# Output and Input Impedance - Incandescent Light Bulb

#### pratto

Joined Dec 10, 2012
36
For years I have struggled to understand the idea of impedance matching. I know the mantra of max power transfer when Output and Input impedances are equal, and that in order for a load to have a purely resistive impedance, the capacitive impedance must equal the inductive impedance. But from a common sense point of view (maybe I should say layman's point of view) I don't get it.

If I added a capacitor in series with a 120v incandescent light bulb, would this be adding a capacitive impedance element ? Would I see less light ?

#### GopherT

Joined Nov 23, 2012
8,012
For years I have struggled to understand the idea of impedance matching. I know the mantra of max power transfer when Output and Input impedances are equal, and that in order for a load to have a purely resistive impedance, the capacitive impedance must equal the inductive impedance. But from a common sense point of view (maybe I should say layman's point of view) I don't get it.

If I added a capacitor in series with a 120v incandescent light bulb, would this be adding a capacitive impedance element ? Would I see less light ?
You wouldn't see more light. Any capacitance in series with an AC power supply will add some resistance. So you will see some dimming (less dimming for a large capacitor and more dimming for a little one).

#### AnalogKid

Joined Aug 1, 2013
10,127
Yes.
Yes.

This is a common trick in very low cost, very well insulated things such as plug in LED nite lights, air fresheners, etc.

A small general purpose green LED is rated for 20 mA continuous current. At 120 Vac, a 6 K resistor would yield 20 mA, AC, rms current. Not the same as DC, but good enough for this example. And hot. The resistor would dissipate 2.4 W, so a 5 W part would get too hot to touch. Using the formula for capacitive reactance, you can calculate that a 0.44 uF capacitor yields an equivalent impedance. Put a 1N4004 across the LED in inverse parallel to prevent reverse voltage, and you have a teeny nite light. Dangerous, and cool-running, but teeny.

The problem with any offline power supply that relies on a series impedance (resistor, capacitor, inductor, whatever) is that the "correct" voltage drop happens only with the right current. If you disconnect the LED, the voltage on the end of the cap rises to 120 Vac. This can be lethal.

There is much more to this topic, and all of it is banned on this forum.

ak

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#### pratto

Joined Dec 10, 2012
36
You wouldn't see more light. Any capacitance in series with an AC power supply will add some resistance. So you will see some dimming (less dimming for a large capacitor and more dimming for a little one).
Ok. I will carefully try it. I take it that I can not use an electrolytic capacitor.

#### WBahn

Joined Mar 31, 2012
27,486
For years I have struggled to understand the idea of impedance matching. I know the mantra of max power transfer when Output and Input impedances are equal, and that in order for a load to have a purely resistive impedance, the capacitive impedance must equal the inductive impedance. But from a common sense point of view (maybe I should say layman's point of view) I don't get it.

If I added a capacitor in series with a 120v incandescent light bulb, would this be adding a capacitive impedance element ? Would I see less light ?
The "mantra of max power transfer when output and input impedances are equal" is commonly recited, and pretty much as commonly used incorrect since it is simply wrong (or not complete).

Consider the following very simple example. I have a 12 V battery, and two 6 Ω resistor, R1 which is the supply resistance and R2 which is the load resistance that you are trying to get as much power as you can into it. So with this set up, you have 1 A of current and so you are dissipating 6 W of power in R2. If that "mantra" was correct, then if I lower R1 from 6 Ω to 2 Ω, then I should reduce the power transferred to R2 since R1 and R2 are no longer matched. But now the total current is 1.5 A and the power transferred to R2 more than doubles to 13.5 W.

So what gives?

The mantra is missing the caveat that it applies to when you match the load resistance to the supply resistance under the condition that you can't change the supply resistance. If you CAN change the supply resistance, then you want to make it as low as possible regardless of what the load resistance is. Let's say that we could make R1 very close to 0 Ω -- THAT is when we get maximum power to R2 of 24 W.

Now, let's say that the source resistance was changed and is now fixed at 2 Ω. With our 6 Ω load resistor we get 13.5 W into it. What if we match it to the source resistance of 2 Ω? We now get 3 A and dissipate 18 W in R2. Can we do better? Sure -- IF we could lower R1. But if it is fixed, then we can't. If we lower the value of R2 to 1 Ω then we get more current in it (now we have 4 A), but the voltage across it drops as R1 hogs more of the voltage and our power to R2 drops to 4 W. But if we increase R2 to 3 Ω, then the voltage across it increases (to 7.2 V) but the current is dropped so much that the power is reduced to 17.3 W.

So don't go making the common assumption of thinking that if change the source resistance to match a load resistance that you have no control over that you will get more power to the load. It only works when you match the load resistance to a source resistance that you have no control over.

Once you make the shift to impedance matching (so you have reactive as well as resistive elements), the same notions and limitations apply, but they are a bit more complicated.

When you have a source impedance that is fixed, meaning you can't do anything about it, then you want load resistance to equal the source resistance but you want the load reactance to be equal and opposite to the source reactance. You to NOT want the load impedance to be purely resistive if the source impedance has reactance.

Don't get yourself confused with the notion of power factor correction, which is very different than the notion of maximum power transfer.

#### GopherT

Joined Nov 23, 2012
8,012
Ok. I will carefully try it. I take it that I can not use an electrolytic capacitor.
Two back to back willl do.
Use a stepped down 12vac power supply so nobody gets hurt and an automotive bulb.

#### pratto

Joined Dec 10, 2012
36
Two back to back willl do.
Use a stepped down 12vac power supply so nobody gets hurt and an automotive bulb.
Back to back. Do you mean + to - to + to - ? or + to . Well, I guess I am asking about the orientation of the common terminals in the back to back configuration. And being that it is in series with the light, is the side closest to the light considered negative ? I would just try out the different permutations, but they tell me electrolytic caps can explode if wired backwards.

#### pratto

Joined Dec 10, 2012
36
When you have a source impedance that is fixed, meaning you can't do anything about it, then you want load resistance to equal the source resistance but you want the load reactance to be equal and opposite to the source reactance. You to NOT want the load impedance to be purely resistive if the source impedance has reactance.

Don't get yourself confused with the notion of power factor correction, which is very different than the notion of maximum power transfer.
That is something that I never heard before (source impedance being an un-balanced reactance). As usual, I try to go back to ground I am more familiar with : when I hear that the amplifier should be connected to an 8 ohm speaker, I don't ever remember hearing anything about the speaker reactance having to be capacitive or inductive.

#### pratto

Joined Dec 10, 2012
36
You wouldn't see more light. Any capacitance in series with an AC power supply will add some resistance. So you will see some dimming (less dimming for a large capacitor and more dimming for a little one).
Two back to back willl do.
Use a stepped down 12vac power supply so nobody gets hurt and an automotive bulb.
Good thinking. I will just use my bench power supply with a 15v AC output with a 12v DC bulb.
Still hoping to hear back on the Electrolytic Cap orientation question.

#### KeepItSimpleStupid

Joined Mar 4, 2014
5,088
That is something that I never heard before (source impedance being an un-balanced reactance). As usual, I try to go back to ground I am more familiar with : when I hear that the amplifier should be connected to an 8 ohm speaker, I don't ever remember hearing anything about the speaker reactance having to be capacitive or inductive.
The frequency matters. Audio doesn't count. When in say the MHz region, it's best to use silver plated copper tubes for wires. More like plumbing.

There is an amplifier spec called damping factor. To get the amplifier output Z = (rated load)/(damping factor).

More education about speakers: http://education.lenardaudio.com/en/05_speakers_3.html So, now you know-- 8 ohms at 400 Hz.

#### Ylli

Joined Nov 13, 2015
1,058
Avoid the electrolytic caps if possible. If you must, you want them in series with opposing polarities. Put a diode across each so that the anode of the diode goes to the - terminal and the cathode goes to the + terminal. You want to prevent reverse voltage on the electrolytic cap.

#### GopherT

Joined Nov 23, 2012
8,012
Good thinking. I will just use my bench power supply with a 15v AC output with a 12v DC bulb.
Still hoping to hear back on the Electrolytic Cap orientation question.
A non-polarized capacitor can be made from an electrolytic by putting two polarized electrolytic negative to negative and then put that in series with your bulb.

For a 12 W bulb at 12v,
You should get a pair of 100u, 330u and 1000u capacitors to do this experiment (electrolytics). Make sure the working voltage Of the caps is 30 or more volts.

A 12V AC source, a 27 w bulb and...
- two 10,000uF caps back to back (2900K bulb temp) - nearly the same as a solid wire.
- two 3300 uF (2700K)
- two 2200 uF (2500K
- two 1000 uF (1600K)
- two 680 uF (1000K)
- two 470 uF (570K)
- two 330 uF (340K) - slightly above room temp

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#### pratto

Joined Dec 10, 2012
36
The frequency matters. Audio doesn't count. When in say the MHz region, it's best to use silver plated copper tubes for wires. More like plumbing.

There is an amplifier spec called damping factor. To get the amplifier output Z = (rated load)/(damping factor).

More education about speakers: http://education.lenardaudio.com/en/05_speakers_3.html So, now you know-- 8 ohms at 400 Hz.
While your response wasn't helpful (it seemed like extraneous info), the link you sent was. After reading the 'Load Impedance' section a couple of times, I think I understand source impedance better.

It isn't a question of the source having an impedance, it is all about specifying a minimum load resistance in order to prevent the source from frying itself by using too much current. I have some more questions, but is that part right ?

#### KeepItSimpleStupid

Joined Mar 4, 2014
5,088
While your response wasn't helpful (it seemed like extraneous info), the link you sent was. After reading the 'Load Impedance' section a couple of times, I think I understand source impedance better.

It isn't a question of the source having an impedance, it is all about specifying a minimum load resistance in order to prevent the source from frying itself by using too much current. I have some more questions, but is that part right ?
An IDEAL voltage source has 0 impedance. An ideal current source has infinite impeadance. Voltage sources and current sources can be "transformed".

Yes, you don't want to fry the power supply, but the load is generally function of time and not constant. A power plant that supplies the home may be 5 MW, but the circuits in the home are protected by a fuse suitable for the wire gauge used.

#### Sensacell

Joined Jun 19, 2012
3,069
Note that a tungsten filament lamp is not an entirely linear resistance!

The filament has a lower resistance when its cold, this could throw your math and measurements out of whack if you assume it's linear.
(this drove me near insane when I was a kid, learning ohms law)

Just for fun, I measured the resistance of a Philips Par 38 lamp I have, it's rated 220 volt, 120 Watts, incandescent.
It reads 33 ohms cold with my meter... doing the math, that's... 1466 watts the instant you turn it on.

This highly nonlinear behavior is a trap that can cause much confusion- the damn lamps should come with a warning label!

"Warning, non-linear resistance"

#### pratto

Joined Dec 10, 2012
36
Avoid the electrolytic caps if possible. If you must, you want them in series with opposing polarities. Put a diode across each so that the anode of the diode goes to the - terminal and the cathode goes to the + terminal. You want to prevent reverse voltage on the electrolytic cap.
And being that it is in series with the light, is the side closest to the light considered negative ?

#### GopherT

Joined Nov 23, 2012
8,012
And being that it is in series with the light, is the side closest to the light considered negative ?
Leave off the diodes!

#### pratto

Joined Dec 10, 2012
36
Leave off the diodes!
And being that it is in series with the light, is the side closest to the light considered negative ?

#### ebeowulf17

Joined Aug 12, 2014
3,306
And being that it is in series with the light, is the side closest to the light considered negative ?
When the caps are back to back, their negative sides face each other, not any other part of the circuit:

Power ... (+cap-) ... (-cap+) ... Lamp

Imagine the two polarized caps have combined into one non-polarized cap with only positive sides:

(+cap--cap+)

#### pratto

Joined Dec 10, 2012
36
When the caps are back to back, their negative sides face each other, not any other part of the circuit:

Power ... (+cap-) ... (-cap+) ... Lamp

Imagine the two polarized caps have combined into one non-polarized cap with only positive sides:

(+cap--cap+)
thanks.