Single ended operation in ADL5380 for DC output

Hi akshay92,

• Attached is the latest version of the schematic for the board which you have in the photo.
• To help troubleshoot, please markup the schematic and send back with any changes you have made.
• Be sure the RF and LO sources are synchronized or there could be residual frequency error due to the internal references of each source. The residual error could cause an AC signal at the baseband outputs.
• To measure true DC you will need to bypass the transformers on the baseband outputs, since the transformers will block DC.
• The baseband differential load impedance should be kept above 200 ohms.
• When connected with the baseband transformers and terminated into 50 ohms single ended, the differential load impedance presented to the device is 450 ohms.

PDF

Best Regards,

David

Hi akshay92,

• Attached is the latest version of the schematic for the board which you have in the photo.
• To help troubleshoot, please markup the schematic and send back with any changes you have made.
• Be sure the RF and LO sources are synchronized or there could be residual frequency error due to the internal references of each source. The residual error could cause an AC signal at the baseband outputs.
• To measure true DC you will need to bypass the transformers on the baseband outputs, since the transformers will block DC.
• The baseband differential load impedance should be kept above 200 ohms.
• When connected with the baseband transformers and terminated into 50 ohms single ended, the differential load impedance presented to the device is 450 ohms.

PDF

Best Regards,

David

Thank you for the timely response. We have gone through the updated schematic and I think it is the same what we have and we have not connected or removed any component.

We are giving trigger to RF and LO at the same time, if there is any other way to make sure if signals are in sync then please guide for the same.

From datasheet page number 34, table 5 it is shown that for the single-ended output, the resistors R2x-R5x needs to be populated for bypassing the output transformer and R13x-R18x needs to be populated for a single-ended mode. Please correct us if anything is written is incorrect. According to our understanding we have to populate registers at R2x-R5x.

Another point which is not clear is about the 200ohm impedance, is the eval board configured for 50ohm per pin? please elaborate a little on this. And I guess since the output being DC we might not require any matching at the output ports, please correct if any mistake in understanding.

Hi akshay92,

You're welcome. Please see below for some more guidance.

• Instrument Synchronization:
  • You should make sure the 10MHz references on the rear panel of the instruments are connected.
  • You can use one instrument as the master by connecting to the 10MHz REF OUT port.
  • Be sure to check on a spectrum analyzer or scope that there is in fact a reference signal coming out.
  • Sometimes you need to enable reference signal in the instrument utility configuration.
  • Then connect this 10MHz REF OUT port to the other instruments 10MHz REF IN port.
  • This will ensure that all instrument internal references are synchronized, eliminating any residual error.

• Bill of Material:
  • Please disregard datasheet figure 102, figure 103, and table 5, as these are out of date.
  • Attached is the latest bill of materials for the board you have.

• Load Impedance Information:
  • From datasheet table 3 pin descriptions:
    • I Channel and Q Channel Mixer Baseband Outputs. These outputs have a 50 Ω differential output impedance (25 Ω per pin). Each output pair can swing 2 V p-p (differential) into a load of 200 Ω. The output 3 dB bandwidth is ~400 MHz.
  • From datasheet page 25:
    • The baseband outputs QHI, QLO, IHI, and ILO are fixed impedance ports. Each baseband pair has a 50 Ω differential output impedance. The outputs can be presented with differential loads as low as 200 Ω (with some degradation in gain) or high impedance differential loads (500 Ω or greater impedance yields the same excellent linearity) that is typical of an ADC. The TCM9-1 9:1 balun converts the differential IF output to a single-ended output. When loaded with 50 Ω, this balun presents a 450 Ω load to the device. The typical maximum linear voltage swing for these outputs is 2 V p-p differential. The output 3 dB bandwidth is 390 MHz. Figure 85 shows the baseband output configuration.
  • Once you bypass the transformers, please follow this guidance and use a differential load impedance of greater than 200 ohms.

PDF

Best Regards,

David

Thank you for such an elaborate explanation. Will do this and get back as soon as possible.

I am making the register changes to bypass transformer balun as written in my previous message and use the differential load impedance as suggested by you. While testing for DC may I get some reference data or input Vs expected output if any observations are recorded at your end to verify proper configuration of the board. Or else please guide on verification of correct configuration.

And if I bypass the balun, according to my understanding please correct me if wrong I will get differential output but I want single ended output so how to achieve that for DC output at 5.81ghz RF and LO.

We also tried to see the inputs of the balun at the I and Q side after giving 5.81Ghz at RF and LO. We observed 2.4 V DC at both I and Q. But this value did not change even after changing the cable length at RF. We were expecting to see a voltage change with respect to the phase change due to cable length change at the RF but at both times we observed same 2.4V at the input of the Balun. Here we have not yet made any change to the eval board. We did testing this way to check the desired DC signal since the balun will block our desired DC signal. Actually we dont think this voltage is due to the signal but it was also present when the RF and LO were OFF and only board was powered on. Please correct us if there is any confusion in our understanding.

Hi akshay92,

• We don't have measurements for your test conditions, but attached is data which was collected on a similar device the ADL5382, at 700 and 1000MHz. You could use the same test methodology for your evaluation.
• We used an oscilloscope to measure phase and two DMM's to measure voltage across the load resistors. The more accurately you measure these two quantities, the more precise estimation of phase between the two paths.
• The 2.4V you measured at the baseband outputs is the common mode voltage. Once you add the differential load resistor, you should see the voltages change as current flows through the resistor.
• You could use an instrumentation amplifier or ADC driver as a buffer to convert the differential voltage across the resistor to a single ended voltage. This has the benefit of adjusting the voltage range and common mode voltage to in line with the next stage in your system (ADC, etc).

XLSX

Best Regards,

David

Thank you for the response. Will do as you have suggested and get back as soon as possible.

Hello Sir,

As you recommended above, for DC output we have to bypass the balun and add load resistor but before doing that we thought to test the output between  IC pin 3 and 4 and pin 15 and 16 directly through Multimeter.

We just added Load resistor of value 500 ohm at R6x and R7x.

So at LO/RF Freq 5.81Ghz and input power 0dbm , We were getting only 770 microV across R6x and for phase difference we changed the length of cable at RF side. so the voltage variation observed was from 500 microV to 1.4mV. 

Is it right what ever we did? or need to Bypass the balun first and then carry out the experiment.  

So can you help us with what exact value of the Load Resistor should be and what is the expected result range and please correct us if any mistake we have done.

Hi akshay92,

• You will need to use the phase sweeping capability of the two RF sources, which are synchronized. Many sources will allow you to sweep the output phase. If the sources do not have that feature, you could use a mechanical phase shifter to sweep through various phase states.
• Also, one additional change you may need is the leave the ADJ resistor open on pin 19. For operation above 5GHz, this will optimize the mixer configuration.

Best Regards,

David

Thank you so much sir. I’ll do the changes as suggested and get back to you if i face any problem.

We are able to see the output on the 500-ohm load connected at the I and Q port after making the resistor changes as suggested by you. We are able to see the change in the DC level when observed on the load resistor but when observed on the single-ended connector positive of I and Q we see a non-changing DC level of 2.4V or around 3V. Please guide on what else changes we have to make to see the output on the single-ended connector.

After 2 days of testing, we are unable to see repetitive results as mentioned above. And no change in results after removing the resistor19. We now see a constant DC even at the load resistors of 500ohm. The current is same, means no damage to the board