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# Getting Starting with AD8302

Hi,

After I reviewed the AD8302's datasheet, i found a paragraph that says:

"There is an internal 10 pF capacitor to ground that sets the maximum corner to approximately 200 MHz.

The corner can be lowered according the formula fHP (MHz) = 2/CC(nF), where CC is the total capacitance from OFSA or OFSB

to ground, including the internal 10 pF."

Then, datasheet also says that a capacitor between OFSA (or OFSB) and Ground is necesary firstly for reducing fHP and deleting the CD frecuencies.

But in my application, I'll use the AD8302 for detecting frecuencies between 1 kHz to 10 MHz, so how can i select the correct value for the capacitors conected between OFSA (or OFSB) and ground?????

My first idea was to choose a 2.2 uF, so according with paragraphs listed above we have

Cc=10 pF (internal capacitance) + 2.2 uF (my chosen value) = 2.20001 uF

fHP =  2 / 2.20001 uF = 909,086.7 Hz

is this idea correct???????

My second doubt, what value is correct for coupling capacitors conected between INPA or INPB???? is 10 uF a correct value??

Regards

• The attached application note  (Operating RF Detectors at Low Frequency) should answer your questions.

AN-691_Low_Freq_Detector_Operation.pdf
• Hi,

Actually I reviewed that application note.... But I'm still without understanding because of they consider the AD8302's performance at low frecuencies... moreover the capcitors conected in INPA,INPB,OFSA,OFSB are the same  (Cc).

Something important is Cc >= 10 uF, then I believe this value is important for coupling capacitors conected to INPA and INPB and this is the reason that I asked if this value is the correct??

But what happen with capacitors conected to OFSA and OFSB that I believe are the most important for setting the fHP????

• Hello Mresau19,

There are two things going on here that are somewhat intertwined:

-     The first is value of the couplling caps used on the INPA and INPB pins.  You want to pick the cap value such that the impedance is small at the                frequency that you are operating at.  In your case the cap will be large because you want to operate at low frequencies.  Keep in mind that the                  frequency corner is actually a combination of the series coupling capacitor (C1 and C5 on the eval board) and the input resistance (fc = 1/(2*pi*R*C)).

-     The second is the value of the de-coupling caps used on the OFSA and OFSB pins (C4 and C6), where each cap is in parallel with a 10pF internal cap.       And the equation f_hp (MHz) = 2/Cc (nF) is given for the lowest frequency of operation for the offset control loop.

So what you must do is have BOTH sets of caps set such that neither one of them inhibits are disired frequency range.  2.2uF series coupling caps should works just fine and would give you an fc ~ 25 Hz; well below where you want to operate.  Also, remember that f_hp for a 10pF cap is 200 MHz (f_hp = 2/.01).  So for a 2.2 uF cap, that would give a corner of 909 Hz (2/2200), NOT 909 kH. I'm pretty sure that's how the equation works.  In your application, where you want to operate down to 1kHz, I would use 10uF decoupling caps, which would put your f_hp = 200Hz.  That way the impedance looking into the pins INPA and INPB stays at 3kOhms because you are operating at well above the f_hp.

So, recommended caps values are:

C1 = C5 = 2.2uF

C4 = C6 = 10uF

J

• Hi,

Thanks for your explanation that was good... Is correct to calculate fc like this???

fc = 1 / 2pi53.2 ohms*2.2 uF = 1383.2 Hz

53.2 ohms corresponds to value in the evaluation board

and 2.2 uF the value of C1  or C5

I ask this procedure because in your explanation says

"2.2 uF series coupling caps should works just fine and would give you an fc ~ 25 Hz"

then, i don´t know how you calculate those 25 kHz.. can you tell me please??

Moreover, you suggest me the caps on INPA and  INPB must be large, but you consider c1,c5=2.2 uF instead 10 uF

My question is what would happen if C1,C5 = 10 uF and C4,C6= 2.2 uF????

ThanksDate: Fri, 4 Mar 2011 18:25:53 -0500

Subject: New message: "Getting Starting with AD8302"

Analog Devices EngineerZone

Getting Starting with AD8302

reply from jdobler in RF Components - View the full discussion

Hello Mresau19, There are two things going on here that are somewhat intertwined: -     The first is value of the couplling caps used on the INPA and INPB pins.  You want to pick the cap value such that the impedance is small at the                frequency that you are operating at.  In your case the cap will be large because you want to operate at low frequencies.  Keep in mind that the                  frequency corner is actually a combination of the series coupling capacitor (C1 and C5 on the eval board) and the input resistance (fc = 1/(2piR*C)). -     The second is the value of the de-coupling caps used on the OFSA and OFSB pins (C4 and C6), where each cap is in parallel with a 10pF internal cap.       And the equation f_hp (MHz) = 2/Cc (nF) is given for the lowest frequency of operation for the offset control loop.   So what you must do is have BOTH sets of caps set such that neither one of them inhibits are disired frequency range.  2.2uF series coupling caps should works just fine and would give you an fc ~ 25 Hz; well below where you want to operate.  Also, remember that f_hp for a 10pF cap is 200 MHz (f_hp = 2/.01).  So for a 2.2 uF cap, that would give a corner of 909 Hz (2/2200), NOT 909 kH. I'm pretty sure that's how the equation works.  In your application, where you want to operate down to 1kHz, I would use 10uF decoupling caps, which would put your f_hp = 200Hz.  That way the impedance looking into the pins INPA and INPB stays at 3kOhms because you are operating at well above the f_hp. So, recommended caps values are:C1 = C5 = 2.2uFC4 = C6 = 10uF I hope this answers your questions,

J

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• Hi,

Thanks for your explanation, it was good but.....

Is correct to calculate Fc like this??

*If C1,C5=2.2 uF-------------------> fc=1/2*pi*53.2 ohms*2.2uF = 1383.2 Hz

where 53.2 ohms is the input resistance considered in eval board and 2.2 uF is only the value of C1 or C5

I ask you this procedure because in your explanation there is a paragraph

"2.2 uF series coupling caps should works just fine and would give you an fc aprox to 25 Hz"

Then i don'nt know how you determine 25 Hz.. can you explain me please???

Moreover, you suggest me the caps on INPA and INPB must be large.... But you consider C1,C5=2.2 uF instead 10 uF

My question, what would happen is C1,C5=10 uF and C4,C6=2.2 uF???

Thanks a lot

• Hi Mresau,

I forgot that there was a 53.2 Ohm resistor shunted to ground on the board...that will raise the fc a bit, but not much.

If you think of what is going on in terms of impedances, at the lowest frequency of operation, 1KHz, a 2.2uF cap will have an impedance of ~72 Ohms and at that frequency the input resistance still looks like 3kOhms at the DUT.  Then there is the C1/C5 cap in series with the 3kOhm input resistance and then that cap and the input resistance are in parallel with the R1/R2 53.2 Ohm resistor.  So looking into the input pin INPA/INPB at 1kHz we would see the equivalent impedance of: Zeq = 53.2 * (3k + 72)/(53.2 + (3k + 72)) = 52.2 Ohms.  So as long as the input resistance (3kohms at these low frequencies) plus the impedance of the C1/C5 cap is big, the voltage across the R1/R2 resistors (the input voltage) will not change much.  So we don't need to worry about the R1/R2 53.2 ohm resistor in our fc calculation.  What we really care about is the voltage right on the DUT, and that is determined by the voltage divider between the resistance of the input (3kohms) and the impedance of the C1/C5 cap.  And so fc = 1/(2*pi*Rs*C) = 1/(2*pi*3k*2.2uF) ~25 Hz.

As to your question, if C1,C5 = 10uF and C4,C6 = 2.2uF then the fc = 5 Hz (not too much change) but the f_hp = 2/2200 = 909 Hz which is right under your frequency of operation...you want to set this corner as low as possible beneath your lowest frequency corner of operation.  That is why I advised using a 10uF cap for C4, C6 which makes f_hp = 200 Hz.  Using 2.2uF would probably work, you would probably get some performance degradation though around 1kHz.

Hope this helps,

Joel

• Hi,

Thank you, I had not considered the input resistence by the way the input resistance is 3 kohms at low frequencies and 50 ohms at 2.7 GHz????

I was understanding before you explain me:

fhp must be the nearest value at 1 kHz because of my lowest frequency is 1 kHz so C4,C6=2.2 uF produces an fhp=909 Hz that is the nearest value to 1 kHz DESPITE  with 2.2 uF, fc = 200 Hz. This ws my idea because for me fhp is more imortant than fcp BECAUSE fcp is produced for simple AC coupling caps. oTHER THING THAT CONFUSE TO ME is the simplest respose for a high pass filter, then if C4,C6 with R1,R2 (eval board) ARE  this type of RC  filter the corner frequency must be 1 kHz isn´t it??? Then this is the reason that for me fhp is more important tanh fcp.

nOTE: i DIND´T CONSIDER THE INPUT IMPEDANCE • Hi,

There are two input resistances/impedances.  Please read the attachment to clear up the confusion.  Zeq1 will pretty much always stay at about 50 Ohms across frequency, but Zeq2 will change with frequency due to Cc and Cin.  You care about Cc at low frequencies (hence making it as large as possible, to make it's impedance as small as possible; you want the impedance of Cc to contribute as little as possible) and you care about Cin at high frequencies (this is the on-chip capacitance that you can't do anything about).

Hope this clears things up,

Joel