FAQ: Calculating the Output Power of DAC/IQ Modulator Combination

Q.
How do you Calculate the Output Power of DAC/IQ Modulator Circuit?
 
A. 
IQ Modulator gain and is commonly specified as dB voltage gain. It is not that meaningful to talk about IQ Modulator power gain since these devices generally have high input impedances on their I and Q input (this would result in infinte power gain). In the case of DACs, we generally do not talk about gain; instead, we specify the full-scale output voltage swing. To complicate matters further, the impedance levels change as you move from the DAC output (usually a 100 ohm load) to the modulator input (usually terminated with 100 ohms on I and 100 ohms on Q), to the modulator output (50 ohm single-ended load).
 
To calculate the modulator output power, let’s do the calculation in the voltage domain. Once we have the final output voltage we will convert it into dBm assuming a 50 ohm load.
 
The spreadsheet below which can be downloaded,  allows calculation of output power based on the DAC's Digital Backoff level[dBFS], the signal's Crest Factor[dB] or peak-to-average ratio, the DAC's full-scale current[mA],  the baseband filter loss[dB],  DAC load resistors and filter  termination resistors and modulator voltage gain dB. As an output, it provides the DAC output level, Modulator input level, and Modulator output level (peak and rms).

*1  RB sets the DAC's full-scale peak-to-peak swing and the DAC's output bias level

*2  Shunt resistor RS scales down the DAC's peak-to-peak voltage without affecting the bias level

*3 The "DAC Peak Back-Off" refers to how close the peaks of the DAC's output signal come to the DAC's full-scale output voltage

*4 The Crest Factor that is inputted sets the backoff level between the DAC Peak Output Level and the RMS Level of the carrier

Therefore:      DAC Fullscale Level = RMS Level + Crest Factor + DAC Peak Back-Off

In general, the signal peaks need to be backed off from the DAC's full-scale level by a few dB. The amount of back-off required depends on the desired/acceptable signal distortion

*5 Modulator Voltage Gain is defined as the dB difference between the voltage level on the I or Q input (generally these are equal) and the RF Output

 

At digital domain, signal is scaled  down to prevent signal distortion from peak-to-average ratio(Crest Factor). And  DAC output swing level is determined by Full scale current and DAC Load Resistors(RB).  So Differential peak-peak Voltage and RMS Voltage at DAC Output is calculated by:
 
 
DAC output level[Vp_p]= 2 x RB  x IFS  X  10^(Digital Back Off[dBFS]/20 ) 
 
DAC output level[Vrms] = DAC output level[Vp_p] /2 x 10^(- signal CFR[dB]/20 )
 
 
This calculation is assumes no filter and no shunt resistor(Rs).
 
 
To ajust the modulator input drive level without adjusting the DAC settings and DAC voltage bias level,  the shunt resistor Rs is adjusted at modulator inputs. Please refer Fig 59 of ADL5375 datasheet , which shows relationship between the AC swing-limiting resistor and the peak-to-peak voltage swing when the DAC load resistors (Rb)  are 50 ohms.  We typically put a 100 ohm resistor. This provides a 100 ohm termination to the filter and also allows us to scale down the DAC's 2Vpp full-scale voltage down to 1Vpp.
 
 
The Modulator input level is calculated by using the value of this shunt resistor along with the  filter loss :
 
 
Modulator Input level [Vp_p] =  DAC output level[Vp_p] x RS /[2 x RB + Rs] x 10^(Filter Loss[dB]/20 )
 
Modulator Input level [Vrms] =  Modulator Input level [Vp_p]/2  x 10^( - signal CFR[dB]/20 )
 
Modulator Input level[dBV] = 20 x Log(Modulator Input level [Vrms])
 
 
The Modulator Output Level in dBV unit  is calculated by adding the modulator voltage gain, and converted to power units at 50 ohm .
 
 
Modulator output level[dBV] =  Modulator Input level[dBV] + Modulator Voltage Gain[dB]
 
 
This dBV value can be converted to dBm by adding 13 (in a 50 ohm system)
 
Modulator output level[dBm] = Modulator output level[dBV] + 13
 
 
The above example shows a +1.08dBm modulator output power for  1Vp_p sine wave input and -2.9dB modulator  voltage gain.  This comes in almost equal to the measured modulator output power  (1.05dBm output power from  ADL5375 at 2140 MHz).
Attachments:
Comments
Anonymous
Parents

  • Dear Sirs

    I think you have a mistake in calculation.

    Modulator Input level [Vrms] =  Modulator Input level [Vp_p]/2  x 10^(signal CFR[dB]/20 )

    What you calculate here is not RMS value but PEAK value.

    If you want RMS value it has to be:

    Modulator Input level [Vrms] =  (Modulator Input level [Vp_p]/2) x (sqrt(2)/2)  x 10^(signal CFR[dB]/20 )

    Also calculation with Full scale current you have to be really careful. Because for example from datasheet of AD9125:

    The DAC full-scale output current (IOUTFS) is nominally 20 mA. The output currents from the IOUT1P/IOUT2P and IOUT1N/ IOUT2N pins are complementary, meaning that the sum of the two currents always equals the full-scale current of the DAC.

    So it means in calculation you have to take into account and

    DAC output level[Vp_p]= 2 x RB  x IFS  X  10^(Digital Back Off[dBFS]/20 )

    leads to:

    DAC output level[Vp_p]= 2 x RB  x (IFS/2)  X  10^(Digital Back Off[dBFS]/20 )

    http://www.analog.com/static/imported-files/application_notes/AN_912.pdf

    Otherwise yours calculation is "improved" by ~6dB (I'm not 100% sure, if I'm wrong just let me know)

    BR

    Igor

Comment

  • Dear Sirs

    I think you have a mistake in calculation.

    Modulator Input level [Vrms] =  Modulator Input level [Vp_p]/2  x 10^(signal CFR[dB]/20 )

    What you calculate here is not RMS value but PEAK value.

    If you want RMS value it has to be:

    Modulator Input level [Vrms] =  (Modulator Input level [Vp_p]/2) x (sqrt(2)/2)  x 10^(signal CFR[dB]/20 )

    Also calculation with Full scale current you have to be really careful. Because for example from datasheet of AD9125:

    The DAC full-scale output current (IOUTFS) is nominally 20 mA. The output currents from the IOUT1P/IOUT2P and IOUT1N/ IOUT2N pins are complementary, meaning that the sum of the two currents always equals the full-scale current of the DAC.

    So it means in calculation you have to take into account and

    DAC output level[Vp_p]= 2 x RB  x IFS  X  10^(Digital Back Off[dBFS]/20 )

    leads to:

    DAC output level[Vp_p]= 2 x RB  x (IFS/2)  X  10^(Digital Back Off[dBFS]/20 )

    http://www.analog.com/static/imported-files/application_notes/AN_912.pdf

    Otherwise yours calculation is "improved" by ~6dB (I'm not 100% sure, if I'm wrong just let me know)

    BR

    Igor

Children
  • in reply to igor.halas

    Hello Igor,

    We can generalize the relationship between V_peak and V_rms for any type of signal with signal CFR. If it is CW(sinewave), CFR is 3 [dB] and V_rms is 1/sqrt(2) * V_peak. Let's drive the equation below.

    Signal CFR [dB]= 10*log(P_peak/P_rms) = 10*log(V^2_peak / V^2_rms)=20*log(V_peak /V_rms)

    V_peak/V_rms = 10^(signal CFR[dB]/20)

    V_rms = V_peak * 10^( - signal CFR[dB]/20) , V_peak = Vp_p/2

                = Vp_p/2 * 10^( - signal CFR[dB]/20)

    There is typo at sign assuming signal CFR[dB] is positive number. I will put negative sign at original one and correct it. And at the DAC output swing level, when we set IFS at 20mA, IOUT1P+IOUT1N is 20mA(when IOUT1P is 20mA, IOUT1N is 0mA, for example). Voltage swing level at each IOUT1P or IOUT1N is RB*IFS*10^(Digital Bask Off[Dbfs]/20) and Voltage[Vp_p] between IOUT1P and IOUT1N is 2 times to each. So the equation is correct.

Related