Kelvin connection in IC design

Table of Contents

1. Introduction
2. The concept
3. 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 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 and LDO:

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

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 lead 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 Kalvin connection:

Unity gain buffer schematic using Kelvin connection

The voltage is now sensed directly at the output of the unity gain buffer 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. The voltage is now sensed at the input of the amplifier, forcing $V_{LDO}$ slightly higher to keep $V_{out} = V_{fb}$.

3. LDO example

Now let's have a look on more complex example - 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 its output $(V_{LDO})$. 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 $V_{LDO}$ 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 PMOS device output from the feedback voltage 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. As $I_q$ is 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.

Let's assume the following:

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

Conventional connection:

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

The feedback voltage error is:

$$\Delta V_{ref} = \frac{V_{out}}{2} /\frac{V_{LDO}}{2} = 6\%$$

Kelvin connection:

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

The feedback voltage error is:

$$\Delta V_{ref} = \frac{V_{out}}{2} / \frac{V_{sense}}{2} = 0.11m\%$$