ZHCSSH5 数据表 | 德州仪器 TI.com.cn

ZHCSSH5A August 2023 – December 2024 LOG200

PRODMIX

7.1.1.2 Error Analysis Example

For an illustration of typical system error for a LOG200 implementation, consider the example use case defined by the following conditions:

Table 7-1 Example Design Parameters

PARAMETER	SYMBOL	EXAMPLE VALUE
Maximum input current	I_max	200µA
Minimum input current	I_min	10nA
Output reference voltage	V_REF	REF165 (1.65V)
Input reference current	I_I2	IREF (1µA)
Supply voltage	V_S	10V (±5V)

Table 7-2lists the major error sources, and the typical values of each under the provided conditions. Typical values are generally the sum of the mean value and one standard deviation. Calculations using these typical values tend to be conservative, as the summation of uncorrelated errors tends to result in a larger compounded total predicted error than the actual total error observed in a real system.

Table 7-2 Example Error Sources

PARAMETER	SYMBOL	TYPICAL VALUE
IREF reference current error	I_{REF_error}	0.3%
REF165 reference error	REF165_error	0.06%
Scaling factor error	K_error	0.15%
Logarithmic conformity error	LCE	0.05%
Logarithmic amplifier output offset error	V_OSO	1.3mV

These error terms are used to calculate actual values, as per the following equations:

Equation 8.

I_{R E F_a c t u a l} = I_{R E F} \times (1 - I_{R E F_e r r o r}) = 1 µ A \times (1 - 0.00 3) = 0.997 µ A

Equation 9.

V_{R E F_a c t u a l} = V_{R E F 165} \times (1 + {R E F 165}_{e r r o r}) = 1.65 V \times (1 + 0.00 06) = 1.65099 V

Equation 10.

K_{a c t u a l} = K \times (1 + K_{e r r o r}) = 250 \frac{m V}{d e c} \times (1 + 0.00 15) = 250. 375 \frac{m V}{d e c}

Begin error analysis by solving for the nominal output voltage at the minimum and maximum currents, without considering error terms. The results are then used to approximate the contribution of the logarithmic conformity error, in mV.

Equation 11.

V_{L O G_nominal_atIm i n} = K \times \log_{10} (\frac{I_{m i n}}{I_{R E F}}) + V_{R E F} = 250 \frac{m V}{d e c} \times \log_{10} (\frac{10 n A}{1 µ A}) + 1.65 V = 1.15 V

Equation 12.

V_{L O G_nominal_atIm a x} = K \times \log_{10} (\frac{I_{m a x}}{I_{R E F}}) + V_{R E F} = 250 \frac{m V}{d e c} \times \log_{10} (\frac{200 µ A}{1 µ A}) + 1.65 V = 2.2253 V

Equation 13.

{L C E}_{atIm i n} = L C E \times (V_{L O G_nominal_atIm i n} - V_{R E F}) = - 0.0005 \times (1. 15 V - 1.65 V) = 0.25 m V

Equation 14.

{L C E}_{atIm a x} = L C E \times (V_{L O G_nominal_atIm a x} - V_{R E F}) = 0.0005 \times (2. 2253 V - 1.65 V) = 0.288 m V

Repeat this exercise, taking into account typical error values as previously calculated, and then determine the difference of the results to calculate the output error at each current level.

Equation 15.

V_{L O G_a c t u a l_atIm i n} = K_{a c t u a l} \times \log_{10} (\frac{I_{m i n}}{I_{R E F_a c t u a l}}) + V_{R E F_a c t u a l} + V_{O S O} + {L C E}_{atImin} = 1.15 21 V

Equation 16.

V_{L O G_a c t u a l_atIm a x} = K_{a c t u a l} \times \log_{10} (\frac{I_{m a x}}{I_{R E F_a c t u a l}}) + V_{R E F_a c t u a l} + V_{O S O} + {L C E}_{atImax} = 2.2290 V

Equation 17.

V_{L O G_e r r o r_atIm i n} = V_{L O G_a c t u a l_atIm i n} - V_{L O G_nominal_atIm i n} = 2.117 m V

Equation 18.

V_{L O G_e r r o r_atIm a x} = V_{L O G_a c t u a l_atIm a x} - V_{L O G_nominal_atIm a x} = 3.767 m V

The output error at a given current level is then expressed as a percentage of the full-scale range as per Equation 19 and Equation 20:

Equation 19.

{E R R O R}_{f u l l_s c a l e_atIm i n} = \frac{V_{L O G_e r r o r_atIm i n}}{V_{L O G_nominal_atIm a x} - V_{L O G_nominal_atIm i n}} = 0.197 %

Equation 20.

{E R R O R}_{f u l l_s c a l e_atIm a x} = \frac{V_{L O G_e r r o r_atIm a x}}{V_{L O G_nominal_atIm a x} - V_{L O G_nominal_atIm i n}} = 0.350 %