How to calculate DC blocking capacitor for RF Attenuator

Thread Starter

kmpres

Joined Jan 24, 2016
12
Hello All,

I'm building an RF attenuator for use in radio repair and restoration projects. It's a simple passive design using seven pi-attenuators in series that should give -1dB, -2dB, -4dB,-8dB, -16dB, -32dB and -64dB of attenuation from the output of my low budget MHS-5200A function generator. The seven switches will give a total 128 dB of attenuation in 1 dB steps, at least that is my intent. I calculated the resistors without difficulty, but now I need to add a series DC blocking capacitor which I assume will be in the neighborhood of 0.01 to 0.1 uF, but I don't know how to calculate it to be sure. Though I've studied it many times, I still have a tenuous grip on how impedance is applied from device to device. I assume, for example, that the input impedance of my attenuator will be 50 ohms as that is the output impedance of my function generator, and the output impedance will be "high" as it will generally be placed in the grids or input stages of whatever radio I'm working on. I really have no idea how "high", nor do I know if that particularly matters in this case. Desired frequencies for -3dB dropoff are maybe 500 Hz to 15 MHz, though those are just guesses. How does one calculate the capacitance, or is it simply a rule-of-thumb kind of thing?

Thanks in advance!
 

crutschow

Joined Mar 14, 2008
23,291
The =3dB low-frequency corner for a capacitor in series with a resistor to ground is 1/(2*pi*RC).
Thus for a low-frequency corner of -3dB at 500Hz and a 50Ω resistance, the capacitor should be at least 1/(2*pi*50*500) = 6.4μF.

0.1μF would give a LF corner of 32kHz.
 

Thread Starter

kmpres

Joined Jan 24, 2016
12
The =3dB low-frequency corner for a capacitor in series with a resistor to ground is 1/(2*pi*RC).
Thus for a low-frequency corner of -3dB at 500Hz and a 50Ω resistance, the capacitor should be at least 1/(2*pi*50*500) = 6.4μF.

0.1μF would give a LF corner of 32kHz.
Thank you, Crutschow, that's just what I needed. I'm wondering now if I should include the audio spectrum in this calculation. Is it normal practice to include that in RF attenuator designs? Also, how does one calculate power dissipation? I'm using 1/4W 1206 resistors and plan an input of 1Vp-p so as to make the resultant waveforms scalable on a standard dB amplitude chart.

Thanks!
 

Thread Starter

kmpres

Joined Jan 24, 2016
12
Obviously the capacitors become pretty large at 20Hz. ;)
Power dissipation is V²/R (where V is the RMS voltage) so for a 1Vpp signal and a 50Ω resistance the power is (0.5*.707)²/50 = 2.5mW.
Thanks. I was not sure which resistance to use. I'm entirely self-taught and the process of teaching myself the ropes is taking longer than expected.
One final question: Do I need to add a terminator anywhere in the line from signal generator to attenuator to DUT?
My circuit works, BTW. Now I need to figure out how to read the output accurately on my new Rigol DSO.
 

DickCappels

Joined Aug 21, 2008
5,873
If your 50 ohm input impedance is truly 50 ohms and that is the impedance of your transmission line or coax cable between the generator and the attenuator then you should not have any trouble. If the generator is also 50 ohms that would help minimize the effects if the attenuator is mismatched.

The passive attenuators I designed need to be terminated in order for the input impedance to be correct. It is likely that yours is that way too.
 
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