Does Stepper driver IC have clamp diodes?

[eem2am]

Hello

Do you think that the NJM2673 bipolar, switch-mode stepper driver IC has clamp diodes on its signal input pins?
(pins: I01, I11, I02, I12, PHASE1, PHASE2)





NJM2673 DATASHEET (njr.com):-
http://www.2shared.com/document/JtaCgPvr/NJM2673.html

Or find datasheet here:
http://semicon.njr.co.jp/njr/hp/pro...ductId=1690&_moveKbn=PRODUCT_DETAIL_MOVE_SPEC






(i.e. protection diodes with anode to pin, and cathode to VCC?)

..by signal pins, I mean that the current magnitude and direction inputs (I01, I11, I02, I12, PHASE1, PHASE2)



Its just that we I have connected the microcontroller outputs directly to the signal pins of NJM2673 with no series resistors.

-Also, there is a switch so that we can switch off the VCC supply to NJM2673 when we are not driving the stepper motors (this saves some power)

-in this state, VCC = 0V, and for a short time, the microcontroller will be putting logic highs into the current magnitude pins (I01, I11, I02, I12)………

……..if such clamp diodes exist, then the microcontroller pins will end up basically driving into short circuits.

 

[(*steve*)]

Look at the absolute maximum ratings and you'll see that the inputs are limited to -0.3V to Vcc.

If you turn off the power and Vcc falls to zero, you will probably sink current into the Vcc via the inputs. This may power the device, or it may cause it to latch up. Either way, not what you have in mind.

It probably won't look like a short to earth unless you connect Vcc to ground as part of your power off sequence (a fairly unusual thing to do)

 

[eem2am]

Hi , thanks,

the thing is, in order to save some power, we actually switch off the VCC supply to this chip ...when we are not driving the steppers........

at these times, the microcontroller outputs on I01, I02 etc are high, and VCC= 0

 

[(*steve*)]

Can you not change the program so that the outputs are low at this time?

 

[eem2am]

...unfortunately the high level input signal (I01 = 5V & I11 = 5V) turns off the current....so this would be the state just before switching off the 5V.....also, the software guy may make a mistake while practising and take the pin high anyway, so i want to protect against this.

 

[(*steve*)]

The best solution may be to put 2 resistors in the lead between the output of the uC and the input to the stepper driver. Between these 2 resistors, connect a schottky diode's anode, with the cathode connected to Vcc.

Select resistors low enough in value that when Vcc is +ve, enough current gets to the driver input, but high enough that when Vcc is 0 only a small current flows (neither enough to damage the uC OR to power the chip itself.

Another solution is to use a device such as a 4066 with one of these devices (4 in a package) between the uC and the driver. Power this chip from the uC power, but use Vcc as the control voltage. Thus will effectively disconnect the 2 devices when the power is removed from the driver chip. The current consumption is VERY low.

The last (and simplest) option is to place a single large value resistor between the uC and the driver. Pretty much the same caveats as for idea #1, with the side effect that it may make the device more prone to latch-up (if it is indeed prone to this)

OK, one more option. See here for a method of translating between 2 voltage levels. Actually, that shows a mosfet. This discusses what I had in mind, and there is a link to the application note in there as well.

I can't tell you which design is best for you. If you are redesigning the board you might go for the 4066. If you're forced to cut tracks and add components, a simple resistor may be your best option.