PBL3853N データシートの表示（PDF） - Ericsson

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PBL3853N

Universal Speech Circuit

Ericsson  

PBL 3853

Making any of the impedances unne-

cessary high will make the circuit sen-

sitive to RFI. All values given here are

approximate and serve as starting enti-

ties only. The final trimming of side tone

network is a cut and try proposition

because a part of the balance lies in the

acoustical path between the microphone

and earphone.

DC-Supply

In general

The most significant feature of this

speech circuit is its ability to draw

current from the line, that is used to a

DC-supply for auxiliary electronics and

for the receiver amplifier on the chip,

under maintaining the line impedance.

This function is boosted by two external

high β PNP transistors T1 and T2, and

internal amplifier and a comparator.

In detail

The circuit is designed such that it

supplies current first into the DC supply

output (priority) so that the micro-

computer which is controlling the auxi-

liary circuits and functions will have

power. The rest of the line current is

going to the speech function supply at

pin 4 (≈0.3mA) and through the transmit-

ter. The speech function needs 4mA

min. to operate to full specification. The

current to the DC supply is set by

expression:

IDC = (I1 x R12 + VD)/R13 where VD

= 0.65V, I1 ≈ 4 - 0.3 = 3.7mA and

controlled with an amplifier through the

voltage between terminals 1 and 9. A

certain increase in the voltage (∆VL)

across the resistor R12 will result in an

equal increase in the voltage across R13

which gives the expression:

∆IL=∆I1 (1 + R12/R13). From this can be

seen that an increasing line current will

also partly increase the DC-supply

current.

The comparator will compare the

voltage at terminal 18 plus 2xVD

(≈ 1.3V), with the voltage at terminal 1,

whether it is higher or lower. The line

current will be distributed as follows: The

comparator will route the line current to

the DC-supply until the set current is

reached after which the exceeding line

current will be divided between the

transmitter and the DC-supply according

to the expression

∆IL = ∆I1 (1 + R12/R13).

Note: That the DC-supply charge current

coming from the line, given by the

expression VD/R13, because of

the constant voltage difference of

VD between the pins 1 and 9,

makes it possible to use the DC-

supply for external electronics at

low line currents even before the

speech function. It can be seen in

the figure 16 that the line voltage

at low line currents is given by

VL = V18 + 2 VD + I1xR12 ≈ 6.0V.

(V18 = 4.5V;I1 small)

The DC-supply level is monitored by a

circuit that will cut the charge current

whenever the line voltage with the

modulated signal reaches a value 2 x

diode drops below the DC level. The

current will be returned via TR2 to

ground thus maintaining the correct

impedance towards the line and making

it possible to transmit a swing to the line

that has lower level than the DC. See

figure 18. When the line voltage without

signal reaches the TR2 monitor level the

charge to the DC-supply will be cut off

whereby the receiver dies because it

draws its current from the DC-supply. In

case that the dc-characteristic is set

such that the current will come first to

the minimum working level (≈IDC + 4 mA)

then the transmitter will die first because

of the IDC priority.

Example:

A payphone is to be designed. According to the specification the minimum line

current is 20mA at 6V inclusive the bridge for the phone to work with all its auxiliary

functions. The auxiliary functions will need as much current as it is possible to draw

from the line and the worst case is naturally at the longest line length. The speech

section of the circuit with the earphone amplifier needs ≈ 4mA for function. In this

case the highest possible IDC with the longest line will be 20 - 4 = 16mA.

IL= I1 +Ipin4 +IDC , see figure 16

or

I1= IL - Ipin4 - IDC

or

IDC= IL - Ipin4 - I1

and

I1 R12 + VD= IDC R13

Values for R12 and R13:

( ) I1 ⋅ R 12 + VD = IDC

R 13

The speech function current consists of two branches I1 and the current to pin 4

which is ≈ 0.3mA thus the current I1 through R12 will be 4 - 0.3 = 3.7mA. VD is

taken to be 0.650V. Choose R12 = 50 Ω to start with. The voltage drop across this

resistor is translated to voltage drop across R13 which in its turn will steal available

voltage from VDC. These values render a R13 = 51.6 Ω.

An increasing available line current will be divided between I1 and IDC as follows:

∆I1 R12= ∆IDC R13

∆IL = ∆I1 + ∆IDC

∆I1 = ∆IL /( 1 + R12/R13 )

or

∆LDC = ∆IL /( 1 + R13/R12 )

∆IL= ∆I1(1+R12/R13)=∆IDC(1+R13/R12)

Simply, when the voltage drop across R13 reaches one diode drop (0.650V) then

the current will be divided between I1 and IDC as

1:(R13/R12) = 1:(51.6/50) ≈ 1:1.

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