LabVIEW
What's the difference between active drive and open collector?
Hi, I am tring to figure out what is the difference between the two digital output driver types, i.e. Active drive(push and pull) and open collector(open drain).
I found two articles about this topic.
Article #1:
http://digital.ni.com/public.nsf/allkb/D9B4018C3E540CE4862570F30068C0DB
Article #2:
http://digital.ni.com/public.nsf/allkb/0C5091E9099059BC86256FC1007947AA
I am confused by something in these articles.
- In the first article, the push-pull circuit is what shown below. And It claims that when Vin is low, the bottom BJT is on and top BJT is off. Why? I think Vin and VE are all 0V, so both of the BJTs are off.
2. In the article #2, the push-pull is different with what’s shown in article #1. And it claims “A push-pull output is a combination of a line driver and an open collector. In the off state it will supply a path to ground and in the on state it will supplyVcc. “
I think this claim is reasonable. So can I say the push-pull circuit in #1 is wrong?
3. In the article#2, it claims
It seems the difference between open collector and push and pull is one is float and the other is not. Here I am not quite clear with the concept “ float” . The article #2 said for the open collector circuit, when in the off state, an open collector will float. But I think in the on state, push and pull will also float. I see no difference between the these two. Could anyone give me some inspriation?
Thanks.
"The lower transistor is a PNP. It will turn off when a positive voltage is on the base, and turn on when a negative value is on the base. It will turn on in this case because the Vin is 0."
Why in this case it will turn on? You said when a negative value is on the base, it will turn on. But here the base is 0V not negative, and the emitter is also 0V, so there is no voltage difference between emitter and base, why it will turn on?
"Float means the voltage value is somewhere in between, probably at a level based on relative charges. If neither transistor is on in either case, the load point is neither pulled high, or pulled low. The output is in an undefined state in between."
I think you are saying for the push and pull, the output is float both transistor are off. But the artcle says for the push and pull, it will never float, as it says"As mentioned above, the Push-Pull configuration does not float, it is always supplying Vcc or ground, which is why it will work best with a Multifunction DAQ card."
And for the open collector(below), there is only one transistor, the article says it will float when the transistor is on, why? I think when the transistor is on, the output is connected to 5V through a resistor(or directly connect to 5V), and if the load is large enough, the voltage on the load will be close to 5V. Why the output is float?
@cli21 wrote:"The lower transistor is a PNP. It will turn off when a positive voltage is on the base, and turn on when a negative value is on the base. It will turn on in this case because the Vin is 0."
Why in this case it will turn on? You said when a negative value is on the base, it will turn on. But here the base is 0V not negative, and the emitter is also 0V, so there is no voltage difference between emitter and base, why it will turn on?
You'll have to look up the physics of transmitters if you want a more precise answer. But it does not have to be actually negative. It just has to be below whatever is the turn on/turn off threshold. Remember, the vast majority of digital circuits are either a positive voltage relative to ground, or at ground. If a PNP transistor actually had to go below the voltage of the emitter (negative relative to ground in this diagram) they would never be practical as a semiconductor device.
"Float means the voltage value is somewhere in between, probably at a level based on relative charges. If neither transistor is on in either case, the load point is neither pulled high, or pulled low. The output is in an undefined state in between."
I think you are saying for the push and pull, the output is float both transistor are off. But the artcle says for the push and pull, it will never float, as it says"As mentioned above, the Push-Pull configuration does not float, it is always supplying Vcc or ground, which is why it will work best with a Multifunction DAQ card."
The article is a little confusing. Both images you attached are both listed as Push-Pull. In the upper image, if Vin winds up as some voltage in between, depending on the threshold voltages for each transistor, then neither transistor will turn on and Ve winds up floating to wherever. In the lower image, there are two inputs, one being the complement of the other. Much like they are in differential pairs of encoders. So if one is "high", the other is "low". and vice versa. There is no "in-between" voltage and thus one of the transistor has to be actively conducting.
And for the open collector(below), there is only one transistor, the article says it will float when the transistor is on, why? I think when the transistor is on, the output is connected to 5V through a resistor(or directly connect to 5V), and if the load is large enough, the voltage on the load will be close to 5V. Why the output is float?
Now you are adding some more images. I'm not sure how accurate you are with the text you are quoting.. When I read the 2nd link http://digital.ni.com/public.nsf/allkb/0C5091E9099059BC86256FC1007947AA it says the open collector, the output will float when the transistor is OFF.
In the left image, there is no 5V on the output line for it to move to. It should be connected to ground when the Input line is high. When the input line is not high, then the output line is disconnected from the ground, and it could float to anywhere. Where in the right and image (which is missing the word output, but I'm assuming it is at the collector of the transistor) it will be pulled up by way of the connection of the external pull-up resistor.
