Re: [Discuss-gnuradio] output level of basic tx is very very low

[Martin Dvh]

Hi all,
>Eric Blossom wrote:
> On Fri, Aug 19, 2005 at 02:56:53AM +0200, Martin Dvh wrote:
> > Hi,
> > What output voltage level should I expect from the basic tx?
> > The docs say it delivers maximum 20 mA into 25 or 50 ohm.
> > At 50 ohm this should result in 1.0 V max.
> > The most I can get out of it is 50 mV
>
> This is what I get too.
>
> > What am I missing here.
> > Or should I suspect a hardware failure (of both dacs, of the reference 
> > voltage, of the basic-tx board) ?
>
> The output signal level on the Basic Tx d'boards is lower than Matt or
> I expected.  We would appreciate any trouble shooting help you can
> provide.  It's possible that the termination isn't correct or has the
> wrong values, etc on the daughterboard.  Please take a look at the
> schematics and let us know what you think.
I think I found a rather obvious faulty termination.
Half of the output transformer is shortcicuited.
The output should be connected to pin 3 (as it is)
And pin 1 OR pin 2 should be connected to ground.
(Depending on the transformation factor you want)
Not Both.
Pin 1 and 2 of the T50 and T51  (ADT1-1WT) are connected to ground.
This way you shortcircuit half of the windings (so the other half also won't 
give much signal)
This also means the ADC sees a much too low impedance.

The best resolution would be to cut free pin 2 and only ground pin 1.
This is a bit hard since for TX_A pin 1 of T50 is connected to ground through 
pin 2.
For TX_B the connections are below the transformer T51 so I would have to 
desolder that one first.

A quick test is just to cut the connection between pin 1 and 2 of T50.
You will have the wrong transformation factor but the shortcircuit will be gone.

Another way is to desolder only the pin 2 of the transformers, lift them up a 
bit and put a bit of plastic under the pin to insulate it.


Besides this I also think the resistors have the wrong values.
If you remove the connection to ground of pin 2, the adc still sees an 
impedance of only 7 to 16 ohm
(dependant on the impedance of the device you connect to the output)
This should be 25 to 50 ohms.

See my attempt below to calculate the impedance the ADC sees (if you remove the 
shortcircuit).
It would be a whole different value if you would also remove the connection of 
pin 5 to ground.
(although I don't know if we should do this)

Greetings,
Martin

Impedance calculations for the basic TX-board of the usrp.
These are to determine Zadc=Z-IOUTP_A= is the impedance ADC_A sees on its 
output.

Z1//Z2 means Zparallel=1/(1/Z1+1/Z2)
Zc=1/jwC
Zr=R
Zl=jwL

w=2*pi*f
pi=3.14159265....
f=frequency in Herz

R32=50 Ohm
C40=10pF
R33=100 Ohm
R34=50 Ohm
C41=10 pF

Ztrans=Transfactor * Zexternal = 0.5* Zexternal

(Ztrans is from one outer input lead to the center input lead of the transformer which is connected to ground. Note that this is half the number 
of winds of the total input number of winds of the transformer.)

Zadc=Z-IOUTP_A= is the impedance ADC_A sees on its output

Zadc=(Zc40 //Zr32) // Ztrans//(Zr33//Zr33)//(Zr34//Zc41)//Ztrans)

The fact that you see R33 twice and Ztrans twice has to do with the fact that 
the middle of the input of the transformer is connected to ground.
Because the input on one half of the transformer will also show up negative on the other side of the input. This also means the impedance which 
is on th eother side will be seen in parallel.

Zc40//Zr32=ZC41//R34= 1/(jwC40 + 1/R32)=R32/(1+jwR32C40)
at f=1 MHz
Zc40//Zr32=R32/(1+j*2*pi*1e6*R32*C40)
          =50/(1+j*2*pi*1e6*50*10e-12)
          =50/(1+j*pi*1e-3) ~~=~~ 50 Ohm

50//Ztrans//50//Ztrans//50
if Zexternal is infinity
Zadc is about 16 Ohm
If Zexternal is 50 Ohm
Ztrans=25 Ohm
Zadc=50//25//50//25//50 = 16 // 12.5
Zadc is about 7 Ohm

Zadc should be between 25 and 50 Ohm according to the datasheet.

It has been a while since I determined impedances of circuits including 
transformers so corrent me if I am wrong.

Martin Dudok van Heel
(Martin Dvh)
nldudok1 at olifantasia,com