Series RC circuits in parallel

Question

Morning all, happy new year!

Mathematical query today, one of simplification:

The circuit

Here we have a parallel circuit of two identical series RC circuits. I.E. both resistors are identical and both caps are identical (this is key).

The question:

What is the equivalent circuit as a single series RC circuit. I.E. what are the equivalent resistor and cap values?

I'm certain there is a solution to this when the parallel circuits are identical. I'm aware that if they weren't there would be oscillation etc, but this isn't relevant in this case...in a way it is just two high ESR caps in parallel.

Many thanks

Simon

rachaelp · Accepted Answer

Hi Simon,

It's exactly like two caps in parallel regardless of their ESR as all caps still have some resistance, even low ESR ones, so this is just what you would see with any bulk decoupling caps in parallel on a power supply.

If you look at the charging of the caps through the resistors when you first switch the power on you can see that if you put double the capacitance on one of the resistors the charge time would double so you have to halve the resistance for it to stay the same. If you look at the current required from the power supply when it's first switched on and the caps are charging you can see the current is equal through both resistors so the total current is the sum of both paths. To make an equivalent with a single resistor you would need it to be halved and then as per my first sentence you'd need to double the capacitance on that single resistor to maintain the same charge time.

Therefore:

R3 = R1 || R2

C3 = C1 + C2

Best Regards,

Rachael

jc2048 · Answer

Rats! A Champion got there before me. My answer was going to be this (I'll post it anyway because it's a different approach): If the two sides are identical, the voltages at the nodes between the resistors and capacitors are identical too. That means you could join those points with a piece of wire and nothing would change - there can't be a current flow in the wire because the end potentials are the same. Your new circuit is then simply two resistors in parallel and two capacitors in parallel. So the top resistor is r/2 and the bottom capacitor is 2 x c. Whilst it sounds very artificial in this question, it's essentially why having a lot of identical small ceramic capacitors spread around a high-speed logic board is such good value for decoupling. The Cs add, the ESRs divide down and the inductances divide down too. [At high speed it's a bit more complicated because you also have to start thinking about things like the rate at which changes propagate across the board - not all the capacitors will be able to participate equally.] If you're interested in this kind of thing, read Johnson and Graham's High-Speed Signal Propagation - it's very good. They're academics who I can recognise as being engineers, if you know what I mean, which is much rarer than you might suppose.

mcb1 · Answer

My immediate thought is that any ripple on the VCC line will have two (identical) paths to go. Since they are the same values, then it must be half the 'impedance'. So the equivalent resistance will be half while caps in parallel double. My punt is (without resorting to google) that R is half and C is twice. Mark

D_Hersey · Answer

There is a general analytic solution to this, start here: How to find the reciprocal of a complex number? | Chemistry Learning

Series RC circuits in parallel

Top Replies