NPN sensor to TTL input

This means that the enable signal has to be able to pull down to below .3 volts while pulling 100 uA from the enable pin to perform the enable function (it is active when it is a logic low), and must be able to pull high to at least 2.4 volts while supplying 100uA to the enable pin to disable it. Voltages in between .3 and 2.4 are not guaranteed to act either as an enable or a disable.

This signal can be provided from any 5 volt powered logic family, or from an active pull down device, like an NPN transistor, whose emitter is grounded, with a pull up resistor connected to a 5 volt supply. The absolute maximum voltage that can be safety applied to the enable pin is 7 volts.

thanks for the reply, can you give me a little bit more direction, as I am not a pro.

Not without learning a lot more about exactly what you are hooking together. For instance, I have no idea what "sunx cx-24 @ 5v supply" means. I gather that you are somehow connecting an L294 with a 20 volt supply and some sort of photo sensor and there might also be a 5 volt supply involved. If you email me a schematic of what you have, I will get back to you tonight.

email sent thanks a million....

--- The SUNX site doesn't seem to have a wiring diagram, right off the bat, for the CX24, but from the information there, if it has a 4 wire connection, it should be something like this: (View in Courier)

+V | +---- +-+----|-+ | C | | | | E | +-+----|-+ | +---- GND

Where +V is the supply voltage, GND is the supply ground, and C and E are the uncommitted collector and emitter of the output NPN transducer.

If it only has three wires out, though, it's probably like this:

+V | +---- +-+----|-+ | C | | | | E | +------|-+ | GND

Where GND is the common ground for the sensor supply and the emitter of the output transducer.

If the first case is true and you want the 20V supply to be isolated from the 5V supply, you should wire your circuit like this:

+5 +5 | | [1k2] +---+-... +20 | |_ | +---------O|E L294 +-+----|-+ | | C | +---+-... |CX24 | | | E | | +-+----|-+ | | | | +20GND +--------------+ | +5GND

Or, if you want the 20V supply and the 5V supply to share a common ground, like this:

+5 +5 | | [1k2] +---+-... +20 | |_ | +---------O|E L294 +-+----|-+ | | C | +---+-... |CX24 | | | E | | +-+----|-+ | | | | +----+--------------+ | | +20GND +5GND

If the CX24 only has three wires coming out, you'll need to hook it up like this :

+5 +5 | | [1k2] +---+-... +20 | |_ | +---------O|E L294 +-+----|-+ | | C | +---+-... |CX24 | | | E | | +------|-+ | | | +--------------+ | | +20GND +5GND

Note also that the input to the sensor needs to be _modulated_ IR

-- John Fields Professional Circuit Designer

thanks for the reply the cx-24 has 3 outputs +, 0, out. Although I am not sure, the 20v is to drive the solenoid. The 5v has a dual purpose, it powers the cx-24 as well as providing a reference voltage to the L294.

is the the reference with Fig. 7a as the test application. I believe that the cx-24 (connects to pin9) should be isolated from the circuit, as pin 9 is supposed to be TTL.

--- No. As shown on the data sheet for the CX-24, its supply voltage must be from 12 to 24VDC. Also, as shown on the data sheet for the L294, its supply voltage is +5V. In order to make the circuits operate properly you need to connect them like this:

+5 +5 +Vs | | | CX-24 [1K2] +----+-----+--+ +---------+ | | Vss Vs | +20>----|+ OUT|---+---O|E | +5 |CX24 | | GND Vi | | | 0 | +----+-----+--+ [R1] +----+----+ | | | | | +------+ +--------+------------------+ | | | | [R2] +5GND +20GND | | +------------+ +Vs is the voltage used to drive your solenoid (It could be the +20V if that's what you want to use), and Vi is the reference voltage input. The voltage developed at the junction of R1 and R2 will determine how much current is allowed to flow into the solenoid. Read the data sheet.

--- You don't understand. The output of the CX-24 is an uncommitted open collector, so you use that to pull down the enable input of the L294 when you want to energize the solenoid. When you don't want to energize the solenoid you turn off the output transistor and allow the voltage on the enable input to rise to +5V. Now, since you only have three connections to the CX-24, the emitter of the output transistor is common to the 20V supply ground, so if you use a 5V supply to run the L294 (and the enable), that power supply ground _must_ be tied to the 20V ground. Consequently, all the grounds are common.

-- John Fields Professional Circuit Designer

you guys make everything sound so easy, I actually have hooked up a different sensor, it does use 5v with NPN output, but I need to use the cx-24, which I have on hand. I'm going to try your circuit today. Let you know. The 5v on the L294 is to produce a voltage reference, which translates to the output current spike I need. There is no mention of a 5v supply for the L294 except for diagnostic

The diodes that they are referring to are regular diodes??

thanks again

--
Not true.  Take a look at page 2 of the data sheet and you\'ll see,
under "ABSOLUTE MAXIMUM RATINGS", that Vss, the logic supply voltage
can be run up to 7V.  

Then look at "ELECTRICAL CHARACTERISTICS" on page 3 and you\'ll see
that the test circuit is run with Vcc = 5V.

The reference voltage is something else again, and is used to set
the peak current into the solenoid.  Look at figure 2 of the data
sheet.

Yes I did note the fast switch diode, that is where I started to have trouble. I knew it wasn't zener, so I opted for a schottky, I suffer from limited resources on short notice.

I actually built the fig. 7a found in the app notes pages. There is no mention of the 5 volt. I have it set-up to try the led, but I didn't hook it up. That is funny because the L295 says it doesn't require externat 5v.

I will hook it into the test tonight.

Sorry for the delay in my response, but I was called away, yesterday.

Thanks for the link to the data sheet and application note:

I can find no where on the data sheet or application note that specifically says that Vss (logic supply) is optional, but application note 7a implies that it is. I think you need to supply Vss with 5 volts if you want the diagnostic output to function, and that output isn't used in application note circuit, 7a.

As long as the sensor supplies only a pull down current path, then it should work with this circuit, with the possible exception that an external pull up resistor might be needed to get a full 5 volt swing on the RC and transistor network going into pin 7. A 10 k resistor between the top of the 5.1 volt zener and pin 9 should do it.

That said, I think the enable pin is not responsible for your L294 failures. I suspect a slow output diode, D2 or some other error in your circuit.

Got the enable ( pin 9 ) volts working fine, cost me a few chips along the way. Down to 3 chips left before I need to order from digi-key. Now everything seems to ok, the diodes that I have are schottky and standard. Would the schottky diodes work, if not please let me know what type I should look for

Schottkys should work fine, as long as they have the required voltage and current rating. One other possible problem occurs if the supply leads are long. Then the wiring inductance can generate large voltage spikes during switching. The ap note does not show any, but there should be a bypass capacitor between the point where D2 ties to pin 1 and the point where Rs ties to D2. Something on the order of 470 uF

35 volt electrolytic would probably work, if it is a low inductance type, like Panasonic type FC or FM.