# Sensing Voltage and Current of an Adjustable Voltage Source

I am currently working on an Adjustable Voltage Source, the Voltage Source is built by a voltage regulator with an OpAmp with and a Darlington transistor for current amplifying as the following diagram shows.

Vin is the output of the voltage regulator, so we can't know for sure the voltage at Vin. But because of the configuration of the OpAmp (Voltage Follower), we can assure that Node A and Vin are at the same voltage.

Given that Node A is equal to Vin, the OpAmp will compensate for the voltage drop across the Shunt Resistor, thus Node B voltage will be Node A Voltage plus the voltage across the shunt resistor.

I've thought of a Differential amplifier between nodes A and B, to get to know the Voltage across the shunt, thus knowing the current flowing across the load and the shunt.

*Suppose 4 Resistor values are the same

With this configuration a microcontroller con sort out the shunt current with Ohm's Law. But the equation system I tried to solve was missing one variable.

It is possible to figure out the voltage and resistance of the load by knowing the current and voltage across the shunt?

If not, which would be the most part efficient way of sensing the voltage and current across the Load?

It is possible to figure out the voltage and resistance of the load by knowing the current and voltage across the shunt?

Yes, if you know the voltage across the shunt, it's a simple V/R=I equation to know the current. Since the input bias current to the op amp is infinitesimal compared to the current flowing through the shunt, the current flowing to the op amp can be neglected.

As far as the voltage goes, you can consider the voltage out to be the same as voltage in, with a few exceptions. Vout can't get as high as the rail, the voltage drop across the darlington pair will limit the voltage to something like 14.5V or 14.7V. The load will also limit the voltage range of vout, the lower resistance, the lower the voltage the circuit will be able to maintain.

For example, the darling ton pair saturates at 1A, and you put a 5Ω resistor for a load, the circuit cannot supply more control energy beyond 1A, so you would be limited to range of 0 to 5V. With a 10Ω load, a range of 0 to 10V.

It will be a good idea to select your components and run a spice AC analysis on the circuit, the capacitance of the transistor pair can change and create oscillations at high frequencies. A good place to start is an RC filter between the negative terminal and Vout. A series resistor on the output of the amp to limit the current is also a good idea.

These circuits also can have problems with inductive loads, if your load is inductive, the feedback loop will need to be compensated for it.

• Thanks, if I only use a single transistor instead of the darlington pair, the capacitance problem disappears? For the current limit resistor should I pick the value given the highest voltage and the highest current? Also I don't fully understand how to get the load resistance and voltage values with the shunt resistance and current values. I fully understand that the current on both resistors are the same since it is in series. But how can I write an expression that defines the Load voltage and resistance in terms of the shunt voltage and current? Commented Jun 13, 2019 at 20:40
• No, the capacitance problem is there, even if you use an nmos, there is frquency compensation problems. I'd pick the highest current you expect, then size the current resistor with that. Vout is equal to Vin, until the current runs out, see the edited post. The amplifier will control the voltage on vout to equal vin, so it will provide more current to counteract the shunt resistor Commented Jun 13, 2019 at 20:56
• meta.stackexchange.com/questions/126180/… Commented Jun 13, 2019 at 20:57

It is better to create a known current source and measure the output voltage to measure the load.

• This can be DC with a sinusoidal signal Vpp = Vdc for unipolar measurements or no DC for bipolar measurements.

• Then you would use complementary Darlington Emitter Followers and a bipolar supply.

• Current sensing is OK and you can use R ratios of 10 to 50 to get voltage gain, depending on your range specs.

However, if using a voltage source, you can sense current and amplify it and also use AC superimposed on DC. But you would have to compute V/I ratio to get R and measure phase to get reactance.