### DCR Current Sensing Explained

There are many ways to measure a dc-dc converterâ€™s inductor current for use in control. *Direct-current resistance* (DCR) is one low loss and cost method for measuring this current, it consists of a series RC network in parallel with the inductor. The RC network has some advantages over other methods: it has lower I^{2}R losses than shunt resistors and is cheaper than transducers. The correct RC time constant will produce a measurable capacitor voltage that matches the inductor current. However, DCR could run into issues at fast switching frequencies where parasitics arise or when the inductor coil heats-up which increase resistance. Both situations disarrange the tuned circuit network making the measurement less accurate. To demonstrate how DCR works, consider in the dc-dc converter in Fig. 1.

**Fig. 1 Buck converter with DCR**

The buck converter circuit was simplified for analysis purposes by replacing the input and output with arbitrary voltage sources as shown in Fig. 2. The input voltage *V _{in}(t)* is a voltage signal produced by switching a dc voltage source. The input voltage passes through an LC filter, where

*L*is the inductor and

_{2}*R*its winding resistance. The inductor and output capacitor produce a filtered output signal

_{2}*V*. The components in parallel with the inductor,

_{out}(t)*R*and

_{1}*C*, is the DCR network, which is chosen in such a way that the voltage

_{1}*v*across

_{C}*C*is proportional to the inductor current

_{1}*i*.

_{L}**Fig. 2 Simplified Buck circuit**

The circuit is further simplified by combining the network into parallel impedance, *Z _{1}* and

*Z*as shown in Fig. 3.

_{2}**Fig. 3 Parallel impedance network**

The following derivation shows the mathematical relationship between *R _{1}*,

*C*,

_{1}*R*, and

_{2}*L*how to choose the RC network.

_{2}The voltage across *C _{1}* is

(1)

and total input current is

(2)

where the total input impedance is

. (3)

The voltage difference term in (1)_{ }replaced with according to (2) gives

. (4)

Looking at the parallel impedance as a current divider, the inductor current is

(5)

Substituting (3) into (4)

(6)

then (5) into (6)

(7)

The DCR capacitor voltage is proportional to inductor the current by a factor *K* when

(8)

Which is the same as

. (9)

Thus when

(10)

(11)

and thus

(12)

To validate this mathematical relationship, the buck converter in Fig. 4 was simulated in Simulinkâ„˘ and the results plotted Fig. 5. The parameter values for the circuit are *E* = 12 V,* R _{1}* = 1 kÎ©, C

*= 1 ÎĽF,*

_{1}*R*= 1 Î©, and

_{2}*L*= 1 mH. The circuit ran at a switching frequency of 100 kHz at 50% duty cycle.

_{2}**Fig. 4. Simulink Buck converter model**

**Fig. 5. Simulated output**

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Grant,

Laplace analysis doesn’t say anything about the dc value itself, so this is incomplete. You might be able to show that any error due to startup dies off exponentially, as it seemed to in the simulation, because the system is strictly passive, but as it stands now, equation 12 is not entirely true. Just put a starting values of 20000 V on the cap and 0 A in the inductor, and clearly equation 12 is wrong at time T0 . I’m not saying you are wrong, just that you cant go back from Laplace domain to time domain like that without considering starting conditions.

-Nick

FALSE: “Laplace analysis doesnâ€™t say anything about the dc value itself”Â

The Laplace Transforms are used specifically for solving differential equations with initial values. The general form for the Laplace transform include both “s” and “dc” terms.Â

e.g. Consider a continuous function x(t) whose derivative has the Laplace Transform: Â L{dx/dt} = sX(s) – x(0) where the initial condition can be considered a dc term.Â Also, results from the final value theorem, where sÂ -> 0 can be considered dc values.Â

NOT POSSIBLE: “Just put a starting values of 20000 V on the cap and 0 A in the inductor, and clearly equation 12 is wrong at time T0 .”Â

You are correct, initial conditions are important parts of a dynamic solution. However, we are taught to ignore them in circuit analysis since they are almost always zero (due to passivity).Â In the proposed circuit, any starting voltage on the capacitor C1 is impossible. A small amount of initial energy in C1 would dissipate through R1, R2, and L2. A small amount of initial energy in L2 would dissipate through the load.Â True, aÂ large jump in capacitor charge (perhaps from a nuclear EMP) would have a short time scale effect;Â it would not in the large time scale.Â

I think the following statement will clear up the confusion:

All LTI systems dx/dt = Ax + BÂ reach the same dc steady-state operating point that is independent of its initial conditions. The same cannot be said for nonlinear systems which have many steady-state points.Â I believe that the circuit is nearlyÂ LTI for small variations in the arbitrary voltage sources Vin and Vout .Â

This is a rather long response, perhaps it willÂ bud into its own blog article.

This blog was… how do you say it? Relevant!!

Finally I’ve found something which helped me. Appreciate it!