A question about a RF Differential RF Amplifier (AD8325)

A question about a RF Differential RF Amplifier (AD8352)

Hi all,
My question is about the matching for a Differential RF Amplifier.
From the AD8352 datasheet, we can see that the input matching circuit is used to provide the input impedance match for a 50ohm system. I think the reason is for maxium power transfer for the amplifier input. When those energy meets the amplifier's high input resistance, the current will be changed to voltage.

But, when AD8352 is connected to a load, the output matching circuit can not be found between the load and amplifier's output impedance. I don't why. Without it, the energy will loss.

BTW, AD8352 input impedance is 3000ohm(+j0.9pF), output impedance is 100ohm(+3pF). The load is 200ohm or 1000ohm. If AD8325 drives AD9445 (a ADC), the AD9445 presents approximately 2 kΩ in parallel with 5 pF/differential load to the AD8352.

If the frequency is not too high, there will be a reflection but it can be neglected because of low output impedance of the amplifier.If you drive this amplifier sufficently
high (because the gain is quite high), the output swing will be rail to rail-or close to- and this will be quite proper level or ADC..
That's why, if the operating frequency is not too high, a matching network is not necessary..
But for input side, a matching network is necessary because input signal will-generally-be low and therefore maximum energy must be transferred into amplifier.

Thank you for your reply. I still have some other questions... Please hlep.
If I use AD8352 to work as a RF gain block. It seems that it just amplifies voltage for its high input impedance (3000Ω) and low output impedance (100Ω), if it is connected to a high-resistance load (e.g. 200Ω or 1000Ω). Is that right ?

Assuming that it can work at 2GHz to as a RF gain block, whether the output matching circuit is needed for this frequency ?
Because I want a 2-GHz Voltage Amplifier (or voltage gain block), whose input and output current can be reduced to “minimum” (almost zero).

jianke said:

Thank you for your reply. I still have some other questions... Please hlep.
If I use AD8352 to work as a RF gain block. It seems that it just amplifies voltage for its high input impedance (3000Ω) and low output impedance (100Ω), if it is connected to a high-resistance load (e.g. 200Ω or 1000Ω). Is that right ?

Assuming that it can work at 2GHz to as a RF gain block, whether the output matching circuit is needed for this frequency ?
Because I want a 2-GHz Voltage Amplifier (or voltage gain block), whose input and output current can be reduced to “minimum” (almost zero).

Click to expand...

If you work 2GHz, you can not work with this amplifier because bandwidths of these kind of amplifiers don't reach to 2GHz..Except some special and custom designed structures but not standard products..
If you wanna drive an ADC at that frequency, you should use low noise RF amplifiers and design of this circuit will take some paying of attention.