ADL5565 EVM Reconfiguration

Hi FrenchyRaoul,

                              To convert the ADL5565 input to differential the input transformer should be removed configure the jumpers for the gain desired. See Figure 31. Please note the input impedance changes with gain. See page 3, or page 6, of the datasheet. 

The output as configured on the evaluation board presents a 200 ohm load and transforms the impedance to 50 ohms single ended. There is 10.8 dB of loss associated with this network.

If you wish, the resistors and transformers can be removed, ac couple one output to ground through a 100 ohm resistor. The other output can be used to drive a 100 ohm load. Keep in mind doing this will result in increased harmonic distortion.

Jim,

Thank you for the reply. I am sorry about the delay, as I did not get an email notification when you answered my question.

I have two followup question. Excuse my ignorance- you said there is a 10.8 dB loss associated with this setup? In other words, if I have the amplifier set up for maximum gain at 15.5 dB, I am only going to get about 5 dB overall gain out of the amplifier? This is with the 200 ohm input impedance and 50 ohm output impedance. I imagine removing the transformer does not much reduce the 10.8 dB loss?

Secondly, I was wondering if you could point me in the direction of the SMA connector part that Analog used on the evaluation board. I would like to use the same part as to make sure I keep the board inputs balanced.

Thank you much for your time, this has been very useful.

EDIT: It always helps to give more information. I am using this amplifier to amplify the output of an ADF4351 EVM, which has two 50 ohm differential outputs. This amplifier serves the dual purpose of amplifying, and going from balanced to single ended.

The RF connector is Johnson PN: 142-0701-851.

The ADL5565 the voltage gain and power gain are not equal.

The 15.5 dB gain is the voltage gain. The power gain for the ADL5565 is 10.3 dB. 

If you are looking at power you would see -0.5 dB of gain.

See the attached file for formulae to convert from voltage gain to power gain

Thank you for your help!

Hey Jim,

When converting the voltage gain to power gain, I did not come up with 10.3dB. Am I missing something?

Hi Thomas,

                    the 10.3 dB of power gain assumes the load is 200 ohms. For optimal performance the part needs to drive a 200 ohm load. I'm assuming you just used the Rout of the amplifier is this correct?

Hi Sir,

May I know that what the mean of there is 10.8 dB of loss associated with this network?