Kelvin connection in IC design

Table of Contents

1. Introduction
2. The concept
3. Kelvin connection in unity gain buffer (example)
4. Kelvin connection in LDO (theory)
5. Kelvin connection in LDO (example)

1. Introduction

Kelvin connection - is a very clever concept, often used for high-accuracy resistance measurements by separating the leads that supply current from the leads that measure voltage. Let's explore how this concept applies to the IC design and helps to create more accurate circuits.

2. The concept

The entire idea behind Kelvin connection is to separate the nodes that are carrying high currents from the sensing nodes to the feedback. Let's have a look on a very simple example - a unity gain buffer:

Unity gain buffer schematic

In this circuit, the error amplifier controls the gate of the PMOS transistor to maintain the voltage at the output equal to $V_{ref}$ by supplying enough current to the load. When this circuit is physically implemented on chip, the routing resistance $R_{trace}$ appears between the output of the PMOS pass device ($V_{force}$) and the actual output of the circuit (i.e. chip pad). This parasitic resistance makes the output voltage $V_{out}$ to be lower than output voltage of an LDO due to the voltage drop:

$$V_{out} = V_{LDO} - I_{L}R_{trace}$$

As we are sensing the feedback voltage from the output of an LDO, we don't see that voltage change caused by $R_{trace}$ which leads to an output error. If the output current is large, i.e. $100mA$ and even if the routing resistance is $\approx 1 \Omega$, the output voltage difference will be:

$$V_{drop} = I_{L}R_{trace} = 100mA*1 \Omega = 100mV$$

To avoid such a large voltage error, we can use a Kelvin connection:

Unity gain buffer schematic using Kelvin connection

In this case, the voltage is now sensed directly at the output and takes the error caused by $R_{trace}$ into account. Since there is no current flowing into the input of the amplifier, there is no longer any $IR_{drop}$ issue. This voltage is seen at the input of the amplifier, forcing $V_{force}$ to be slightly higher to keep $V_{out} = V_{fb} = V_{ref}$.

3. Kelvin connection in unity gain buffer (example)

For a unity gain buffer circuit, let's assume the following:

• $V_{ref} = 1V$
• $V_{out} = 1V$
• $I_{load} = 100mA$
• $R_{trace} = 1\Omega$

Unity gain buffer example

Conventional connection:

$$V_{force} = V_{ref} = 1V$$

Due to the resistance of the trace, the output voltage is:

$$V_{out} = V_{force} - I_{L}R_{trace} = 1V - 100mA*1\Omega = 0.9V$$

Since $V_{fb} = V_{force}= V_{fb}$, the loop is locked, while the error of $0.1V$ is present at the output:

$$\Delta V_{out} = \frac{V_{out}}{V_{ref}} = \frac{0.9V}{1V} = 10\%$$

Kelvin connection:

Since now force and sense are separated and no current is flowing to the input of the amplifier, we can see that even though $R_{trace}$ is still present in the feedback path, there is no $IRdrop$ hence no voltage error. This means that $V_{fb} = V_{out}$ and the voltage difference has to be adjusted by the amplifier.

When the loop is locked, we will observe the following:

$$V_{force} = 1.1V$$ $$V_{out} = V_{force} - I_{L}R_{trace} = 1.1V - 100mA*1\Omega = 1V$$

There is no voltage error at the output in this case, thanks to the Kelvin connection.

4. Kelvin connection in LDO (theory)

Now let's take a look at more complex circuit - a low-dropout regulator (LDO), which contains a resistive feedback:

Low dropout regulator (LDO)

In this case, using a conventional feedback connection LDO regulates the voltage at the output of the pass device $V_{force}$. When the load current is flowing through the routing resistance $R_{trace}$ it causes the output voltage of the LDO to deviate from $V_{ref}$ by $V_{drop} = I_{L}R_{trace}$. Since the feedback is derived from $V_{force}$ and not from $V_{out}$, the feedback loop (and error amplifier in particular) doesn't see that deviation of the $V_{out}$ from $V_{ref}$.

The solution is to separate the feedback voltage from the PMOS device output by having two outputs - $V_{force}$ and $V_{sense}$:

Low dropout regulator (LDO) schematic using Kelvin connection

Let's say, $V_{force}$ is set to be equal to the $V_{ref}$ by the loop, then:

$$V_{out} = V_{force} - I_{L}R_{trace}$$

Considering the same trace resistance in the $V_{out}$ to $V_{sense}$ path, the voltage drop will be:

$$V_{drop2} = I_{q}R_{trace}$$

, where $I_q$ is the quiscent current, defined by the total feedback resistance ($R_{fb1} + R_{fb2}$). As $I_q$ is usually very small compared to the load current, the voltage drop will be also very small, delivering almost entire value of the true output voltage to the feedback. We can now say that true output voltage value is fed back into the loop.

5. Kelvin connection in LDO (example)

For an LDO circuit, let's assume the following:

• $V_{ref} = 0.9V$
• $V_{out} = 1.8V$
• $I_{load} = 100mA$
• $R_{trace} = 1\Omega$
• $R_{fb1} = R_{fb2} = 50k\Omega$

Conventional connection:

LDO example (conventional connection)

$$V_{force} = V_{ref} = 1.8V$$ $$V_{out} = V_{force} - I_{L}R_{trace} = 1.8V - 100mA*1\Omega = 1.7V$$

The feedback voltage is:

$$V_{fb} = V_{force} \frac{R_{fb2}}{R_{fb1} + R_{fb2}} = 1.8V \frac{50k\Omega}{50k\Omega + 50k\Omega} = 0.9V$$

In this case, $V_{fb} = V_{ref}$, while the output of an LDO is 1.7V instead of 1.8V, producing an error at the output:

$$\Delta V_{out} = \frac{V_{out}}{2V_{ref}} = \frac{1.7V}{1.8V} = 6\%$$

Kelvin connection:

LDO settling using Kelvin connection

If the force and sense paths are different:

$$V_{force} = V_{ref} = 1.8V$$ $$V_{out} = V_{force} - I_{L}R_{trace} = 1.8 - 100mA*1\Omega = 1.7V$$ $$V_{sense} = V_{out} - I_{q}R_{trace} = 1.7V - 18\mu A*1 \Omega = 1.699982V$$

The feedback voltage is:

$$V_{fb} = V_{sense} \frac{R_{fb2}}{R_{fb1} + R_{fb2}} = 1.699982V \frac{50k\Omega}{50k\Omega + 50k\Omega} = 0.85V$$

Since $V_{fb}$ is not equal to $V_{fb}$, the loop will drive $V_{force}$ to become $1.799982V$, which will make $V_{sense}$ to be equal to:

$$V_{sense} = V_{out} - I_{q}R_{trace} = 1.799982V - 17 \mu A*1 \Omega = 1.799965V$$

So that the feedback voltage becomes:

$$V_{fb} = V_{sense} \frac{R_{fb2}}{R_{fb1} + R_{fb2}} = 1.799965V \frac{50k\Omega}{50k\Omega + 50k\Omega} = 0.9V$$

In that case, the output voltage error when the LDO's loop is locked will be:

$$\Delta V_{out} = \frac{V_{out}}{2V_{ref}} = \frac{1.799982V}{1.8V} = 0.001\%$$