Circuits - Astable Multivibrator

[Toby_Obie]

Hi,

I've started to study electronics and currently looking at "Astable Multivibrator"

I've looked this up on wikipedia and presented with the illustration and explanation (see below)

The question I have with it is with capacitors C1 and C2,

It states that :

"The right side of C1 (and the base of Q2) is being charged by R2 from below ground to 0.6 V."
also
"R4 is charging the right side of C2 up to the power supply voltage (+V)"

On the diagram, the right side of C1 is negative and the right side of C2 is positive

My question is how are different sides being charged ? I though capacitors only charged one particular side (negative) ?

Have I overlooked something ? (btw this is brand new material)

Thanks

Basic mode of operation

The circuit keeps one transistor switched on and the other switched off. Suppose that initially, Q1 is switched on and Q2 is switched off.

State 1:

* Q1 holds the bottom of R1 (and the left side of C1) near ground (0 V).
* The right side of C1 (and the base of Q2) is being charged by R2 from below ground to 0.6 V.
* R3 is pulling the base of Q1 up, but its base-emitter diode prevents the voltage from rising above 0.6 .
* R4 is charging the right side of C2 up to the power supply voltage (+V). Because R4 is less than R2, C2 charges faster than C1.

When the base of Q2 reaches 0.6 V, Q2 turns on, and the following positive feedback loop occurs:

* Q2 abruptly pulls the right side of C2 down to near 0 V.
* Because the voltage across a capacitor cannot suddenly change, this causes the left side of C2 to suddenly fall to almost −V, well below 0 V.
* Q1 switches off due to the sudden disappearance of its base voltage.
* R1 and R2 work to pull both ends of C1 toward +V, completing Q2's turn on. The process is stopped by the B-E diode of Q2, which will not let the right side of C1 rise very far.

This now takes us to State 2, the mirror image of the initial state, where Q1 is switched off and Q2 is switched on. Then R1 rapidly pulls C1's left side toward +V, while R3 more slowly pulls C2's left side toward +0.6 V. When C2's left side reaches 0.6 V, the cycle repeats.

 

[KataKoniK]

Hi there,

Great question! It's important to note that capacitors can be charged on either side, depending on the direction of current flow. In this circuit, the capacitors are being charged by the current flowing through the resistors and transistors. Let's break down the process step by step:

1. In State 1, Q1 is switched on and Q2 is switched off. This means that the right side of C1 (which is connected to the base of Q2) is being charged by the current flowing through R2. This current is coming from below ground (0V) and is gradually increasing the voltage on the right side of C1 to 0.6V. At the same time, R4 is charging the right side of C2 up to the power supply voltage (+V). This means that the voltage on the right side of C2 will be higher than the voltage on the left side, which is connected to ground (0V). This creates a potential difference across the capacitor, causing it to charge.

2. When the voltage on the right side of C1 reaches 0.6V, Q2 turns on. This creates a positive feedback loop, causing the left side of C2 to suddenly fall to almost -V (well below 0V). This sudden change in voltage causes Q1 to switch off, and R1 and R2 work together to pull both ends of C1 toward +V, completing Q2's turn on. The voltage on the right side of C1 will not rise very far due to the B-E diode of Q2, which prevents it from exceeding 0.6V.

3. This brings us to State 2, where Q1 is switched off and Q2 is switched on. In this state, R1 rapidly pulls C1's left side toward +V, while R3 more slowly pulls C2's left side toward +0.6V. When the left side of C2 reaches 0.6V, the cycle repeats.

In summary, both sides of the capacitors are being charged in this circuit, depending on the direction of current flow. This is how astable multivibrators work, constantly switching between two states and creating a pulsing output. I hope this helps clarify things for you. Keep up the good work with your electronics studies!