double BTS7960 H-bridge: how to obtain DCC phase reversal?

[earckens]

Given:
1. A BTS7960 half H-bridge mounted in a pair on a module (IBT-2) to form a full H-bridge.
2. a controller to detect overcurrent (Pro Mini, input on A0) to shut the bridge by setting output on DIO3.
3. a DCC signal that is fed straight into pin 1 (R_PWM) and inversed into pin 2 (L_PWM).
4. the Pro Mini controller will be programmed with an extra output pin that will be used to control either feed the signal as it is now into the IBT, or enables the inversing of this signal: inversed into pin 2 of the IBT ,and straight into pin 1 of the IBT.

Pinout on the IBT:
pin 1: RPWM
pin 2: LPWM
pin 3: R_EN
pin 4: L_EN
pin 5: R_IS (right current sense)
pin 6: L_IS
pin 7: Vcc
pin 8: GND

Question:
I want to get rid of this transistor and use instead a 74LS04 hex inverter and a 74LS86 quad XOR gate for this action. How do they need to be wired to allow this "reversing" or "straight" functionality, given just one output from the controller?

Background: this schematic needs to be adapted for 3 more independent IBT's, hence each of this logic IC's will be fully used.
I realise too that 3 more current sense inputs will need to go to the Pro Mini.

 

[rjenkinsgb]

Use a single LS04 inverter to give the true and complement signals from the source.

Then add a 74LS86 section in line with each PWM drive signal.

With the second input of the 86 low, the signals are straight through. With that input high, both are inverted, effectively swapping over the two.

 

[earckens]

Which IC do I place first?

 

[rjenkinsgb]

You could do it the other way around with one less gate...

Feed the main signal though a single LS86 and call the output the non-inverter drive, then add an LS04 section after that to give the inverted drive.

Taking the other input of the LS86 high would invert both.

 

[earckens]

I dont understand JRW: the XOR is needed to have an output either inverted or straight by toggling one of its inputs, while the other input gets the signal. The XOR output goes both to the IBT (R_PWM) and the inverter input.
The inverter is still needed to receive the XOR output for an inverted signal to the IBT (L_PWM).
How can this be done with one gate less?

 

[rjenkinsgb]

Like this:

Invert input low; "true" output follows the signal input & invert is the complement.

Invert input high; "true" output is inverse of the input & invert is still the complement of true.

(I found a neat gate simulator, here, and used it for the diagramsL
https://circuitverse.org/simulator )