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STLC1512

Appendix D - Headroom Calculation for Switching

The headroom for switching can be determined from the numbers in Table 4. The switching headroom is 0.70

V at low currents (i.e. while on the low supply rail) and 0.85 V at high currents (i.e. while on the high supply rail).

The most difficult number to arrive at is the voltage that will appear at the pins PAOP1,2 and PAON1,2. This is

a combination of the input voltage, the line impedance and the losses in the transformers.

For a 100

Ω load the maximum signal on the line will be 10.7 V. Since we are generating an active 100Ω output

impedance the voltage on the line for any other load is given by:

(EQ D.1)

There are various losses in the transformers that can be modeled as resistors. To calculate the effect of these

losses we must know the current through the load which is given by:

(EQ D.2)

The loss through the line transformer can be modeled as a 2.6

Ω resistor. There is also a drop across the two

10

Ω reference resistors. Therefore to determine the voltage at the output of the switched transformer we have:

(EQ D.3)

At this point there is some additional current that flows through the hybrid balance network. This current flows through

a resistance that is equivalent to 1270

Ω. Therefore the current flowing out of the switched transformer is:

(EQ D.4)

The switched transformer has losses that can be modeled as a 3.6

Ω resistor and has a 4:1 turns ratio. Therefore

the voltage at the primary side of the transformer is given by:

(EQ D.5)

combined to calculate the required headroom to drive a certain impedance.

(EQ D.6)

can be rearranged to calculate the switching threshold. The headroom can be determined from the drop across

the diode from the low supply and the low current drive capability of the amplifier given in Table (0.70V).

(EQ D.7)

be scaled to the nominal value of 10.7V (full scale) to determine a switching threshold based on the full scale

level of the signal.

The headroom calculation is worst at maximum line impedance. There is also a supply rail requirement for the

high (5.0V) supply which is based on being able to supply enough current to drive an 80

Ω line impedance. This

is not a trivial calculation and has been based on simulations. The possibility exists that the requirements on the

minimum supply voltage may be able to be reduced in the future.

V

li ne

2 10.7

()

Z

o

100

Z

o

+

-----------------------







=

I

l oad

V

line

Z

o

------------

=

V

sw txout

V

line

20

2.6

+

()I

loa d

+

=

I

sw txout

I

loa d

V

sw tx ou t

1270

----------------------

+

=

V

PA Ox

V

sw tx out

3.6 I

sw txout

()

+

4

---------------------------------------------------------------

=

V

PA Ox

V

n

2

------

Z

o

20

2.6

3.6

Z

o

20

2.6

++

1270

----------------------------------

1

+





++

+

Z

o

100

+

---------------------------------------------------------------------------------------------------











=

V

hea droo m

V

supply min

0.70

V

dio de

–

–

=