Question about long RC time?

[samy555]

Hello all
In this circuit:

from: http://stab-iitb.org/wiki/Mobile_Bug
I worked some calculations like:

I didn't understand why the designer made the (R1*C4) time very large (220 seconds), that is it take the DC voltage at pin #3 of CA3130 IC a 220 sec to increase from 0 volt to 0.522 volt?
Can you please explain to me the benefit of that.
thanks

 

[Milad-D]

The time constant at node 3 is determined by the parallel combination of R1 and R2 and C4. So its time constant is much smaller than what you have stated.

 

[chuckey]

I think there is an error in the circuit. The + input of the opamp is decoupled with the 100MF. The active (-) input is connected to this point via C 3, so at any frequency where the reactance of C3 is low, the + and - inputs of the amp are shorted together. So the input is only going to respond at frequencies where the C3 reactance is high (>100K), so its a "bass amp" !! The input is coupled via a 22PF capacitor, which won't pass bass frequencies. I would try to connect the - end of C3 to earth, so the + input is solidly bypassed to RF.
Frank

 

[samy555]

The time constant at node 3 is determined by the parallel combination of R1 and R2 and C4. So its time constant is much smaller than what you have stated.

Did you mean that C4 charging is done through the parallel combination of R1 and R2, and discharging through R2?

I think there is an error in the circuit. The + input of the opamp is decoupled with the 100MF. The active (-) input is connected to this point via C 3, so at any frequency where the reactance of C3 is low, the + and - inputs of the amp are shorted together. So the input is only going to respond at frequencies where the C3 reactance is high (>100K), so its a "bass amp" !! The input is coupled via a 22PF capacitor, which won't pass bass frequencies. I would try to connect the - end of C3 to earth, so the + input is solidly bypassed to RF.
Frank

Unfortunately, I did not understand
If it is possible to repeat the idea in an easier way
This video shows that the circuit is working perfectly:
https://www.youtube.com/watch?v=ewXShfRy-rM

To everyone: What is the purpose of having R1, R2 and C4 network?
thanks alot

 

[Milad-D]

1) to make the DC voltage at node 3 equal to R2/(R2+R1)*12~0.5 V
2) The capacitor C4 is used to make this node ground at high frequency.

"Did you mean that C4 charging is done through the parallel combination of R1 and R2, and discharging through R2? "

No, it will be charged and discharged by the parallel combination of the two resistors.

 

[schmitt trigger]

To everyone: What is the purpose of having R1, R2 and C4 network?
thanks alot

Is to bias the noninverting input(+) tp about 0.521 volts DC. That is it. By coincidence it will take about 9.5 seconds to charge, but it is not relevant to the operation. It is not 220 seconds, the actual equivalent charging resistance is the parallel of 2.2M and 100k.

I have see the circuit, and I agree with chuckey: the circuit, as designed does not work. There are way too many reasons.
And yes, I've seen the video. That it appears to work is a fluke, triggered by parasitics rather than straight operation.

 

[samy555]

2) The capacitor C4 is used to make this node ground at high frequency.

The transmission frequency of mobile phones ranges from 0.9 to 3 GHz.
At 1 GHz XC4=1.6 micro ohm
At 10 MHz XC4=160 milli ohm
There is no big difference. I don't think that C4 is to ground the node at high frequency. Why not to say that CA3130 operate at freq up tp 15MHz, where the recieved signls range from 0.9 to 3 GHz and C4 do the delay process?

No, it will be charged and discharged by the parallel combination of the two resistors.

C4 will charge when closing the main switch. When it will be discharged?

Is to bias the noninverting input(+) tp about 0.521 volts DC. That is it. By coincidence it will take about 9.5 seconds to charge, but it is not relevant to the operation. It is not 220 seconds, the actual equivalent charging resistance is the parallel of 2.2M and 100k.

If the delay time was by coincidence and has no role in the work of the circuit, does it mean that C4 can be indispensable?

I have see the circuit, and I agree with chuckey: the circuit, as designed does not work. There are way too many reasons.

Please tell me in a simplified manner what is the most important of these reasons

Thank you very much

 

[schmitt trigger]

The transmission frequency of mobile phones ranges from 0.9 to 3 GHz.

Please tell me in a simplified manner what is the most important of these reasons

Thank you very much

At those frequencies, the selected components and the layout are vital. Ground planes, stripline tuned circuits, impedance matching, neutralizing parasitic elements....... the list goes on and on.

The photo in the the attached link shows a protoboard with long wires.

 

[samy555]

At those frequencies, the selected components and the layout are vital. Ground planes, stripline tuned circuits, impedance matching, neutralizing parasitic elements....... the list goes on and on.

The photo in the the attached link shows a protoboard with long wires.

Yes, it is as you mentioned, but the circuit does not work at microwave frequencies, it picks up only a portion of the signal power, CA3130 operates at a frequency of only 15 MHz.
You did not answer my question about C4?

thanks

 

[schmitt trigger]

"CA3130 operates at a frequency of only 15 MHz."

That is precisely my point......mis-interpretation of data sheets.

15 Mhz is not the operating frequency, it is only the unity gain crossover frequency....google the term.

And the spec sheet shows that would be only true with a compensation capacitor of zero pF...the schematic shows 47 pF, which would lower that value to 4 Mhz.

But even then, let's assume that the CA3130 could indeed work at 15 Mhz.....the phone's carrier frequency is 900 to 3000 Mhz, orders of magnitude higher. It still won't detect anything at the frequencies that a mobile phone is transmitting. It will pick up powerline harmonics, AM and short wave radio, SMPS noise.....but no cell phone transmissions.

Now...C4 is to clean the DC bias applied to the non inverting input. But its value is waaaaaaay too large, 1 uF would be enough.

There are many other mistakes like that.

- - - Updated - - -

"CA3130 operates at a frequency of only 15 MHz."

That is precisely my point......mis-interpretation of data sheets.

15 Mhz is not the operating frequency, it is only the unity gain crossover frequency....google the term.

And the spec sheet shows that would be only true with a compensation capacitor of zero pF...the schematic shows 47 pF, which would lower that value to 4 Mhz.

But even then, let's assume that the CA3130 could indeed work at 15 Mhz.....the phone's carrier frequency is 900 to 3000 Mhz, orders of magnitude higher. It still won't detect anything at the frequencies that a mobile phone is transmitting. It will pick up powerline harmonics, AM and short wave radio, SMPS noise.....but no cell phone transmissions.

Now...C4 is to clean the DC bias applied to the non inverting input. But its value is waaaaaaay too large, 1 uF would be enough.

There are many other mistakes like that.

 

[samy555]

"CA3130 operates at a frequency of only 15 MHz."

That is precisely my point......mis-interpretation of data sheets.

15 Mhz is not the operating frequency, it is only the unity gain crossover frequency....google the term.

And the spec sheet shows that would be only true with a compensation capacitor of zero pF...the schematic shows 47 pF, which would lower that value to 4 Mhz.

But even then, let's assume that the CA3130 could indeed work at 15 Mhz.....the phone's carrier frequency is 900 to 3000 Mhz, orders of magnitude higher. It still won't detect anything at the frequencies that a mobile phone is transmitting. It will pick up powerline harmonics, AM and short wave radio, SMPS noise.....but no cell phone transmissions.

Now...C4 is to clean the DC bias applied to the non inverting input. But its value is waaaaaaay too large, 1 uF would be enough.

There are many other mistakes like that.

- - - Updated - - -

"CA3130 operates at a frequency of only 15 MHz."

That is precisely my point......mis-interpretation of data sheets.

15 Mhz is not the operating frequency, it is only the unity gain crossover frequency....google the term.

And the spec sheet shows that would be only true with a compensation capacitor of zero pF...the schematic shows 47 pF, which would lower that value to 4 Mhz.

But even then, let's assume that the CA3130 could indeed work at 15 Mhz.....the phone's carrier frequency is 900 to 3000 Mhz, orders of magnitude higher. It still won't detect anything at the frequencies that a mobile phone is transmitting. It will pick up powerline harmonics, AM and short wave radio, SMPS noise.....but no cell phone transmissions.

Now...C4 is to clean the DC bias applied to the non inverting input. But its value is waaaaaaay too large, 1 uF would be enough.

There are many other mistakes like that.

I am convinced
Thank you

 

[FvM]

Although I basically agree with the previous comments about the circuit, I think the bread board design in the original link might explain why the circuit exposes a certain succeptebility to GSM transmitters despite of RF-wise shorted OP inputs, simply through rectifying the RF picked up by the "antenna" wires connecting the components. The range of 1 to 1.5 m reported in the links sounds more likely than elsewhere claimed 10 m. I guess, most people already experienced nearby transmitting mobile phones picked up by telephones, CD players or TVs.