AD9951 Amplitude

Hi,

I removed 0Ω resistance after transformer(DIFF ⇒ SE), then each output of AD9951 shows 500mVpp before Trans and  1Vpp after Trans.

Best Regards,

ttd

PDF

Hi,

I also find this somehow mysterious. Will have to setup a bench test next week for us to verify this one. We're still waiting for some components to arrive. 

Also can you verify the new output given on your latest comment? It seems like the one you sent is just the same as the first one. 

Regards,

Jules

Hi Jules,

Is there any update so far?

Best Regards,

ttd

Hi there,

Apologies for responding late. Still waiting for the components to arrive. I will send you an update once we will start the bench test. 

All the best,

Jules

Hi Jules,

Thank you for your reply.

I understand your situation.

Best Regards,

ttd

Hi Jules,

Do you have any update so far about your test bench?

Best Regards,

ttd

Hi ttd,

It's been a while. Apologies for the late update. The parts will arrive this week. As soon as it arrives, I will give you another update once I commence the bench test.

All the best,

Jules

Hi ttd ,

I will now start the bench test. I will let you know once I get some results and will get back to you.

All the best,

Jules

Hi Jules,

Thank you for your reply.

I'm looking forward to see your resules.

Best Regards,

ttd

Hi ttd ,

Upon bench testing, we were able to generate these waveforms.

For reference, we were using this circuit (AD9954 Eval Board) which is somehow similar to the circuit you have given.

For the full schematic diagram of AD9954 Eval Board, here the link.

Based on this schematic diagram, we measured to different points.

Point A = Pin 1

Point B = Pin 3

Point C = Pin J2

Point D = Pin J1

Using jumpers to pins 1 and 4, here’s the output of J2 (unfiltered output)

Channel 1 is Pin 3

Channel 2 is Pin 1

Channel 3 is J2

Using jumpers to pins 1 and 2, here’s the output of J1 (filtered output)

Channel 1 is Pin 3

Channel 2 is Pin 1

Channel 3 is J1

And comparing these results from your original circuit to the bench test we did today, it is somehow similar or approximately having close values. To answer your question, yes, the amplitude is similar to the results of the bench test that we did today. Your initial results are the expected amplitude that you should also be getting.

For comparison here’s the results of our bench test today:

All the best,

Jules

Hi Jules,

Thank you for your bench test.

It seems like my results.

I assumed 50Ωx10mA = 500mVpp, but the level is actually low. I feel that it is less than the original 10mA full scale depending on the load, please tell me the reason. If the amplitude is not exactly determined, it is very difficult to estimate the amplitude before board development.

Best Regards,

ttd

Hi ttd ,

Just want to ask the value of your Rset?

Regards,

Jules

Hi ttd ,

Just want to ask the value of your Rset?

Regards,

Jules

Hi Jules,

DAC Rset=3.92kΩ for 10mA.

Best Regards,

ttd

Hi ttd ,

With 10mA into 50Ω yields 500mVpp, as expected. The problem is that the 50Ω load connected to the DUT has other circuitry connected (namely, the transformer). If you disconnect the transformer you should see 500mVpp across the DUT load resistor. However, 500mV is on the cusp of violating the DAC compliance voltage, so you may not see a full 500mVpp output (should be close, though). With transformer connected, things get complicated in a hurry.

Regards,

Jules

Hi Jules,

Thank you for your reply.
Is the following understanding correct?

--------------
Since the compliance voltage (= 0.5V) is barely enough, it is NG if various load circuits are attached.
The solution is to set the compliance voltage to 0.25V (= 25Ω).
--------------

Also, is it better to use a buffer amplifier instead of a transformer?
Since the maximum frequency this time is about 13MHz, I think that an operational amplifier is okay.

Best Regards,

ttd

Hi ttd ,

The maximum full-scale output current of the combined DAC outputs is 15 mA, but limiting the output to 10 mA provides the best spurious-free dynamic range (SFDR) performance. Given that 0.5V is on the cusp of violating the DAC compliance voltage, using a lower load resistor will help.

A transformer by itself can change the voltage of a signal but does not change the amount of power. For a given increase in voltage, there is an equivalent decrease in the current.  The power remains constant.   In fact due to resistive losses the amount of power is actually a bit less.  An amplifier, on the other hand, can increase the power of a signal by using the input voltage to control a more powerful source. In this specific application, I suggest using the transformer since it is used for impedance matching.

All the best,

Jules