Open Collector Logic Chips

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Open Collector Logic Chips - Simply Put

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Simply Put

5 жыл бұрын

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Пікірлер: 24

@seyhuncelik9584 Жыл бұрын

Great explanation. Darn KZfaq never recommends the best KZfaqrs out there.

@ajaykamath1853 Ай бұрын

Thanks mate for this simple and easy to understand explanation.

@aleshkovalev 4 ай бұрын

Thanks mate. Best explanation i found in a hour or so. Now i got it.

@GLuft3 6 ай бұрын

Thank you for explaining the WHY.

@banban8481 2 жыл бұрын

Wow, just wow, very clear and good explanation. Thank you very much!!

@carrocesta 10 ай бұрын

a master piece session! thanks dude

@fredkester6130 2 жыл бұрын

Thank you very much. This way I understood that I can connect to a Arduino output a transistorbase and then a counter device to the collector . Arduino, emitter and counter together to ground. The counter I put in NPN mode. This means the channel of the counter is internally via a pullresistor connected to it's plus voltage (18 V). When the arduino output is low, the channel has high voltage. When arduino is high , the transistor conducts and the channel reads 0 volt. It works/counts

@006hetzkin5 5 жыл бұрын

Thank you, that was an awesome tutorial!

@birdybird9819 2 жыл бұрын

Thank you for this explanation. It is what I was looking for.

@TheHereticAnthem20 5 жыл бұрын

This video saved my life. I can't thank you enough!

@florian65 4 жыл бұрын

Thanks helps a lot. I like your Headphone :D

@user-xc8vw5oq8v 6 ай бұрын

Tnx so much.

@TOMTOM-nh3nl 11 ай бұрын

Thank You

@giusepperinaldi9411 4 жыл бұрын

You are very clear... . This is a great video! I would like to use an Arduino's open collector output with a pullup resistor, coupled with a capacitive sensor. I am using a capSense library that takes two pins. On the first one sends out pulses and on the other reads it back. The rx pin is also connected to a metal plate and works as a touch sensor. I would like to have the following result: - switch open: 5v on the input pin; - switch closed: 1)same pulse of the output pin (the input would match the output); or 2)0v on the input pin; I cannot figure out how to make it. Your help would be very appreciated.

@giladsivan3735 2 жыл бұрын

Great Video! THANK YOU

@hardeepsingh-le7xh 2 жыл бұрын

very nice information but it would be good if we get in written form so we can read until we understand ..i mean we dont move to next word until we understand previous words...good information..

@ZENERVOLTAGE Жыл бұрын

This helped me a lot thank you. However, I'm using a 74LS09 (quad AND gate IC, each with 2 inputs)...and all 4 outputs are open collector. I want to power an LED as the load. The datasheet says that the max LOW output current is 8mA. From your explanation and what I've read online, when the transistor is "OFF", current flows through the resistor and also the load. Therefore for an LED, the pull-up resistor has to satisfy the current limit for an LED and also the max output LOW current limit too. For a 5V supply, I chose a 1K resistor, which gives 5mA to the LED when transistor is off...and therefore 5mA through the transistor when it's ON and conducting. My question is; Is there a way to have a separate pull up resistor, so that I can supply my LED load with a higher current...say 20mA??? Or are the load and pull-up resistor always linked in the above manner? Many thanks

@aviandragon1390 Жыл бұрын

Generally the "sink" and "source" current are different values when you're talking about logic devices. If your device is not capable of either sinking or sourcing enough current to illuminate an LED then you will need to use either a buffer or inverter IC (your simplest options) or include a transistor (BJT or FET) in your design to drive the load instead. Don't forget to add appropriate resistors in either case.

@johnortiz9789 5 жыл бұрын

Your content is great! Thanks alot

@anyphonline4005 4 жыл бұрын

Can i replace this "open collector/drain with pull up resistor" as a protection diode for relay? I have couole of relays to be switched one by one and need protection diode for each to avoid the excessive reverse emf once turned off. Seems this one is much better. Thanks.

@simplyput2796 4 жыл бұрын

Transistors are not diodes (though sometimes one of the base junctions is used as one due to switching characteristics), and trying to reverse-volt the collector (which is what the inductive emf would do) would just drive the BJT backwards (and if it's a logic chip, would fry the chip). For shunting the emf of an inductive load, use a diode. There are plenty to choose from, based on the magnitude of back voltage to deal with, total power dissipation, and switching frequency.

@anyphonline4005 4 жыл бұрын

So still need to put the diode even the high induced current after turned off will just flow directly to the ground? BTW: the negative side of my coil is directly connected to an " TTL level open drain with pull up resistor" abd which i think it uses the bus that you mentioned in the video. And as you said, it doesn't care how high the voltage it was but will just check that its working properly. So my understanding is that the high current will just sink in to the ground and the pullup resistor will keep the voltage up to 5V (my supply, anyways my pull up resistor is 4.7kohm). So im thinking that putting diode to shunt across the coil is useless since we have that open drain to sink the high current to the GND. Please check if my understanding is correct. Thanks for your help

@TheNat2341 4 жыл бұрын

where would my led out bulb go?

@simplyput2796 4 жыл бұрын

A logic chip isn't designed to carry the current an LED takes, so you're better off buffering the chip output and using that to drive the LED, unless your spec sheet says the chip can do it.