# Hi! i want to calculate the Rinput and Routput of this Active Filter.

1. Nov 7, 2013

### Fovakis

Hi! i want to calculate the Rinput and Routput of this Active Filter.

How can i do that?

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2. Nov 7, 2013

### Staff: Mentor

I believe you mean Zin and Zout, right?

Do you want to calculate those impedances, or measure them with your simulator? What frequency range do you want this circuit to operate over?

3. Nov 7, 2013

### Fovakis

yeah exactly this ! :)

So the frequency bandwidth is DC to 17khz (audio)! Before this filter there is a double balanced ring diode mixer that downconvert a signal of 30MHz to a baseband DC to 17khz. The mixer's output (ideally) is 50ohm.

RF front....---Mixer-----Active Filter--Demodulator (DSP). I want to calculate the total gain of the stages of receiver but in the filter i don't know the Zinput and Zoutput to convert the voltage gain=20log(Vout/Vin) to power gain. I need to know what is the Zout and Zin. After the Active filter there is a DSP kit i think it's Zinput is about 10k maybe not sure...

4. Nov 7, 2013

### Staff: Mentor

Well Zout is easier to find, since it is dominated by the low output impedance of the LM358, along with the series cap. Look in the LM358 datasheet for Zout numbers.

For the Zin, use the "virtual ground" property of the first opamp, and ground the input circuit component at the opamp's - terminal. Connect the right side of C5 to ground through the equivalent output impedance of that first opamp. That should give you a mixed RC circuit that you can analyze with KCL.

5. Nov 7, 2013

### AlephZero

A practical way to find Zin (which will be frequency-dependent, in general) is:

1. Disconnect the filter input from the rest of the circuit
2. Short circuit the filter output
3. Connect an ideal voltage source (i.e. a sine wave) ito the input and measure the input current.

If the output of the filter is the output of an op-amp (U1B in your diagram?) then for practical purposes the output impedance is the same as the op-amp output impedance, unless the filter's feedback loop has a very low impedance for some strange reason. Check the op-amp datasheet if you really need the value, but unless the filter is driving a low impedance input (which is probably a bad design idea anyway) the exact value isn't important.

EDIT: or do what Berkeman said - the two posts had exactly the same time stamp before I edited this one!

6. Nov 7, 2013

### Staff: Mentor

Great minds think alike!

7. Nov 7, 2013

### Fovakis

Hi AlephZero ! thanks! i am a begginer so please more simple the explanations if you can :(

First of all i understand that Zout is Zoutput of Operation Amplifier + pluse Zof capacitor=1/jωC where C=10^(-6)Farad. (basically i have a dual fet OPA2608 not the LM358 that is at the schematic).

What do you mean the filter input? what components consisting that?

8. Nov 7, 2013

### Staff: Mentor

The input to the filter is the SMA connector on the left...

9. Nov 9, 2013

### Fovakis

can i have more help please..

10. Nov 10, 2013

### Staff: Mentor

@Fovakis. The input impedance is not a fixed value, it varies with frequency following a complicated equation. What do you need to know about it? What is the important consideration? Instead of trying to calculate it, it would be much easier to ask your simulator to "measure" the input impedance at various frequencies to allow you to plot a graph of impedance over the range of frequencies of interest. Are you sure that the component values on your schematic will give the filter response that you need? Have you asked your software to plot the passband?

@berkeman. I see no virtual earths in this schematic.

11. Nov 10, 2013

### Fovakis

yes you right about that. what simulator can i use for that? Is there any tool or toturial to find out?

Also, if i want to calculate the Zin for f=20khz. it would be much better to do the maths i guess.

I am starting, first a Capacitor with Zc=1/j2πfC1..8/j Farads. Is it right to do with capacitors? then i must add this with the voltage diviver ? i will have a complex number for Zin right?

12. Nov 10, 2013

### Staff: Mentor

BTW, you won't get any DC through this filter with those 3uF caps in the way. Are you wanting its passband to extend right down to DC (i.e., are you wanting a low-pass filter)?

13. Nov 10, 2013

### Fovakis

Yes it is a low pass filter the first stage. the second is an amp. :tongue:

14. Nov 10, 2013

### Staff: Mentor

I presumed the KiCad that you drew the schematic on?

You'll have to ask others for recommendations, otherwise. I can't offer advice on PC software.

I had concluded that it was beyond your ability, at this stage.

Zin will be a different complex number for every frequency. The equation is very complicated.

15. Nov 10, 2013

### Staff: Mentor

So what modifications are you going to make to it, to turn it into a filter that passes DC? It has just ratcheted up a notch in complexity, because the circuit is shown operating off a single power supply. It won't do well preserving the DC component of the input.

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon2.gif [Broken] You need to get it clear from the start: do you need its response to include DC?

Last edited by a moderator: May 6, 2017
16. Nov 10, 2013

### Fovakis

http://i1284.photobucket.com/albums/a578/fovos1/SISOFILTE_zps56bc4977.png

I don't want this filter to PASS any dc signal , because at the next stage there is a demodulator. So i need to cut the dc signal that maybe leakage from other stages to the input of my active filter. My circuit is operating with a single power supply (DC). I use the single power supply to operate my operational amplifiers. please check my photo above. this is the right filter. thanks for your patience

Last edited by a moderator: Nov 10, 2013
17. Nov 10, 2013

### Staff: Mentor

As you already have the filter constructed (and I presume, tested) it should be a simple matter to use a signal generator and a couple of probes to take measurements to determine the input impedance vs frequency graph?

18. Nov 10, 2013

### Staff: Mentor

Yes, you are correct. I misspoke.