# Resistor role in a high pass filter

Discussion in 'General Electronics Chat' started by boogie, Dec 23, 2010.

1. ### boogie Thread Starter New Member

Jan 10, 2010
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0
Hi guys

looking at the diagram and explanation on a capacitive HPF in the filters chapter, everything makes sense and is understood. A capacitor is placed in series with the load, so far so good (As in the below image):

However in every other source on the internet, there is always a resistor placed in parallel to the output (which I assume is the where the load will be), as in the following image:

My question is why do I need the resistor? the first diagram made more sense.
If for example the load will have much higher impedance, then the resistor will almost completely shunt the signal and if the load will have much lower impedance then it will completely shunt the resistor.
So why is the resistor there?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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This resistor represents the load. But sometimes we use extra resistor in case when we disconnect the load then this extra resistor provide a path to GND.
Prevent input from Floating. And then we use Rx>Rload

And remember that we can look at filters as a voltage dividers.
For example a low-pass filter consists a resistor and capacitor connected in series. The output voltage is taken from capacitor.
This circuit should be treated as a voltage divider where the role of the variable resistance is take by capacitor. For low frequencies the capacitor has a high reactance so almost the entire input voltage is transferred to the output.
When input frequency increase the reactance of a capacitor decreases and thus causes output voltage to decrease.
At frequency at which reactance of a capacitor is equal to the value of resistor the output voltage is equal Vout=0.707*Vin
And this happens at frequencies equals

$F=\frac{1 }{2*\Pi*R*C}$

The divider's voltage ratio Vout/Vin of a voltage divider that contain two resistors doesn't depend on frequency, because the resistance of resistors does not change with frequency.
In our case the divider's voltage ratio change with frequency.
Because R is unchanged but capacitor reactance Xc is change with frequency

Last edited: Dec 23, 2010
3. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
I'm not sure I understand ... especially the voltage divider part. why would I need one for a filter? Let's start with the low pass filter with a resistor before the cap:
The resistor comes first and there's some loss of voltage on it - i.e - we lose some signal - what is that good for? what would happen if we get rid of it? on low frequencies the capacitor would have high reactance and most of the signal would go to the load (exactly what we want). On high frequency signals the capacitor would have very low reactance and will shunt most of the signal away from the load (again exactly what we want). The only situation I can think of, where the resistor would be useful is if the output (load) had extremely low impedance or was shorted - and then the resistor would help prevent a short-circuit.
The same goes to the high pass filter - the only reason I can think of why we need a resistor is to prevent a short circuit.
Am I correct?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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This is "Voltage" low-pass filter.

And what I mean by "Voltage" is that filter is connect to ideal voltage source, and Rload = ∞.
And we have voltage phase shift, and voltage is attenuated
And in reality we choose R << Rload so there is no problem with extra attenuation.

And you give as example of a "current" filter.
The current phase is shift and current is attenuated.
So this circuit cannot be connect to voltage source.
Becaues you have to remember that voltage is the same across all branches in a parallel circuit.

5. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
Jony13 - thanks for your help but I'm even more confused now
How did phase shift get into the picture now? and what does it have to do with the resistor? resistors don't cause phase shift. The capacitor can cause a phase shift but in your diagram the signal goes to Vout before it reaches the capacitor....
I also don't understand what you meant by voltage filter and current filter.

I still don't understand why we need the resistor...

6. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
bumping this up since i still don't understand.
Can anyone explain to me what would happen if I lose the the resistor and just use the capacitor before the load.
What is the resistor good for?

Thanks

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
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First this classic low-pass filter.

And in ideal case we assume RL = ∞
So Vout is equal to voltage on capacitor.
And all the components in a series connection carry the same current

Vout = I * Xc = Vin * Xc/( R + Xc) = Vin* Xc/Z
So the gain is equal :

$K = \frac{Vout}{Vin} = \frac{Xc}{Z} = \frac{\frac{1}{\omega *C}}{\sqr{ R^2 + (\frac{1}{\omega *C})^2}}=\frac{1}{\sqr{R^2*\omega^2 *C^2 +1}$

And know if we remove R1 we end-up with this circuit

And now because all components are connected in parallel Vin = Vout (voltage is the same across all branches in a parallel circuit)
So this circuit no longer work as a filter because Vin = Vout thanks to KVL.

The only think that is change in this circuit is current.

I1 = Ic + Ir

The load current is constant and equal:
Ir = Vin/R and this current will not chance with frequency.
And Ic will increase with frequency. But this will not make any change in Ir current.

To make this circuit to work, we need replace the Vin voltage source with a current source

And know this circuit work as you describe here

Do you understand now ?

Last edited: Nov 13, 2015
8. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
Thank you so much for your very detailed response.
THANK YOU - THANK YOU!!!
The low pass filter makes perfect sense and is much clearer now.

However my initial question about the resistor in the high pass filter still remains.
What is it used for? You said that it's used to provide a path to ground in case the load is disconnected and prevents the signal from floating.

I have two questions:
1. If the load is disconnected, then there will be an open circuit and no current will flow in the circuit, so where is the problem? there is no current to go to ground.
2. what does it mean for a signal to be "floating"

Once again - thank you so much for taking the time to answer my questions so thoroughly. I appreciate it a lot.

9. ### #12 Expert

Nov 30, 2010
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7,489
Referring to your first post, first drawing, point 2 to point 0 as the voltage measuring place:

On a very simple level, any RC filter depends on the impedance of the capacitor changing as the frequency changes. Changes impedance compared to what? Compared to the resistor.

Pretend the capacitor is a variable resistor and you will see that without a resistor to ground (point 0), it doesn't have any effect. When the resistor to ground is there, you can calculate the voltage out as if you are using a voltage divider circuit, but the capacitor changes impedance as the frequency changes so you get a different voltage out for every frequency.

I hope this helps.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
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If we talk about the ideal the high-pass filter is identical to the low-pass filter except the components have been reversed, with the capacitor in
series with the input, and the resistor in parallel with the output signal.
Note that in this case, the output signal is the voltage drop across the resistor. Once again, the capacitor and the resistor form an AC voltage-divider network.

But in ideal cases Rload = ∞
So this R1 resistor represents R_load.

As I said earlier in some diagrams we can see that extra resistor is add.
But in this case we talking about AC-coupling.
http://en.wikipedia.org/wiki/Capacitive_coupling#Use_in_analog_circuits
And we put this extra resistor to provide path to ground.
Without this extra resistor one plate of a capacitor is connect to some DC voltage, and the second plate of a capacitor is floating.
But is always good to have "steady state" even if load is disconnect.
This additional resistor helps to prevent "turn on thump" when we try to connect the load.

And sometimes we add this "extra" resistor simply becaues R_load has a very high resistance (almost ideal case).
So we need this resistor to determine the cut-off frequency.

11. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
Thanks again for the explanation.
I understand the logic when R-load is very high.
I'm a beginner so I don't quite understand the entire coupling thing ... I'll try to read up on it.
By "steady state" did you mean that the reason we have the resistor is to discharge the capacitor when we have an open circuit?
Also, in your first reply you said that Rx > Rload. But don't we want to use a lower resistance for Rx so we will be able to attenuate the resistance in case of a very high Rload? so we can determine the cutoff frequency?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
1,122
Yes
It is all depend on situation and what designer want to achieve.
If R_load is relative small (and RL can be disconnect) we can use Rx>>R_Load. And when RL>>Rx then Rx determine Fc.