## Resistance Measure

### CIRCUIT

##### DMM_R_MEAS1.CIR
Download
the SPICE file

How do you measure resistance when the dual-slope integrator measures only voltage? This topic explores the ratio method.It harnesses the fundamental operation of the dual-slope integrator which creates a digital word using a ratio of an unknown voltage to a known reference voltage. We'll take advantage of this mechanism to find a ratio of an unknown resistance to a known resistance.

### RESISTOR RATIO

The unknown resistance (Rx) is simply placed in series with a Reference Resistor (Rref). A bias voltage(VB)drives the series resistors causing a current to flow. What do we know about the ratio of V_Rx to V_Vref? It should be equal to the resistor ratios

= Rx / R_Ref

### DUAL SLOPE INTEGRATOR

How do you connect these voltages to the dual-slope integrator inputs? As you might have guessed, V_Rx connects directly to the signal input, while V_Rref connects to the reference voltage input.

A quick recap of the dual-slope integrator might help here. Basically, the integrater has two phases

You typically use a counter to set T1 (fixed) and
measure T2 (variable).

The bottom line? The ratio T2 / T1 equals Vx / Vref so you could write

or, in terms of resistance

As an example, if you designed a DMM with T1=10ms, then for Rref=10k and a measured T2=7.5ms, you can deduce a measured resistance of

= 7.5k ohms

### MEASURE IT

Enough talking, let's start simulating. VB creates a bias voltage for the resistor string. RB = 1k provides an additional series resistance to limit the current for low level measuremenbts (<100 ohm). Let say you've selected the 1 k ohm range. This means of course you need to set R_REF = 1k. Finally, set the unknown resistance to something like RX=900 ohms.

= Tfall - 10ms

For RX=900 ohms, where is the falling edge of V(5)? Uing the cursor, you find Tfall = 19.01ms which implies T1 = 19.01-10.0 = 9.01ms. The resistance is easily calculated as

= 1k *
9.01ms/10ms

= 901 ohms

A respectable result - nearly a 0.1% measuremt accuracy.

What happens when the ersistance gets too smalll? Try measureing RX near 100 ohms or 10 ohms. How accurate is the measurement? Might be best to switch to lower range. To do this, change R_REF to 100 or 10 ohms. How much did the accuracy improve?

### SPICE FILE

DVM_RES_MEAS1.CIR * * VB VB 20 0 9VDC RB 20 21 1k * R_REF R_REF 21 22 1K * * RX - UNKNOWN RESISTOR RX 22 0 10 * * DIFF AMP - REFERENCE, R_RES E_REF 12 0 21 22 +1.0 * * DIFF AMP - UNKNOWN RES, RX E_RX 11 0 22 0 -1.0 * * CONTROL: VCNTRL=0 S1=ON, S2=OFF * VCNTRL=1 S1=Off,S2=ON VCNTL 15 0 PWL(0MS 0V 10MS 0V 10.01MS 5V) * * INTEGRATOR S1 11 1 15 0 SWB S2 12 1 31 0 SWA R1 1 2 100K C1 2 3 0.10UF IC=0V XOP1 0 2 3 OPAMP1 * * COMPARATOR (ZERO CROSSING DETECTOR) XCMP1 3 0 5 COMP1 * * AND GATE, S2 = ON IF VCNTRL AND XCMP1 OUTPUTS ARE HI VCC 30 0 DC 5V R31 30 31 10k D31 31 15 D1N4148 D32 31 5 D1N4148 * * * SUBCIRCUITS AND MODELS *********************************** * .SUBCKT COMP1 1 2 5 * TERMINALS: 1-INPUT+, 2-INPUT-, 5-OUTPUT * DIFF AMP EDIFF 3 0 VALUE = { V(1) - V(2) } * FREQUENCY RESPONSE RP1 3 4 500 CP1 4 0 1000PF * LIMITER EOUT 5 0 TABLE {V(4)} = (-0.5MV 0V) (0.5MV, 5V) .ENDS * * * OPAMP MACRO MODEL, SINGLE-POLE WITH 15V OUTPUT CLAMP * connections: non-inverting input * | inverting input * | | output * | | | .SUBCKT OPAMP1 1 2 6 * INPUT IMPEDANCE RIN 1 2 10MEG * DC GAIN=100K AND POLE1=100HZ * UNITY GAIN = DCGAIN X POLE1 = 10MHZ EGAIN 3 0 1 2 100K RP1 3 4 100K CP1 4 0 0.0159UF * OUTPUT BUFFER AND RESISTANCE EBUFFER 5 0 4 0 1 ROUT 5 6 10 .ENDS * .MODEL SWA VSWITCH(VON=5 VOFF=0 RON=1 ROFF=1e12) .MODEL SWB VSWITCH(VON=0 VOFF=5 RON=1 ROFF=1e12) * * DIODE .model D1N4148 D(Is=0.1p Rs=16 CJO=2p Tt=12n Bv=100 Ibv=0.1p) * * ANALYSIS ************************************************* .TRAN 25US 20MS UIC *.TRAN 0.1US 2000US UIC* .PROBE .END