Precision Ultra Low Power High-Side Current Sense

Thread Starter

hoyyoth

Joined Mar 21, 2020
217
Dear Team,

I am going through this article from AD .

1662228360016.png
I have some questions about this circuit.

1)May I know how the Vout Equation is derived

2)May I know the use of M1 and M2

3)May I know the use of REF and D1

4)Why the GND pin of Opamp is lifted uisng M2
 

Ian0

Joined Aug 7, 2020
6,293
Dear Team,

I am going through this article from AD .

View attachment 275561
I have some questions about this circuit.

1)May I know how the Vout Equation is derived

2)May I know the use of M1 and M2

3)May I know the use of REF and D1

4)Why the GND pin of Opamp is lifted uisng M2
REF, D1 and M2 produce a supply for the LTC2063 that is referred to the positive supply
The op-amp's purpose in life is to keep its inputs at the same voltage.
Therefore, the voltage across Vin is equal to the voltage across Isense.
The current through Rin = Isense*Rsense/Rin
The same current flows through Rdrive (because it is virtual earth)
You should be able to sort it out from there.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
217
Dear Team,

May I know the intention of putting two 49.9 ohm resistors on both the inputs of Opamp.
How that value is arrived.

Regards
HARI
 

crutschow

Joined Mar 14, 2008
30,788
May I know the intention of putting two 49.9 ohm resistors on both the inputs of Opamp.
How that value is arrived.
The top resistor value determines the gain of the op amp, so it's value depends upon the gain you want.

The bottom one is just added to balance the input impedance (not really needed here).
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
217
Thank you..
Is it a good idea to put a resistor in between the opamp output and gate of the MOSFET (M1).
Here Opamp is driving a capacitive load
 

Ian0

Joined Aug 7, 2020
6,293
Thank you..
Is it a good idea to put a resistor in between the opamp output and gate of the MOSFET (M1).
Here Opamp is driving a capacitive load
Actually, it isn't because the other end of the capacitor (which connects to the source) doesn't connect to an AC earth, so the op-amp sees the load as the capacitor in series with the source resistance.
 

crutschow

Joined Mar 14, 2008
30,788
Actually, it isn't because the other end of the capacitor (which connects to the source) doesn't connect to an AC earth, so the op-amp sees the load as the capacitor in series with the source resistance.
It does see the gate-drain Miller capacitance, so a small gate resistor might be a good idea.
 

crutschow

Joined Mar 14, 2008
30,788
My simulation of the circuit shows a large transient voltage across the op amp power terminals during the application of power, nearly equal to the supply voltage, so that could zap the op amp.
The circuit may need to be modified to prevent that.
 

Ian0

Joined Aug 7, 2020
6,293
My simulation of the circuit shows a large transient voltage across the op amp power terminals during the application of power, nearly equal to the supply voltage, so that could zap the op amp.
The circuit may need to be modified to prevent that.
That 10uF should be across the op-amp supply, not effectively in series with it.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
217
Dear Team,

Thank you.
1662356432386.png
May I know how this passive circuitry maintains a V+ - V- =Supply voltage of LTC2063.

My analysis is given below

Here V+ will be always Vin.
The voltage at the node N will also be Vin- Vref. This voltage minus one diode drop will come at V-.
R2 is used for biasing Vref
Please correct me if I am wrong.

May I know the role of M2 .

Regards
HA
 

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crutschow

Joined Mar 14, 2008
30,788
M2 drops the voltage from V- to ground.
Ignore the diode as it's not active during normal operation (It apparently is there for startup).
V- is thus Vin - Vref + Vgs where Vgs is the gate-source voltage needed to turn on M2 and carry the circuit current (Vgs is a negative voltage).
 

MrAl

Joined Jun 17, 2014
9,546
Dear Team,

I am going through this article from AD .

View attachment 275561
I have some questions about this circuit.

1)May I know how the Vout Equation is derived

2)May I know the use of M1 and M2

3)May I know the use of REF and D1

4)Why the GND pin of Opamp is lifted uisng M2
Hello,

Vout is derived though the mechanism of two balances, one of current and the other of voltage.

The voltage balance comes from the op amp always attempting to keep the inverting and non inverting inputs at the same voltage, that's a way of looking at voltage mode op amps in the linear mode and that's a way they are often used. When the voltage balances, the op amp stops decreasing the M1 gate voltage which keeps the M1 drain source current at a set level. That level of current also goes through Rdrive and that develops a voltage Vout.

The current balance come from the M1 source current attempting to balance the current through Rin. When the current is balanced the voltage at the inverting terminal no longer changes so the circuit is in complete equilibrium with a certain current through Rdrive and thus a certain voltage at Vout.

As the input sense current changes, the balancing changes and thus the current through Rdrive changes and because of those balances the new Vout takes on a new value that is closely related to the sense current.

To derive the equation for Vout, simply put these two balances into the equation such that each component involved is used and the mechanics of the balancing is portrayed as per the various circuitry laws like Kirchhoff's Laws.

You might be able to understand how all this works if you first look at a bipolar transistor version of this kind of high side current sensing without the M2 transistor. It's basically the same.

Side note:
Yeah i wondered about that capacitor too because if you apply 90v to the circuit with the cap discharged, a typical starting condition, the cap takes some time to charge which means the op amp gets 90v at start up. Kind of silly :)
The diode is meant to help that but the reference diode impedance can be too high at startup at least for a short time which means the cap can not charge up instantaneously.
 

MrAl

Joined Jun 17, 2014
9,546
Dear Team,

Thank you very much.

May I know What the diode actually doing at startup.
The diode is there to help get the minus power supply input terminal of the op amp up to about 4.8 volts less than the input source voltage which can be as high as 90 volts. If the op amp can only take 5v and you apply 90v you blow out the op amp.
The diode, in the original circuit, was there to help the 10uf cap charge quickly, but i would not have relied on that so it may be a mistake anyway. In the circuit with the 10uf cap removed and placed elsewhere, the diode does the same thing but should be much faster because of much less capacitance, or it may not be needed at all anymore.

One way to derive the output Vout is as follows.

The voltage at the inverting input of the op amp:
vn=Vin-Iout*Rin
and the voltage at the non inverting input of the op amp:
vp=Vin-Iload*Rsense
and when the op amp is in the linear mode we have very nearly vn=vp so:
Vin-Iout*Rin=Vin-Iload*Rsense
and subtracting the right hand side we get:
Iload*Rsense-Iout*Rin=0
and solving for Iout we get:
Iout=(Iload*Rsense)/Rin
and since Vout is Iout times Rdrive we end up with:
Vout=Iout*Rdrive
which comes out to:
Vout=(Iload*Rdrive*Rsense)/Rin
and now with Rdrive=5000 Ohms and Rsense=0.1 Ohms and Rin=49.9 Ohms we get:
Vout=10.02*Iload

One of the details left out above is how the op amp output controls the mosfet M1.
Because of the large internal gain of the op amp the output of the op amp will go to whatever voltage is needed to drive M1 such that the current through the source (and thus the drain) is equal to the current through Rin. That means there will actually be a small difference between vn and vp but it will usually be small enough to ignore and call zero so the equations are simpler. Without that simplifications, we'd have to know the exact internal gain and the transconductance of the mosfet M1 which makes it more difficult with little accuracy to gain compared to most component tolerances. It's interesting to go through that too though if you want too but you will still have to assume some things about the mosfet that may not be the same with every mosfet of that part number anyway, so the simplified derivation is usually good enough.
As i said before though, if you look at a non floating circuit like this that uses a bipolar transistor you will gain more insight.

I can guess a few other things here pending some tests.
It may be that the circuit was intended to be used with slowly varying Vin that powers the load so it may start up at 5v and gradually ramp up to 90v if needed. That would help to ensure the op amp never gets a voltage supply that is too high.
The other guess is that the output of the op amp is more like a ramp because of the integrating effect of the mosfet gate source and so the output current would be similar to a regular integrating circuit with an op amp and capacitor feedback. This can be verified (or just the opposite) in a simulation.

In closing i have to say that i never liked floating voltage sources just as much as i never liked high voltages bipolar transistors with their emitters and collectors connected in series to create a single transistor with an even higher yet CE voltage rating. Nonetheless i have done both. One very memorable occasion was when one of my CRT TV sets blew out one day and i had to test it with a transistor i didnt have, so i used several 200v bipolar transistors with CE's in series and some resistor divider drive, and i was able to test it to make sure it would work if i bought a new genuine very HV transistor which at the time was very expensive. Without that i would have had to guess if it would work with a new transistor. It turned out that it was that transistor and a blown rectifier diode that killed the TV set.
 
Last edited:

crutschow

Joined Mar 14, 2008
30,788
One other caveat to note:
Since the MOSFET Vgs subtracts from the op amp supply voltage, an M2 Vgs greater than 2.4V when operating in the circuit, will reduce the op amp supply voltage below its 1.7V minimum required.
This means the max Vgs(th) of M2 should be ≤2V, or a REF reference with a higher voltage must be used.

The BSP322P shown in the circuit does meet that 2V criteria.
 
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