Electrical – Calculate Current Regulation – DRV8825

current-limitingdriverstepper motor

I am driving a bipolar hybrid stepper motor rated 1A with a DRV8825 at 12 V.

About current regulation the datasheet says: 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.

Current Regulation

To achieve IFS = 1.25 A with RSENSE of 0.2 Ω, xVREF should be 1.25 V.

In my circuit, RSENSE is 0.1 Ω. 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?

Best Answer

The answer is yes, accuracy can be degraded. The test engineers at TI left you that note to let you know that if you operate the VREF below 1V then the circuit will not preform like the equation that they provided. This is probably because they didn't put a super accurate opamp\comparator in silicon to measure the current (in figure 6) and it has some kind of common mode range or offset.