Adding more capacitors to an AMP

[HollowState]

Frank,
I have some of those very same Rifa capacitors with the same date code. I got them from eBay out of Florida. I had to buy some M5x10 pan head screws for them. A four year shelf life is a riduclously conservitive short amount of time. Rifa makes a good capacitor. Go ahead and use them. Just bring the voltage up slow for the first time, then they'll be fine.

Victor

[f1802llk]

Thanks for the information. I feel comfortable to use the caps now. While thinking about how best to fit them into my amp, I came upon an idea. Instead of putting two caps in series with a pair of dividing resistors as is normally done when V is over the rated voltage of C1:

----------- V
| |
C1 R1
| |
---------
| |
C2 R2
| |
--------- ground

how about just use one cap:

----------- V
| |
C1 R1
| |
---------
|
R2
|
--------- ground

This way, by making R1 bigger than R2, I can actually make the voltage across C1 and R1 closer to the rated voltage of C1. Would this work?

-Frank

[HollowState]

Would this work?

From a strictly voltage point of view across the capacitor, yes it would control the DC. But otherwise it is not a good thing to do. This would make the power supply "spongy" and raise it's internal inpedance. Regulation and base would suffer greatly. I would only do this if I were stranded on a desert island with only one capacitor left. Capeesh?

Victor

[f1802llk]

Hi Victor: I was guessing something must be wrong with that scheme of mine as I did not see anything like that anywhere, but could not figure out why. You said it would make the power supply "spongy", why? I would've thought otherwise. With two caps in series, the overall capacitance is less. By getting rid of one, I thought I'd have increased the stored energy, thus making the supply more stable...

Frank

[HollowState]

Originally asked by f1802llk:

You said it would make the power supply "spongy", why?

The quick answer is that by adding a series resistor you current limit the supply.

If you think about how the capacitor works it becomes clear. Under no load conditions, the capacitor charges to full voltage and remains charged. But as the output of the supply is loaded down, the capacitor is tasked to supply current. It's job is to fill in the voids between the half sine pulses coming from the rectifier. This happens at a 120 cycle rate for full wave recitfication. (60 cycle for ½ wave)

By adding a resistor in series with it the peak current available is reduced because the time constant is increased. The time it takes for one charge/discharge cycle to happen. The higher the resistance, the longer the time constant becomes. This results as a reduced charge/discharge level between cycles, which translates to less power available to the load and a softer supply. Further, usful power will be wasted in that series resistor.

Generally one time constant period is required to charge a capacitor to 63.2% of Es. This can be found by: t (in seconds) = R (in ohms) x C (in farads) Five time constant periods are required for C to be considered full charged. The opposite is true for discharge. (Decreasing to 36.8% in one time period.)

Here's a simple analogy:

Think of the capacitor as if it were a two stage water pump. First filling then pumping out the water through a hose. By adding a resistor it's like reducing the hose diameter which will limit the water (current) flow over a given length of time.

With two caps in series, the overall capacitance is less. By getting rid of one, I thought I'd have increased the stored energy, thus making the supply more stable...

At a quick glance this might seem true, but it isn't. The added resistance, and time constant, makes it not so. Actually, a larger capacitor also produces a longer time constant for a given charge level. For this reason a point of diminishing return is reached as the value increases beyond a certain capacitance when all things are considered. Although I do not think that point has been reached with the values discussed here.

To prove it to yourself, you might use simple ohm's law to select two resistors and try it. You will not damage anything in so doing, but I'll wager you won't like the resulting performance.

Hope this makes things a little clearer.
Victor

[ray_moth]

Victor's explanation is great. The way I think of it is that when there is a sudden increase in the demand on current, it should be filled virtually instantaneously by the reservoir cap of the PS. That requires the series resistance of the cap to be as low as possible.

[f1802llk]

I got it now. Thanks everyone.

Frank