Precision Ultralow Power High Side Current Sense LTC2063

Potential answers:

1) As you would expect from an op-amp,

the output of the LTC2063 does whatever

it has to do in order to make the voltages

on its (+) and (-) inputs be identical.

Assume the (+) and (-) currents into

the op-amp are zero. In this case,

the (+) input to the op-amp is Vload,

the voltage that appears across the

box labelled "load" (why? since

the (+) input is zero current,

there is no voltage drop across R1).

And so, the LTC2063 does whatever it

has to do to make the voltage on its

(-) input also be Vload. And so,

the V=IR equation applied to Rin

implies the current Iin going through

Rin must be Iin = (Vin - Vload)/Rin.

Where does this current go?  It can't

go into the (-), and so it must go

through transistor M1.  And so, the

LTC2063 sets its output voltage to

be the value that makes M1 pass

the current Iin.  This Iin current

then passes through Rdrive, yielding

Vout = Iin*Rdrive = (Rdrive/Rin)*(Vin-Vload).

However, applying V=IR to Rsense yields:

Vin-Vload = Isense*Rsense.

and if we plug that into the previous

equation we get:

Vout = Iin*Rdrive = ((Rsense*Rdrive)/Rin)*Isense.

Plugging in the resistor values yield Vout = 10.2*Isense.

To answer 2) and 3):

Vin can be up to 90V, but the LTC2063 can only handle

5.5V max across its power rails.  M2, REF, D1, and

associated passives form a circuit to "float" the

minus voltage rail of the LTC2063 to keep the difference

across the power rails to be under 5.5V.

Thank you for the nice explanation.I clearly understood the answer.

If you don't mind could you please elaborate your statement "

Vin can be up to 90V, but the LTC2063 can only handle

5.5V max across its power rails.  M2, REF, D1, and

associated passives form a circuit to "float" the

minus voltage rail of the LTC2063 to keep the difference

across the power rails to be under 5.5V."

How this circuitry always maintains 5.5V across OPAMP

The two diodes connect the positive rail

and the negative rail of the LTC2063.

The REF diode is a zener diode: the

voltage across this diode won't exceed

4.096V (thus its name). The other

diode is a normal diode, let's say

it's forward voltage is 0.7V.

So, 4.096V + 0.7V = 4.796 volts. So,

the voltage across the rails of the

LTC2063 can never be more than 4.796V,

which is below the 5.5V absolute maximum.

The other parts (C1, R2, M2, and the

10uF capacitor) act to coordinate the

"floating" of the negative LTC2063

rail to Vin - 4.796V when the power

to the circuit is turned on, and

"de-floating" the negative rail

back to ground when the power turns

off. The 10uF capacitor charges via

the two diodes, and discharges via

M2.

There are more modern op-amps in

the "Power by Linear" catalog that

can handle voltage rails of 100V

or more, which makes this floating

scheme unnecessary.  I'll leave it

to one of the Analog Devices employees

to recommend which one to use.

Hi John,

Thank you

Sorry for the delay in my response.

From the schematic it is clear that Vin is directly coming in contact with opamps V+.

During turn on will this damage the Opamp.

Reagrads

HAri

The absolute rating is for the difference between the positive rail

and the negative rail. The floating circuitry ensures that as

Vin ramps up from ground, the negative op-amp rail also ramps up from

ground, to ensure the differential voltage between the rails does

not exceed the absolute rating.  So, if Vin ramps from 0 to 90V,

when Vin hits 10V the minus rail is (10V - 4.796V), when Vin hits

20V the minus rail is (20V - 4.796V), and when Vin finally gets

to 90V, the minus rail is (90V - 4.796V).  And in this way the

absolute rating is never exceeded during the ramp-up.