Reactance Reminder

[PRR]

[gcwills]

Thanks PRR, that is certainly a useful shortcut for calculating reactance of a cap   - I guess a similar approach can be taken to calculate the reactance of an inductor. Cheers

[jjasilli]

I looked it up; as to inductance, checkout this; it's an inverse thing:  http://www.allaboutcircuits.com/vol_2/chpt_4/3.html


Cap performance is a valuable, but complex topic.  Rarely do we have a sole cap in a circuit, which circuit would be of any practical use.  What we find is an R-C network.  The cap is usually in conjunction with: a resistance, or an impedance, or both.  What is the result of that?  I don't know, but will re-read this, and similar, article(s):  http://www.allaboutcircuits.com/vol_2/chpt_4/3.html

[PRR]

Inductor is just 2 * Pi * Freq * Henries

A 10 Henry choke at 100Hz is 2 * 3.14 * 100Hz * 10H = 6 * 100Hz * 10H = 6,000 Ohms. (Perhaps 6,283 Ohms, but we never know the actual inductance of iron-core chokes with such precision.)

At 200Hz the same choke is obviously 12,000 Ohms.

The problem with caps is that the reactance needs a 1/x computation, and most practical caps are micro-values. Knowing that 1uFd at 1KHz is 159 Ohms gets you closer to an answer quicker, less brain-pain.

It is also good for second-guessing the numbers you get from slide-rule or on-line calculators.

[jjasilli]

But what happens when the cap is in an R-C network?  IOW, does the network circuit alter the (effective) "resistance" value of the cap?  Does it matter if he cap is "seeing" the network resistance, vs. the resistor coming 1st in the network? 


Any simple, short explanation would be helpful.

[HotBluePlates]

But what happens when the cap is in an R-C network?  IOW, does the network circuit alter the (effective) "resistance" value of the cap?  Does it matter if he cap is "seeing" the network resistance, vs. the resistor coming 1st in the network?

No, it all works the same.

Let's assume our sample RC network is a B+ dropping resistor of 5kΩ and a filter cap of 20uF. The Reactance Reminder PRR posted says "Divide 159 by the product of frequency in kHz and capacitance in uF" (to add a little translation).

Well, what is the applied frequency? For B+ in the U.S., is is 60Hz rectified to 120Hz pulses, or 0.12 kHz. 159 / (0.12*20) = 66.25Ω. So the 120Hz ripple sees a voltage divider when applied to that network of 66.25Ω / (5kΩ+66.25Ω), or a reduction to ~1/76th the amount of applied ripple.

But what happens when the cap is in an R-C network?

The practical problem you're asking about is "What happens when considering an R-C network, where many different applied frequencies could be present?"

Let's consider a 0.01uF coupling cap and a 220kΩ grid (to ground) resistor, leading into a tube grid. We are interested in what's happening over a frequency spectrum; let's pick 100Hz (0.1kHz), 1kHz and 10kHz to be representative.

159 / (0.01 * 0.1) = 159kΩ
159 / (0.01 * 1)    = 15.9kΩ
159 / (0. 1 * 10)   = 1.59kΩ

For each of these frequencies, the 0.1uF cap presents some amount of reactance, but because it is reactance it changes depending on the frequency. It doesn't matter if 1, 2 or 8,000 different frequencies are present, they all see the cap as a different level of obstruction. And the R-C network of the cap & resistor form a voltage divider network, same as a voltage divider made of 2 resistors, or an L-C network, etc.

The voltage dividers reduce the output to the grid as shown below:
100Hz   ->    220kΩ / (159kΩ+220kΩ) = ~0.6 * Input voltage
1kHz     ->    220kΩ / (15.9kΩ+220kΩ) = ~0.9 * Input voltage
10kHz   ->    220kΩ / (1.59kΩ+220kΩ) = ~0.99 * Input voltage

You already know the coupling cap and following grid reference resistor are a high-pass/low-cut network. If it's useful to you, now you have a shortcut to directly calculate the reactance of the cap, perhaps for use in a voltage divider equation to see exactly how much a given frequency is being cut.

[jjasilli]

Thanks for these helpful examples!