Current Drain on ADC

Question

I am looking at a MCP3008 datasheet and it says that it has a maximum current drain of 150uA, and a typical current drain of 100uA.

I am building a voltage divider to knock down 96V to 3V. As this is for a solar car, I would like to reduce the amount of current and power used. I would like to be able to read a maximum voltage of 115V, and a minimum of below 75V, and be able to waste the least amount of power nessassary. How does the current drain factor into my calculations?

My calculations are as follows:

• Using ohms law with E=115V I=0.00015, I get 76666 ohms
• Using a voltage divider I get R1 = 76666 and R2 = 2053.55

Are my calculations correct? Can I knock more power out of this circuit?

Answer 1

If you want to express resistor values in 5 significant digits you should at least start with the right voltage:

\$ \dfrac{115\mathrm{\,V} - 3\mathrm{\,V}}{150 \mathrm{\, \, \mu A}} = 746667 \, \Omega\$,

and the other resistor will be 20 k\$\Omega\$.

Anyway, the values you used will get you 1.5 mA, not 150 \$\mu\$A. And even with 10 times higher resistor values you won't get 3 V, since the input current to your ADC is actually a resistor parallel to the 20 k\$\Omega\$.

\$ \dfrac{115\mathrm{\,V} - V_{OUT}}{746.667 \, k \Omega} = \dfrac{V_{OUT}}{20 \, k \Omega} + 150\, \mu A \$

Or \$V_{OUT}\$ = 78 mV. (1.5 mA would leave 2.7 V.)

Don't feed the divided voltage directly to the ADC, but buffer it first using a voltage follower opamp. Then the 150 \$\mu\$A won't load the divider, and you can safely use higher resistor values, > 1 M\$\Omega\$.

edit (see Telaclavo's comment)
I didn't look at the datasheet at first, I trusted the 150\$\mu\$A was a correct value. ADCs do often have a low input impedance. However, checking the datasheet, Telaclavo is right, the 150\$\mu\$A refers to the reference input (page 3). On page 4 you can see the input leakage current for the analog input: 1\$\mu\$A maximum. That means the ADC has to have a buffer for the input signal.
Anyway, this changes things fundamentally. Suppose you choose the resistors so that they'll draw 100\$\mu\$A: 1.12M\$\Omega\$ + 30k\$\Omega\$.

\$ \dfrac{115\mathrm{\,V} - V_{OUT}}{1.12M \Omega} = \dfrac{V_{OUT}}{30 k \Omega} + 1 \mu A \$

Then \$V_{OUT}\$ = 2.97V. That's for a worst case input current of 1\$\mu\$. If you would use an LF411 as buffer, which has an input current of 50pA, the output would be 3V for both the typical leakage current of 1nA, and the maximum of 1\$\mu\$A.

Answer 2

The calculations are wrong, but you are also using in the divider the same leakage current of the ADC. Now you have about a 100% error, but you want something better.

So if you want to save power, it would be better to use a current ten times bigger (1 to 10 mA), so the error due to ADC leakage will be smaller. And you can consider also the leakage in the equation, so reducing the error.

Consider a circuit like this:

You want 3V over R2, so you'll have 112V over R1; at 1 mA, you need a 112KOhm resistor.

With this circuit, you will still have an error depending on the leakage of the ADC (measure it for the best accuracy) (1uA over 1 mA is about 0.1% error), but the consumption is limited, as 1mA in a car application is acceptable.

Note that in this case, the power dissipated by R1 is 112 mW, which is close to the limit if you use 1/4 W resistors; consider using 1/2 W ones.

But also the buffer solution suggested by Federico is good: if you can, use an amplifier.

Update according with real datasheet values

Since leakage is only 1 uA, you can also scale down the current to 100 uA while having 1% error, which is good enough considering the use of 1% resistors.

Answer 3

As the other guys suggested, adding an op-amp as a follower is quite a good idea, but you can do just as well by adding a small capacitor, about 0.1uF.

The ADC has a small internal sample capacitor which is charged up to the input voltage during the sample period.

If you have 2-channel an oscilloscope, it is well worth carrying out the following experiment. Put one channel on the analog input of the ADC (AC coupling, reasonably high gain), and put the other channel on the CLK input to the ADC (DC coupling, 2v/div).

What you may see is that at the sample period, the ADC input droops slightly as the internal capacitor is charged. If the signal hasn't come back up to its normal level by the end of the sample period, then you are not going to get an accurate reading, and will likely also get a noisier reading.

The reason it droops is that you don't have enough current from those high value resistors to charge the internal capacitor during the short sampler period.

You may also see a problem if you are using an amplifier with high gain as the ADC input (for example in strain gauge applications). The sudden appearance of the cap can cause an instability in the OpAmp, which will appear as a little blip:

You have two options, firstly you can lengthen the sample period, as I have done in the image above. This is actually very easy to do, because the sample period can be placed between two consecutive SPI bytes, so the pause comes naturally.

#define SPI_XCEIVE(out, in) {SSPBUF = out; while(!(SSPSTAT&0x01)); in = SSPBUF;}

typedef union
{
    int16u  word;
    int8u   byte[2];
}union16;

union16    mcp_result_union = {0};

void mcp320x_read(int8u chan) __wparam
{
    overlay int8u control_byte_0;

    control_byte_0 = B8(00011000) | chan;                   // Set up control byte:
                                                            // b4 = start bit
                                                            // b3 = single ended
                                                            // b2:0 = channel number

    SPI_ChipSelect(MCP320x_CS_PORT);                    // BEGIN


    SPI_XCEIVE(control_byte_0, control_byte_0);             // Send control byte. Ignore the return value.

    SPI_XCEIVE(0, mcp_result_union.byte[1]);                // Get the sample back. Put it straight into its destination.
    SPI_XCEIVE(0, mcp_result_union.byte[0]);                // Get the sample back. Put it straight into its destination.

    mcp_result_union.word    >>= 1;                         // Shift the sample into the correct position
    mcp_result_union.word    >>= 1;                         // (doing it in two stages is actually faster)
    mcp_result_union.byte[1]  &= 0x0F;                      // and mask off the unused bits.

    SPI_ChipDeselect(MCP320x_CS_PORT);                  // END

}

Alternatively, you can simply add a small capacitor between the ADC input pin and ground. This will act as as local charge supply which will prevent any noticeable signal droop during the sample. This way, you should be able to have pretty high value resistors.

If you want to save even more power, then think about the ADC sample rate. If it's very slow (say once a second), then you might consider powering down the whole ADC during this period. Make sure to check the datasheet for the required power on timing.