My atari 2600 has no sound, which is a common problem with an easy fix, just replace 2 capacitors. The original capacitors are 820pf polystyrene axial capacitors, and these are a bit tricky to find. Could I just replace these with any other kind of capacitor like an electrolytic or ceramic one? DO they do the exact same thing or are they different?
Electrical – Are polystyrene capacitors equal to electrolytic ones
capacitor
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This effect is due to the effects of parasitic characteristics of the device. A capacitor has four basic parasitics:
Equivalent Series Resistance - ESR:
A capacitor is really a capacitor in series with the resistances of its leads, the foil in the dielectric, and other small resistances. This means that the capacitor cannot truly discharge instantly, and also that it will heat up when repeatedly charged and discharged. This is an important parameter when designing power systems.
Leakage current:
The dielectric is not ideal, so you can add a resistance in parallel with your capacitor. This is important in backup systems, and the leakage current of an electrolytic can be much greater than the current required to maintain RAM on a microcontroller.
Dielectric Absorption - CDA:
This is usually of less interest than the other parameters, especially for electrolytics, for which leakage current overwhelms the effect. For large ceramics, you can imagine that there is an RC circuit in parallel with the capacitor. When the capacitor is charged for a long period of time, the imagined capacitor acquires a charge. If the capacitor is rapidly discharged for a brief period and subsequently returned to an open circuit, the parasitic capacitor begins to recharge the main capacitor.
Equivalent Series Inductance - ESL:
By now, you shouldn't be too surprised that, if everything has capacitance as well as nonzero and non-infinite resistance, everything also has parasitic inductance. Whether these are significant is a function of frequency, which leads us to the topic of impedance.
We represent impedance by the letter Z. Impedance can be thought of like resistance, just in the frequency domain. In the same way that a resistance resists the flow of DC current, so does an impedance impede the flow of AC current. Just as resistance is V/R, if we integrate into the time domain, impedance is V(t)/ I(t).
You'll either have to do some calculus, or buy the following assertions about the impedance of a component with an applied sinusoidal voltage with a frequency of w:
\$ \begin{align} Z_{resistor} &= R\\ Z_{capacitor} &= \frac{1}{j \omega C} = \frac{1}{sC}\\ Z_{inductor} &= j\omega L = sL \end{align} \$
Yes, \$j\$ is the same as \$i\$ (the imaginary number, \$\sqrt{-1}\$), but in electronics, \$i\$ usually represents current, so we use \$j\$. Also, \$\omega\$ is traditionally the Greek letter omega (which looks like w.) The letter 's' refers to a complex frequency (not sinusoidal).
Yuck, right? But you get the idea - A resistor doesn't change its impedance when you apply an AC signal. A capacitor has reduced impedance with higher frequency, and it's nearly infinite at DC, which we expect. An inductor has increased impedance with higher frequency - think of an RF choke that's designed to remove spikes.
We can calculate the impedance of two components in series by adding the impedances. If we have a capacitor in series with an inductor, we have:
\$ \begin{align} Z &= Z_C + Z_L\\ &= \frac{1}{j\omega C + j\omega L} \end{align} \$
What happens when we increase the frequency? A long time ago, our component was an electrolytic capacitor, so we'll assume that \$C\$ is very much greater than \$L\$. At first glance, we'd imagine that the ratios wouldn't change. But, some trivial (Note: This is a relative term) complex algebra shows a different outcome:
\$ \begin{align*} Z &= \frac{1}{j \omega C} + j \omega L\\ &= \frac{1}{j \omega C} + \frac{j \omega L \times j \omega C}{j \omega C}\\ &= \frac{1 + j \omega L \times j \omega C)}{j \omega C}\\ &= \frac{1 - \omega^2 LC}{j \omega C}\\ &= \frac{-j \times (1 - \omega^2 LC)}{j \omega C}\\ &= \frac{(\omega^2 LC - 1) * j)}{\omega C} \end{align*} \$
Well, that was fun, right? This is the kind of thing you do once, remember the answer, and then don't worry about it. What do we know from the last equation? Consider first the case where \$\omega\$ is small, \$L\$ is small, and \$C\$ is large. We have, approximately,
\$ \begin{align*} \frac{(small * small * large - 1) \times j}{small * large} \end{align*} \$
which is a negative number (assuming \$small * small * large < 1\$, which it is for practical components). This is familiar as \$Z_C = \frac{-j}{\omega C}\$ - It's a capacitor!
How about, second, your case (High-frequency electrolytic) where \$\omega\$ is large, \$L\$ is small, and \$C\$ is large. We have, approximately,
\$ \begin{align*} \frac{(large * small * large - 1) \times j}{small * large} \end{align*} \$
which is a positive number (assuming \$large * small * large > 1\$). This is familiar as \$Z_L = j \omega L\$ - It's an inductor!
What happens if \$\omega^2 LC = 1\$? Then the impedance is zero!?!? Yes! This is called the resonant frequency - It's the point at the bottom of the curve you showed in your question. Why isn't it actually zero? Because of ESR.
TL,DR: Weird stuff happens when you increase the frequency a lot. Always follow the manufacturers' datasheets for decoupling your ICs, and get a good textbook or take a class if you need to do high speed stuff.
If I could get a ceramic capacitor at the capacitance of 10uF and within my voltage requirements, which from my initial searches I can, what problems would I experience if I were to change, if any?
Some circuits (like some linear regulators, for example), require a certain minimum ESR from the output capacitor, which could cause the circuit to oscillate when using a ceramic but not with an electrolytic.
In a precision circuit, a ceramic might not be preferred due to microphonics, but in those cases you probably wouldn't want an electrolytic either.
Otherwise, ceramics are generally preferred. They'll have lower ESR, they're not polarized, they need less voltage de-rating, and so on.
Finally, when searching SMD footprint standards, the common packages seem to be 0402, 0603, and 0805, where they increase in physical size respectively, but also power rating, which suggests to me I should use as large of a package as possible
Usually you choose the smallest package you can get away with because you want to fit as much circuit as you can in the smallest footprint.
Also, for ceramics, the larger sizes (1210 and higher) can have reliability issues because they can be cracked if the board flexes.
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Best Answer
Polystyrene caps are no longer made. Their major disadvantage is the low melting point of the film, so they can't be machine soldered, only hand soldered.
Back in the bad old days, they were used when a high tolerance was required. If that was the reason these were polystyrene, then any plastic film capacitor of the tolerance would be suitable.
The main specification advantage of polystyrene is its low dielectric absorption, which only really shows up in professional instrumentation-type sample-and-hold applications. Where this is absolutely necessary, use polyphenylene sulphide (PPS) or PTFE dielectric film capacitors to replace it. Unfortunately, both these types are tricky to come by, and expensive.
Polypropylene film is very nearly as good as those, and much more available.
Polyester film is the standard film capacitor dielectric, widely available and cheap, and still very good performance. It's worth trying this type first, to see whether it does do the job. You can buy these in high tolerance values.
You won't find electrolytic caps that small. Don't use ceramic either.