## To calculate the current regulation of DRV8825

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leeana
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### To calculate the current regulation of DRV8825

I am driving a bipolar hybrid stepper motor rated 1A with a DRV8825 at 12 V.
About current regulation the datasheet says:
9.2.2.2 Current Regulation
In a stepper motor, the set full-scale current (IFS) is the maximum current driven through either winding. This quantity depends on the xVREF analog voltage and the sense resistor value (RSENSE). During stepping, IFS defines the current chopping threshold (ITRIP) for the maximum current step. The gain of DRV8825 is set for 5 V/V. To achieve IFS = 1.25 A with RSENSE of 0.2 Ω, xVREF should be 1.25 V.

In my circuit, RSENSE is 0.1Ω. Following, in order to get IFS of 1.0 A, xVREF should be 0.5 V.

However, when I set the reference voltage to 0.5 V, I can measure only 0.5 A in both windings in full-step mode. Through experimentation I found that I get 1.0 A with a reference voltage of 1 V - which should give 2 A following the equation above.

On page 4, the datasheet says the VREF input voltage should be 1 - 3.5 V with the following note:
(2) Operational at VREF between 0 to 1 V, but accuracy is degraded.
Can it really be so inaccurate just slightly below 1 V, or am I getting something wrong here?

Benj
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### Re: To calculate the current regulation of DRV8825

Hello Leeana,

Your maths looks right to me.

I wonder if by current they mean no load (short circuit) current though this would be odd.

Or maybe as you say the accuracy starts to improve at Vref voltages greater than 1V?

Maybe there is a minimum value the RSense resistor can be though this would also be strange as a lot of applications would have a potentiometer to vary the output current?

Looking at the datasheet the max current for the ISense pin is 3uA, at 3uA the voltage on the pin would be 0.000003A * 0.1Ohm = 300nV