Short high frequencies to ground

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
Hi all,

In chapter 5 of Make: Electronics (experiment with 555 timer and high/low pass filters) the author says the Ac to Dc adapter might introduce some hum. He writes you can reduce it somewhat by putting a 1000 uF capacitor between Vcc and ground.

I understand the capacitive reactance is small for high frequencies and therefore those interfering frequencies (probably any non DC?) will be shorted to ground and no longer cause hum in the speaker. I wonder however why this shorting to ground is not a problem. In the end those frequencies are electricity as well and creating a short sounds like something you wouldn't want. Is it because those frequencies are "low energy" and therfore harmless?

Where will their voltage be dropped? Across the internal battery resistance?

Many thanks,
Dirk
 

Picbuster

Joined Dec 2, 2013
1,026
Hi all,

In chapter 5 of Make: Electronics (experiment with 555 timer and high/low pass filters) the author says the Ac to Dc adapter might introduce some hum. He writes you can reduce it somewhat by putting a 1000 uF capacitor between Vcc and ground.

I understand the capacitive reactance is small for high frequencies and therefore those interfering frequencies (probably any non DC?) will be shorted to ground and no longer cause hum in the speaker. I wonder however why this shorting to ground is not a problem. In the end those frequencies are electricity as well and creating a short sounds like something you wouldn't want. Is it because those frequencies are "low energy" and therfore harmless?

Where will their voltage be dropped? Across the internal battery resistance?

Many thanks,
Dirk
Hi Dirk,

first thing to do is to draw a schematic. ( don't forget the internal resistance)
Add pulses to the DC circuit drawings.
Simplify the drawings and go back to the working of a capacitor as described in your physics book.

This will answer all your questions.

Good hunt.

Picbuster
 

MrChips

Joined Oct 2, 2009
22,870
In the case of a 1000μF capacitor after the rectifier diodes in a DC power supply, we think of the capacitor as a storage of charge, called the reservoir capacitor. We look at it from the perspective of trying to hold the voltage constant rather than trying to remove high frequencies.

You are not going to be able to remove all of the line frequency signal. Your objective would be to reduce the AC ripple voltage to an acceptable level.

See this:
Unregulated Power Supply Design
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
Hi,

Thanks for the quick responses.

I understand how the low pass filter works and learned a lot about it by looking at the voltages with an oscilloscope. The cutoff frequency is where the voltage across the resistor and capacitor are equal. Higher frequencies shift towards the resistor: their voltage increases across the resistor and becomes lower across the capacitor. For lower frequencies the reverse takes place which is why they "pass"... they're not dropped (or at least less) across the resistor.

In case of my question: I'm trying to find out where the higher frequencies are being dropped. Somewhere across some component inside the power adapter? I have a somewhat related question with a schematic, I'll try to post it later on ....

Dirk
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
By the way: indeed, the capacitor seems more applicable as a storage device here. But the author must have had a reason to mention it (to filter humming sounds)?!
 

MrChips

Joined Oct 2, 2009
22,870
Hi,

Thanks for the quick responses.

I understand how the low pass filter works and learned a lot about it by looking at the voltages with an oscilloscope. The cutoff frequency is where the voltage across the resistor and capacitor are equal. Higher frequencies shift towards the resistor: their voltage increases across the resistor and becomes lower across the capacitor. For lower frequencies the reverse takes place which is why they "pass"... they're not dropped (or at least less) across the resistor.

In case of my question: I'm trying to find out where the higher frequencies are being dropped. Somewhere across some component inside the power adapter? I have a somewhat related question with a schematic, I'll try to post it later on ....

Dirk
It is a misconception to think that the higher frequencies are being dropped somewhere else.
The higher frequencies are attenuated across the component with the lower reactance.

Here is basic voltage divider circuit:

1596640039002.png

Vout is Vin attenuated.

In this case,

Vout = Vin x Z2 / (Z1 + Z2)

Since Z1 and Z2 are frequency dependent, Vout is a function of frequency. That is the source of the high frequency attenuation.
 

dl324

Joined Mar 30, 2015
12,426
the author says the Ac to Dc adapter might introduce some hum. He writes you can reduce it somewhat by putting a 1000 uF capacitor between Vcc and ground.
The 1000uF capacitor you referred to is a storage capacitor to smooth out the peaks and valleys from the rectified AC; it's usually called a filter cap.

There's a way to calculate an appropriate value for the filter cap. From a National Semiconductor Voltage Regulator Handbook (1980):
clipimage.jpg

The allowable amount of ripple depends on how much head room you have for the voltage regulator.

Here's an illustration of ripple from https://www.electroschematics.com/capacitor-input-filter-calculation/
1596642112196.png
I understand the capacitive reactance is small for high frequencies and therefore those interfering frequencies (probably any non DC?) will be shorted to ground and no longer cause hum in the speaker.
That's only for an ideal capacitor. A 1000uF electrolytic capacitor is going to have some parasitic resistance and inductance (from leads and wires). While capacitance decreases when frequency increases, inductance increases. Because of that, ceramic capacitors are used to decouple high frequency noise because they have lower parasitic inductance.

We don't usually think of hum as high frequency.
Across the internal battery resistance?
If you're using a battery, you don't have any hum to filter. Hum is caused by the 60/120 Hz (or 50/100 Hz) ripples from the rectified AC.
 
Last edited:

bogosort

Joined Sep 24, 2011
596
I understand how the low pass filter works and learned a lot about it by looking at the voltages with an oscilloscope. The cutoff frequency is where the voltage across the resistor and capacitor are equal. Higher frequencies shift towards the resistor: their voltage increases across the resistor and becomes lower across the capacitor. For lower frequencies the reverse takes place which is why they "pass"... they're not dropped (or at least less) across the resistor.
You're on the right track. Consider what happens if you remove the capacitor and replace it with a wire to ground:

1596729225489.png

With your scope connected between Vout and ground, you'll measure 0 V. (If you try this, use at least a 1 kΩ resistor to keep the current low.) With your scope between Vin and ground, you'll measure Vin. Thus, we can say that all of Vin is being "dropped" by the resistor R1. Another way of conceptualizing this is that -- after R1 -- all of the current is shunted to ground. Since no current reaches whatever is connected to Vout, no voltage can be formed there.

Now let's remove the shunt to ground:

1596730821067.png

Assuming that whatever you connect to Vout has a much higher resistance than R1 -- which is certainly true for an oscilloscope probe -- the voltage measured at Vout will be very close to the voltage at Vin. In the "voltage drop perspective", we'd say that the vast majority of Vin is being dropped by the scope, and a fraction of Vin is being dropped by R1. In the "current flow perspective", we'd say that all of the source current reaches the scope, albeit at a slightly reduced magnitude because of R1.

Suppose we add another resistor where the shunt used to be:

1596730394307.png

We saw what happens when R2 is zero (a piece of wire): all of the current goes to ground, producing 0 V at Vout. We also saw what happens when R2 has an enormous resistance value (air): all of the current goes to the scope, producing Vin at Vout. It should be clear that the value of R2 controls how much current reaches Vout versus how much current gets shunted to ground. Therefore, we can get any value we want at Vout in the range [0, Vin] by varying the resistance of R2. As others have mentioned, the formula for this is \[ V_{out} = V_{in} \left( \frac{R_2}{R_1 + R_2} \right) \] Mathematically, this formula tells us that as \(R_2 \to \infty \), the value of Vout approaches Vin; likewise, as \( R_2 \to 0\), the value of Vout approaches 0.

What happens if we replace R2 with a capacitor? That is to say, what happens if we replace the resistance of R2 with the reactance of a capacitor?

1596735917153.png

Reactance, which has units of ohms, is like frequency-dependent resistance. In the case of capacitors, reactance is inversely proportional to frequency. For a capacitor with capacitance value of C, and an applied voltage of frequency f, its reactance is given by the following formula: \[ X_C = \frac{1}{2 \pi f C} \] For any given capacitance value, as frequency goes up, capacitive reactance -- and, hence, the capacitor's resistance -- goes down. Likewise, as frequency goes down, the capacitor's opposition to current increases.

Let's put some numbers to this to get a quantitative sense of the effect. Suppose we're using a 1000 μF capacitor as your book suggests. If Vin is a DC source (0 Hz), the capacitor's reactance approaches infinity; in other words, it acts like an open circuit (third image above) and lets all the current through to Vout. If Vin is an extremely high frequency, say 1 GHz, then the capacitor's reactance is about 159 billionths of an ohm; in other words, for very high frequencies, the capacitor acts like a piece of wire (first image above), and we'll find 0 V at the output.

Just as we could vary the resistance of R2 to set the voltage at Vout, we can play with values of C to change the frequency content of the voltage at Vout. For example, suppose our input signal is a combination of 5 VDC (0 Hz) and a 1 kHz sinusoid: \[ V_{in} = \sin(2 \pi 1000 t) + 5 \] As far as the DC component of the input is concerned, the capacitor is an open circuit and the 0 Hz current goes straight to the output. However, to the 1 kHz component, the capacitor looks like approximately 0.159 Ω -- a tiny resistor -- and so almost all of the 1 kHz current gets shunted to ground. In other words, we've effectively filtered the higher frequency component from the output.

In case of my question: I'm trying to find out where the higher frequencies are being dropped. Somewhere across some component inside the power adapter?
Suppose R1 is 1 kΩ, C1 is 1000 μF, and Vin is a pure 1 kHz signal with an amplitude of 5 V. From the voltage drop perspective, we'd measure 5 V between Vin and ground, and we'd measure almost 0 V between the capacitor and ground. Thus, practically all of the voltage was dropped across R1. From the current flow perspective, C1 acts like a shunt to ground for the 1 kHz current.

On the other hand, if we make the value of C1 much smaller -- say 100 nF -- then about half of the 1 kHz voltage will be dropped across the resistor and half across the capacitor. Thus, we'd measure about 0.5 V at Vout (which is the same as the voltage across the capacitor). This is because with the much smaller capacitance, the capacitor no longer acts like a piece of wire to the 1 kHz signal; it acts like a resistor of about 1.59 kΩ.
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
Hi,

Thanks to you all for taking the time to help me out here!

It is a misconception to think that the higher frequencies are being dropped somewhere else.
Correct, I phrased it wrong. Thanks for sorting that out!

If you're using a battery, you don't have any hum to filter.
:)Very true, something I overlooked.

From the current flow perspective, C1 acts like a shunt to ground for the 1 kHz current.
If R1 would not be present and C1 (1000 uF) would be connected directly between the terminals. Say you still have your 5VDC and a 1kHz sinusoid. The impedance of C1 would still be 0,159 Ohm for the sinusoid. If the amplitude would be big (e.g. 2 volts up, 2 volts down w.r.t. 5VDC), this could lead to quite some current continuously going through C1. Correct? In such situation this shorting to ground would be bad, no? So I guess it’s okay as long as the sinusoid‘s amplitude is rather small....

Regards, Dirk
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
This is the experiment I was referring to earlier: (from Make: Electronics)

E4CB57B3-3EB4-4F2E-BDA3-F4425463DD5F.jpeg
When you press button C4 you’re attenuating the lower frequencies. The oscilloscope clearly illustrates this:

01A1A12B-516A-406A-A7C8-D79964820807.jpeg

Red is across the capacitor, yellow across the speaker. They add up to a square wave:

D47FFEFD-5B35-4E80-876C-67EF3294A1B3.jpeg

When the filter is active I can still see the original square wave when I scope at the lm386 output.

Now when I turn off this filter and enable switch C3 things change. The filter works and attenuates higher frequencies. But I can’t find where the higher frequencies are attenuated. I mean across which component... At C7 you only see the lower frequencies. But when I scope at the lm386 input or 555 output I still only see those lower frequencies. Is it attenuated then across some internal resistor in the 555? I don’t really get it :)

Thanks!
Dirk
 

MrChips

Joined Oct 2, 2009
22,870
You are making the same mistake again!
There is no hidden resistor in the circuit.
Review the voltage divider circuit I posted before. Study how a voltage divider works.

1596809245700.png

Z1 and Z2 are reactances which are frequency dependent.

If Z1 gets large, Vout is reduced.
If Z2 gets small, Vout is reduced.

Vout = Vin x Z2 / (Z1 + Z2)

What you are showing is an experimental setup to demonstrate the effect of different reactances on speaker output.


L and C impedance.jpg


SPK1 receives the full output with no attenuation.

SPK2 receives output with high frequencies attenuated. This is because the reactance of L1 increases with frequency.
Reactance XL = ωL

SPK3 receives output with low frequencies attenuated. This is because the reactance of C4 increases as frequency decreases.
Reactance XC = 1 / ωC

BTW, there is no C4 button in your example circuit. C3 and C4 refer to capacitors, not switches.
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
Indeed, I meant the S3 switch/button.

I do understand how a voltage divider works, both for DC and AC. I'll make a drawing to illustrate the case I was talking about. I forgot some details now that I reread my post. It's not one of the situations you drawed above.

Dirk
 

Thread Starter

- Dirk -

Joined Aug 8, 2019
14
20200807_211746.jpg

This is the circuit I investigated. I've only drawn the part starting at the lm386 output. I didn't build the part which connects the inductor, that was something I wanted to test later on. To keep things simpler.

The 100 uF capacitor is in parallel with the speaker. I know the 220 uF and 100 uF capacitors form a voltage divider but when I compare the voltage across them...the shape is almost exactly the same. I was expecting to see the attenuated higher frequencies somewhere.

I'll play with some more frequencies... maybe that will make things clearer on the scope.

Dirk
 

MrChips

Joined Oct 2, 2009
22,870
The question is, do you know what low frequencies and high frequencies look like on an oscilloscope?
I suspect that you don't.

Disconnect the 100μF capacitor. It will only confuse the issue.
Now you want to see the difference without the 2.2uF in circuit and with the 2.2uF in series with the loudspeaker.
What should the signal look like at the loud speaker on the oscilloscope?

A capacitor in series is a differentiator. It responds to changes in the signal and fails to respond to constant voltages.
So in response to a square-wave signal you should see the derivative of a square wave. (How much calculus do you remember?).

Differentiator
1596829634840.png
 
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