Confusion on relating AC ground and current direction in AC analysis

I can understand why in DC analysis we remove capacitors since the caps do not pass DC in steady state. This is easy to conclude because in reality the DC component of a current do not pass through a capacitor.

But in AC analysis there is a concept called "AC ground", which means we basically ground the Vcc to the real ground. I'm having hard time to understand this. Because in real the AC component of the current does not flow upwards to the Vcc(? not sure). How come then the Vcc is grounded as if the AC component of the current will flow into the Vcc?

Let's look at the below example transformation for AC analysis:

Above on the left, if the circuit is in linear mode and if a sinusoidal small vbe applied to its base, there will be an ic collector current which is composed of a DC offset Ic0 plus an AC component call it ic. The total collector current will look like this:

Considering the above total collector current ic, what can we say about the current direction? Does AC part of the current flow into the Vcc back and forth or not(in reality)? Because in AC analysis we short the Vcc ground as if the current flows into Vcc back and forth. I hope I could articulate where I'm confused.

• AC component of the current flows in both directions but Iac pp cannot exceed the DC current unless it is driving a coil tied to V+ then it can stored reactive energy based on Q =Z(f){Re/Im} with some cap at resonance. – Tony Stewart Sunnyskyguy EE75 Apr 10 '18 at 0:41

What you have to understand is, a voltage source is a short with a specific voltage. From an AC viewpoint, a static voltage does not exist, so the only thing that remains is the short. That's why you can assume the top of a voltage source is also GND if the bottom is.

If you need it more visual, replace your voltage source with a real big capacitor.

• I'm wondering "in reality" does the AC component really pass through the Vcc and reach the real ground. – cm64 Apr 9 '18 at 23:47
• Yes! There's a really big capacitor at the output of your mains adapter. (It also works with batteries, though.) – Janka Apr 9 '18 at 23:48
• What if its a battery the Vcc? – cm64 Apr 9 '18 at 23:48
• Okay so regardless of the type of the Vcc source "in reality" the AC always penetrates into the power supply plus terminal(?) – cm64 Apr 9 '18 at 23:50
• In reality all sufficiently high-frequency AC is shorted through the internal capacitance of the power supply. Being it a mains adapter or a battery. A corner case is charging an accumulator with a rectified, but not smoothed 100Hz source. In these cases you had to calculate it. – Janka Apr 9 '18 at 23:52

You are not confused. There is a thing called 'ground-bounce' where a ground pin and/or trace echo's some of the current back, usually on IC's that have sharp rising and falling edges at the outputs.

Then there is 'noisy' ground which is similar but a constant event. This can come from a poorly filtered switch-mode power supply, or a brush-type motor on the same circuit.

Both of these act as a current source (as noise) from the ground to Vcc and any components connected to them.

It has to be severe to upset logic IC's but analog IC's are very sensitive to ground based noise. A lack of bypass capacitors is sometimes to blame, or poor grounding by using traces too narrow for the peak current they must pass.

AC analysis: From every current and voltage is in calculations subtracted its average value (=DC component). That's possible if we consider the distortion in the circuit to be so low that the circuit can be considered to be linear. It's a good assumption even in transistors if peak to peak changes in currents and voltages are only few percents of the average. We use it, because the calculations are simpler when they can be done separately for DC and the AC signal.

The actual currents and voltages must have their DC components because the physics of the components need it. Omitting the averages is purely a calculation technique. The directions must in AC analysis be considered as "accordant growing directions". If the momentary value of ib (drawn into the base of the transistor) is increasing, then at the same moment vbe and ic must grow.