Design Description

This circuit demonstrates how to convert a voltage-output, current-sense amplifier (CSA) into a current-output circuit using the Howland Current Pump method and operational amplifier (op amp). Furthermore, this circuit demonstrates how to design two separate circuits to measure two separate, but redundant supplies powering one load.

Design Notes

1. The Getting Started with Current Sense Amplifiers video series introduces implementation, error sources, and advanced topics for using current sense amplifiers.
2. Choose precision 0.1% resistors to limit gain error at higher currents.
3. The output current (I_OUT) is sourced from the VS supply, which adds to the I_Q of the current sense amplifier.
4. Use the V_OUT versus I_OUT curve ("claw-curve") of the CSA (INA240A3) to set the I_OUT limit during maximum power. If a higher signal current is needed, then add an op amp buffer to the output of the current sense amplifier. A buffer on the output allows for smaller R_OUT.
5. For applications with higher bus voltages, simply substitute in a bidirectional current sense amplifier with a higher rated input voltage.
6. The V_OUT voltage is the input common-mode voltage (V_CM) for the op amp.
7. Offset errors can be calibrated out with one-point calibration given that a known sense current is applied and the circuit is operating in the linear region. Gain error calibration requires a two-point calibration.
8. Include a small feed-forward capacitor (C_SET) to increase BW and decrease V_OUT settling time to a step response in current. Increasing C_SET too much introduces gain peaking in the system gain curve, which results in output overshoot to a step response.
9. Follow best practices for printed-circuit board (PCB) layout according to the data sheet: place the decoupling capacitor close to the VS pin, routing the input traces for IN+ and IN– as a differential pair, and so forth.

Design Steps

1. Choose an available current-sense amplifier (CSA) that meets the common-mode voltage requirement. For this design the INA240A3 is selected.
   • Note that choosing the most optimal CSA for the system requires balancing tradeoffs in CSA offset, CSA gain error, shunt resistor power rating and thus total circuit design could require multiple iterations to achieve the satisfactory error over the entire dynamic range of the load.
2. Determine the maximum output current (I_{SET_100%}) and maximum output swing (V_{O_ISYS_MAX}) of the INA240A3. Use the output current vs output voltage curve in the data sheet. For this design, choose the maximum I_SET to be 850 µA with a maximum output swing of {Vs – 0.2V} = 4.8V = V_{O_ISYS_MAX}.
3. Given the ADC full-scale range (V_{ADC_FSR} = 1.8V), the number of sources to measure (N = 2), and the maximum CSA output current when the source is at 100% power (I_{SET_100%} = 850µA), calculate the maximum allowable R_OUT which converts signal current to signal voltage for ADC. For this design R_OUT = 1020 Ω is selected.
   $\begin{array}{c}{I}_{OUT\_I_{SYS\_MAX}}=\text{T}\text{o}\text{t}\text{a}\text{l}\text{s}\text{i}\text{g}\text{n}\text{a}\text{l}\text{c}\text{u}\text{r}\text{r}\text{e}\text{n}\text{t}\text{f}\text{r}\text{o}\text{m}\text{a}\text{l}\text{l}\text{N}\text{c}\text{h}\text{a}\text{n}\text{n}\text{e}\text{l}\text{s}\text{w}\text{h}\text{e}\text{n}\text{s}\text{y}\text{s}\text{t}\text{e}\text{m}\text{/}\text{l}\text{o}\text{a}\text{d}\text{c}\text{u}\text{r}\text{r}\text{e}\text{n}\text{t}\text{i}\text{s}\text{a}\text{t}\text{i}\text{t}\text{s}\text{m}\text{a}\text{x}\text{i}\text{m}\text{u}\text{m}\text{(}\text{304}\text{-}\text{A}\text{)}\text{.}\\ I_{SET1\_100\%}=\text{S}\text{i}\text{g}\text{n}\text{a}\text{l}\text{c}\text{u}\text{r}\text{r}\text{e}\text{n}\text{t}\text{f}\text{r}\text{o}\text{m}\text{I}\text{N}\text{A}\text{240}\text{A}\text{3}\text{c}\text{h}\text{a}\text{n}\text{n}\text{e}\text{l}\text{1}\text{w}\text{h}\text{e}\text{n}\text{S}\text{o}\text{u}\text{r}\text{c}\text{e}\text{1}\text{i}\text{s}\text{a}\text{t}\text{100}\text{\%}\text{p}\text{o}\text{w}\text{e}\text{r}\text{(}\text{152}\text{-}\text{A}\text{)}\text{.}\\ V_{ADC\_FSR}=V_{OUT\_I_{SYS\_MAX}}<1.8V\\ I_{OUT\_I_{SYS\_MAX}}=I_{SET1\_100\%}+I_{SET2\_100\%}=I_{SET\_100\%}\times N\\ V_{OUT\_I_{SYS\_MAX}}=I_{OUT\_I_{SYS\_MAX}}\times R_{OUT}\\ \therefore R_{OUT}<\frac{V_{OUT\_I_{SYS\_MAX}}}{I_{OUT\_I_{SYS\_MAX}}}=\frac{1.8V}{850\mu A\times 2}=1058.82\Omega \\ \to R_{OUT}=1020\Omega ,0.1\%\\ \to V_{OUT\_I_{SYS\_MAX}}=1.734V<1.8V\end{array}$
4. Using the following system of equations, we can solve for the minimum allowable R_SET. For this design, R_SET = 3610 Ω is selected.
   $\begin{array}{c}{V}_{OUT\_I_{SYS\_MAX}}=I_{OUT\_I_{SYS\_MAX}}\times R_{OUT}\\ V_{OUT\_I_{SYS\_MAX}}=V_{O\_I_{SYS\_MAX}}-V_{SET\_100\%}\\ V_{SET\_100\%}=I_{SET\_100\%}\times R_{SET}\\ \therefore R_{SET}\ge \frac{V_{O\_I_{SYS\_MAX}}-I_{SET\_100\%}\times R_{OUT}\times N}{I_{SET\_100\%}}\\ \therefore R_{SET}\ge \frac{V_{O\_I_{SYS\_MAX}}}{I_{SET\_100\%}}-\left(R_{OUT}\times N\right)=3607.06\Omega \\ \to R_{SET}=3610\Omega ,0.1\%\end{array}$
5. Using the following system of equations, solve for the maximum allowable shunt resistor. For this design, choose R_SHUNT = 200 µΩ.
   $\begin{array}{c}{V}_{SET1\_100\%}=R_{SET}\times I_{SET1\_100\%}=3610\Omega \times 850\mu A=3.0685V\\ V_{SHUNT\_100\%}=\frac{V_{SET1\_100\%}}{Gai{n}_{INA240A3}}=\frac{3.0685V}{100\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$V$}\right.}=30.685mV\\ R_{SHUNT}\le \frac{V_{SHUNT\_100\%}}{I_{SOURCE\_100\%}}=\frac{30.685mV}{152A}\\ \therefore R_{SHUNT}\le 201.88\mu \Omega \\ \to R_{SHUNT}=200\mu \Omega ,1\%\end{array}$
6. Check that the common-mode voltage (V_CM) and output voltage (V_{O_TLV9061}) of the TLV9061 are in the operational region when the circuit is sensing the minimum required 5% source current. The TLV9061 device is a rail-to-rail-input-output (RRIO) op amp so it can operate with very small V_CM and output voltages, but A_OL will vary. Testing conditions from the data sheet for CMRR and A_OL show that choosing V_{OUT_5%} ≥ 40mV provides sufficient A_OL when circuit sensing minimum load current.
   • If a lower operational V_CM is needed, then consider providing a small negative voltage source to the negative supply pin to extend the range of the op amp or current-sense amplifier.
     $\begin{array}{c}{V}_{O\_MIN\_TLV9061}=40mV\\ V_{SHUNT\_5\%}=5\%\times I_{SOURCE\_MAX}\times R_{SHUNT}=7.6A\times 200\mu \Omega \\ \therefore V_{SHUNT\_5\%}=1.52mV\\ V_{OUT\_5\%}=V_{SHUNT\_5\%}\times Gain\times \frac{R_{OUT}}{R_{SET}}\\ \therefore V_{OUT\_5\%}=42.94mV>V_{O\_MIN\_TLV9061}\end{array}$
7. Using the following equations, calculate and tabulate the total, worst-case RSS error over the dynamic range of the source.
   $\begin{array}{c}R{E}_{MAX\_P}=\text{M}\text{a}\text{x}\text{P}\text{o}\text{s}\text{i}\text{t}\text{i}\text{v}\text{e}\text{R}\text{e}\text{l}\text{a}\text{t}\text{i}\text{v}\text{e}\text{E}\text{r}\text{r}\text{o}\text{r}=\frac{V_{OUT\_MAX}-V_{OUT\_TYP}}{V_{OUT\_TYP}}\\ R{E}_{MAX\_N}=\text{M}\text{a}\text{x}\text{N}\text{e}\text{g}\text{a}\text{t}\text{i}\text{v}\text{e}\text{R}\text{e}\text{l}\text{a}\text{t}\text{i}\text{v}\text{e}\text{E}\text{r}\text{r}\text{o}\text{r}=\frac{V_{OUT\_MIN}-V_{OUT\_TYP}}{V_{OUT\_TYP}}\\ E_{RSS}=\sqrt{{e_{V_{OS\_CSA}}}^2+{e_{V_{OS\_OPA}}}^2+{e_{R_{SHUNT}}}^2+{e_{Gain\_CSA}}^2+{e_{R_{OUT}}}^2+{e_{R_{SET}}}^2}\\ V_{OUT\_TYP}=I_{SOURCE1}\times R_{SHUNT\_TYP}\times G_{TYP}\times \frac{R_{OUT\_TYP}}{R_{SET\_TYP}}\\ V_{OUT\_MAX}=\left[\right(I_{SOURCE1}\times R_{SHUNT\_MAX}+V_{OS\_CSA\_MAX})\times G_{MAX\_CSA}+V_{OS\_OPA\_MAX}]\times \frac{R_{OUT\_MAX}}{R_{SET\_MIN}}\\ V_{OUT\_MIN}=\left[\right(I_{SOURCE1}\times R_{SHUNT\_MIN}-V_{OS\_CSA\_MAX})\times G_{MIN\_CSA}-V_{OS\_OPA\_MAX}]\times \frac{R_{OUT\_MIN}}{R_{SET\_MAX}}\end{array}$
   $\begin{array}{c}{T}_{MAX}={80}^{o}C\\ \Delta T_{MAX}={80}^{o}C-25°C=55°C\\ R_{SHUNT}=200\mu \Omega ,0.1\%,\text{175}\frac{ppm}{°C}\\ V_{VS}=5V;V_{CM}=12V\\ V_{OSI\_OPA}=\pm 2mV\\ V_{OS\_OPA\_CMRR}=|V_{OUT} - 2.5V|\times {10}^{(-80dB/20dB)}\\ V_{OS\_OPA\_MAX}=V_{OSI\_OPA}+V_{OS\_OPA\_CMRR}+\Delta T_{MAX} \times \left(530\frac{nV}{°C}\right)\\ V_{OSI\_CSA\_MAX}=\pm 25\mu V\\ V_{OS\_CSA\_CMRR\_MAX}=|12V-V_{CM}|\times {10}^{\raisebox{1ex}{$(-CMR{R}_{MIN}$}\!\left/ \!\raisebox{-1ex}{$20dB$}\right.)}=0\\ V_{OS\_CSA\_PSRR\_MAX}=|5V-V_{VS}|\times PSR{R}_{MAX}=0\\ V_{OS\_Drift\_MAX}=\Delta T_{MAX}\times \left(\frac{\Delta V_{OS}}{\Delta T}\right)=55°C\times \left(250\frac{nV}{°C}\right)=\pm 13.75\mu V\\ V_{OS\_CSA\_MAX}=V_{OSI\_MAX}+V_{OS\_CMRR}+V_{OS\_PSRR}+V_{OS\_Drift}\\ V_{OS\_CSA\_MAX}=\pm 38.75\mu V\\ e_{V_{OS\_CSA}}=\raisebox{1ex}{$V_{OS\_CSA\_MAX}$}\!\left/ \!\raisebox{-1ex}{$V_{SHUNT\_IDEAL}$}\right.\times 100\\ e_{V_{OS\_OPA}}=\raisebox{1ex}{$V_{OS\_OPA\_MAX}$}\!\left/ \!\raisebox{-1ex}{$V_{SET\_IDEAL}$}\right.\times 100\\ e_R=e_{R_{TOLERANCE}}+e_{R_{DRIFT}}\\ e_{R_{SHUNT}}=1\%+\Delta T_{MAX}\times TC=1\%+55°C\times \left(\text{175}\frac{ppm}{°C}\right)\times {10}^{-4}=1.963\%\\ e_{R_{SET}}=e_{R_{OUT}}=1\% +55°C\times \left(\text{50}\frac{ppm}{°C}\right)\times {10}^{-4}=1.275\%\\ e_{GAIN\_CSA\_25C}=\pm 0.2\%\\ e_{GAIN\_Drift\_CSA\_MAX}=\Delta T_{MAX} \times \left(2.5\frac{ppm}{°C}\right)\times {10}^{-4}=\pm 0.01375\%\\ G_{MAX}=G_{TYP}\times \left(1+e_{25C\_MAX}+e_{Drift\_MAX}\right)=100 \frac{V}{V}\times \left(1.002138\right)=100.2138 \frac{V}{V}\\ G_{MIN}=G_{TYP}\times (1- e_{25C\_MAX }- e_{Drift\_MAX})=100 \frac{V}{V}\times \left(0.997862\right)=99.7862 \frac{V}{V}\end{array}$
8. Plot the total error as a function of load current
   

Design Simulations

DC Simulation Results

The following graph shows a linear output response for load currents from 5A to 304A.

AC Simulation Result – I_LOAD to I_OUT (V_OUT) circuit gain