how to calculate blocking capacitor?

[matise]

hi.
i have this max2650 amplifier and i want to know what value i need for its blocking capacitor.
this is what the datascheet says:


As shown in the Typical Operating Circuit, the MAX2650 is easy to use. Input and output series capacitors may be necessary to block DC bias voltages (generated by the MAX2650) from interacting with adjacent circuitry. These capacitors must be large enough to contribute negligible reactance in a 50Ω system at the minimum operating frequency. Use the following equation to calculate their minimum value:

cblock = 53,000/f = (pf)

where f (in MHz) is the minimum operating frequency.


im to new to this to understand how to calculate that.
if my freq. is 18Mhz...
will it be 53/18 and 2.9 pF for capacitor?

or:
(53000/18000000 = 2.9 -03 is 2.9 nF)?
what is the correct calculation? 

[Zero999]

The capacitor should be big enough to pass the lowest frequency of interest.

The cut-off frequency, i.e. the point where the amplitude is reduced by a factor of 0.707 can be calculated using the following formula:

F_C = 1/(2πRC)

Where R is the load resistance and C is the capacitance.

Rearrange for C:
C = 1/(2πF_C R)

Choose a cut-off frequency of half the lowest frequency of interest, round the capacitor value up to the nearest E3 value and the capacitor will be more than big enough.

[capt bullshot]

This particular 53,000pF means 53nF. In other parts of the world (including where I live) it'd mean 53pF.

So your minimum capacitor value calculates to 53/18 nF (2.9nF, same result as Hero999 for 3 Ohm reactance). In practice, a 10nF or 100nF will do the job well.

[vk6zgo]

hi.
i have this max2650 amplifier and i want to know what value i need for its blocking capacitor.
this is what the datascheet says:


As shown in the Typical Operating Circuit, the MAX2650 is easy to use. Input and output series capacitors may be necessary to block DC bias voltages (generated by the MAX2650) from interacting with adjacent circuitry. These capacitors must be large enough to contribute negligible reactance in a 50Ω system at the minimum operating frequency. Use the following equation to calculate their minimum value:

cblock = 53,000/f = (pf)

where f (in MHz) is the minimum operating frequency.


im to new to this to understand how to calculate that.
if my freq. is 18Mhz...
will it be 53/18 and 2.9 pF for capacitor?

or:
(53000/18000000 = 2.9 -03 is 2.9 nF)?
what is the correct calculation?

The first answer is incorrect.

The second answer is correct, but they assumed you would, by using the frequency in MHz, get a result for C in pF of approx 2900 pF which (in your case),you would convert into nF.

By expressing f in Hz, instead, your result was in uF, which you converted to nF,still getting the same answer.

[matise]

thank you all for answers!