an ideal source anyways. You would still turn off the power supply by turning off the ideal source in the same way and leave everything else (ie. the resistor) unchanged).

Actually I am also a bit confused by the phrase like "power supply impedance to ground" as in crutschow's reply (as well as in other contexts talking about impedance "from power to ground"). My understanding of the DC power supply, take battery as an example, is that the battery is the "power supply", and it has two terminals, cathode and anode, and the cathode terminal is usually regarded as GND. So the impedance of the power supply (the battery) is the impedance from the cathode to the anode (or v.v.). Can I interprete the phrase "from power to ground" as a simplified form of "from cathode to anode", in this case?

No. The impedance of the power supply (also called the internal impedance) has nothing to do with the circuit itself. It is integral to the power supply. An actual source can be modelled using an ideal source with a internal impedance (resistor, capacitor, and inductors). Usually the inductor and capacitor effects are just ignored and only a resistor is used. In this case it is just called internal resistance. Voltage sources have the resistor in series with the ideal voltage source. Current sources have the resistor in parallel with the ideal current source.

On the other hand, "power supply impedance to ground" refers to the overall impedance of the circuit between the two terminals of the power supply and may include the internal impedance of the power supply (depending on how you want to define it).

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hi,
A point to be aware of, the term 'ground' is often misused.

Ground suggests that the connection is linked to the Earth connection.

In your book examples the 'ground' term should be replaced by 'common' or '0V'.

Its the 'common' reference to which signals/voltages are measured.

So you have: "common emitter, common collector and common base" transistor configurations.

IMO the term 'ground' should only be used on a circuit diagram to indicate a connection to Earth.

hi,
A point to be aware of, the term 'ground' is often misused.

Ground suggests that the connection is linked to the Earth connection.

In your book examples the 'ground' term should be replaced by 'common' or '0V'.

Its the 'common' reference to which signals/voltages are measured.

So you have: "common emitter, common collector and common base" transistor configurations.

IMO the term 'ground' should only be used on a circuit diagram to indicate a connection to Earth.

Or neutral. Usually when we say "ground" we are almost ALWAYS actually talking about common, 0V, or neutral (which are all different, but correct, names for the same thing). Ground, however, is not the same thing, technically. Instead of using the 3 proper synonyms that exist, we tend to use a fourth term which is technically wrong.

(Would you expect anything less of the same people who use negative charge for electrons? So that charge physically flows from negative to positive instead of posiitve to negative? Which also results in positive charge in current flow actually being the absence of the negatively charged particles?)

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Or neutral. Usually when we say "ground" we are almost ALWAYS actually talking about common, 0V, or neutral (which are all different, but correct, names for the same thing). Ground, however, is not the same thing, technically. Instead of using the 3 proper synonyms that exist, we tend to use a fourth term which is technically wrong.

(Would you expect anything less of the same people who use negative charge for electrons? So that charge physically flows from negative to positive instead of posiitve to negative? Which also results in positive charge in current flow actually being the absence of the negatively charged particles?)

hi DK,
I wouldnt suggest 'neutral', as this is used in mains terms.

I agree some of the terminology we use is at best ambiguous, it comes from a time when the technology was determined by empirical means.

One of the older terms 'static electricity' is misleading, if you stand too close to a static source, you will find out very quickly, it 'aint static'. DC power supplies normally have a large decoupling/filter capacitor connected to ground. Therefore for AC signals the power supply impedance to ground is very low and it is thus considered to be at "AC ground". Normally AC and DC ground are the same point.

After reading DK's explanation, I feel a bit clear on your reply now. In this case, if we "cut" the AC loop path at the two terminals of the voltage power supply, we get two partial circuits:
- the power supply itself (with its internal impedance etc), and
- the rest (including load, and bypassing/decoupling caps, which are mostly available as you mentioned).

So by "power supply impedance to ground" (for AC signal), you meant the impedance of the second part (i.e., not including the power supply part). and since its assumed there are bypassing caps, so the AC impedance can be ignored. I hope I understand it correctly Thanks again for all.
/bruin

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