L298 data sheet

Question

In the data sheet of L298 its written that "Turn-On and Turn-Off : Before to Turn-ON the Supply Voltage and before to Turn it OFF, the Enable input must be driven to the Low state" I am trying to design a circuit with logic gates that will fulfil the requirement (not to leav it to the user). But I keep ending up with circular probems.

Any Ideas?? Thanks!!

Answer 1

I am in a bit of a quandary here.
I always tell everybody to follow the recommendations of the manufacturer.

At the same time I know that the L298 has been used on many circuit boards, sold in the ten-thousands for Arduino, Raspberry-Pi etc. AFAIK none of those board follow that recommendation and they all seem to survive again and again.

R1 and D1 remove the enable as soon as Vs goes away. (Power off).
D3 and C1 hold the power a bit longer. The value of C1 depends on how much load there is. 100uF might be sufficient. R2 and R3 pass the enable only if the power is above a certain level (The exact voltage depends on the type of AND gate you use, HC, HCT, LS).

^{simulate this circuit – Schematic created using CircuitLab}

Now I am going to say something unprofessional: if it would be my circuit, I would not bother.

Answer 2

I had a look at the datasheet. I presume the risk is that of "shoot-through" the bridge transistors on one side of the motor or the other during power-up. Pulling the enable pin low would fix that.

On power-up it would be simple to add an RC circuit so that the enable can't go high for a half-second or so after power-up. The problem is that it would keep the EN voltage high while powering-off.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Enable safety circuit.

The Electrical Characteristics table on page 3 quotes

• V_en = L at 1.5 V max
• V_en = H at 2.3 V min but upsets my plan by stating that the max is VSS.

If it were not for that last twist I would recommend the R1/R2 divider to hold the EN pin at adequate voltage once the supply is up and get EN below 1.5 V as the supply voltage drops.

Sorry I can't be more definite.