Best NPN Transistor and Pull-Up Resistor Value for Velleman VM147 Open Collector Output

I have a Velleman VM147 - Up/Down Counter Panel Meter Module with Preset (see attachment). The User Manual is relatively easy to understand. The module works fine.
It has an open collector output that activates when the user preset value matches the count on the illuminated indicator at the front of the module.
If you want you can use the module's open collector output with a relay (coil max. 100mA), a protection diode (IN4148) and a 1K resistor, and here Velleman shows you how to do that.

BUT I WANT TO USE THE OPEN COLLECTOR OUTPUT WITH A LOGIC DEVICE. Here Velleman's Help Sheet (see other attachment) tells me I can do this using an NPN transistor and a pull-up resistor.

But what NPN transistor? And what value resistor?

Two very simple questions for an experts, no doubt. But this hobbyist would really appreciate the advice of somebody who knows for sure, to be sure to be sure.
Michelle O'Brien

So the answer is, it depends on what logic you are trying to drive.  Most logic devices have low to modest current draw, so the transistor can be a fairly small general purpose device like a BC549, 2N2222 or 2N3904.  As for the pull-up resistor, this tutorial gives a good explanation and worked example: https://www.electronics-tutorials.ws/logic/pull-up-resistor.html

Michelle as Martin says above, it depends what you are trying to drive.An Open Collector output implies that they are giving you the Collector of a transistor that is brought out to a pin.  It's not connected to anything else, hence the Open.The nice thing about this is that you can use it for whatever you want,  you're talking about Logic, so I'll first assume CMOS, the inputs of which do not load a source much.However you can't just leave inputs floating (unconnected) with CMOS.  You need to drive them either high or low.When your open collector transistor is switched off (most of the time) it acts like an open circuit to ground, and when it is on (when the counts match) is will act like a short-circuit to ground.  Since you're talking about logic, I've assumed a 5-volt supply (though there is no reason you could not use +12 from the counter supply, especially with CMOS).  You need to put a Pull-Up Resistor from your supply to the open collector of the transistor, as shown below (resistor R).  It's called a pull-up resistor as it pulls the voltage on the collector up to the supply voltage if the transistor is open-circuit (not conducting).  For Cmos, 10K would be fine for R.
The output at pin 4 will then be at 5V (if you don't load it much) normally and will go down to almost 0V when the counts match and the transistor is switched on. At standard logic levels that's a 1 normally and a 0 when the counts match.  If that does not suit you, make that first logic gate an inverter, or a NAND or NOR with the inputs tied together.If you are using TTL logic (74 or 74LS) you can probably dispense with R and connect the input straight to pin 4, as TTL has built-in pull-ups. If you're using 74HC/HCT you'll still need the pull-up R.As you've probably seen in the Velleman manual here, you can drive a relay directly (or a LED or a light bulb) so that implies a fairly beefy transistor on the counter, and they say max 100mA, but be wary of going too near that, you don't want to burn out that transistor on the counter (it looks like a through-hole type but would probably still be fiddly to replace).  I'd probably use the Pin 4 output to drive another transistor (NPN or PNP depending on what you want to do) to minimise the load on your Pin 4 transistor in the counter if you want anything more than about 50mA.As always if I've been telling you stuff you know already, apologies, it is very difficult to know what level to assume, so I usually start low.  And if you need any more advise then just yell out. Cheers // David

"Michelle as Martin says above, it depends what you are trying to drive."
Hi GentlemenI want to use it to turn a stepper motor on and off via a BigEasyDriver.
The UP/DOWN count on the VM147 is driven by the pulses coming from the Stepper/BigEasyDriver combination.
So, the UP/DOWN count hits the preset number that the user has set up on the VM147
... which "fires" the open output collector
... which turns off Enable on the BigEasyDriver
... which brings the stepper to a halt.

That's the general idea, for now at least.

 So your cue now, Gentlemen, to maybe tell this hobbyist she's "doing in all wrong"?

Michelle

Michelle,If what you want is to drive the enable input of the BigEasyDriver, then the circuit David showed you will work. The only question is if you need to invert the signal to have it be the correct polarity to drive the enable.You could use a logic gate to invert the signal as David suggested or a small transistor like one of the ones Martin suggested. You will need a pull-up resistor on both sides of the transistor if you do that.
Elizabeth

Thank you, Elizabeth. That sounds as though I might be doing something RIGHT - which makes a welcome change for me in my explorations in electronics.
I'm thinking of using one of the six inverting buffers and TTL drivers on a  4049 chip to feed into Enable on the BigEasyDriver so that when the open collector output on the VM147 goes HIGH the feed into Enable  on the BigEasyDriver goes LOW. Would that be right?
Michelle

David wrote: "As you've probably seen in the Velleman manual here, you can drive a relay directly (or a LED or a light bulb) so that implies a fairly beefy transistor on the counter ..."
David, I can see three transistors in all on the VM147 (which comes from Velleman as a pre-assembled module with no helpful schematic included to indicate to the amateur keen  to learn what component is what, where it goes on the board, and how it's linked to the other components).
The three transistors I can see are :1  x  BC332-252  x  BC557BI'm guessing that the one "in charge" of the open collector output side of things is the BC332-25. But as I say, that's only my guess.

Michelle

Michelle,Yes, I think you've got it right.Elizabeth

Hi Michelle...Sorry, was meaning to reply before now but been busy...Since the BC557 is a PNP the open collector output must as you say be the BC332.  A BC447 would also work if you do ever need to replace it.I got the datasheet on the Big Easy Driver here, and  I learned:ENABLE: Logic Input. Enables the FET functionality within the motor driver. If set to HIGH, the FETs will be disabled, and the IC will not drive the motor. If set to LOW, all FETs will be enabled, allowing motor control.So your counter board pulls the output LOW when your preset count is reached, but the Big Easy board needs a HIGH to stop it driving the motor.  You're right that you could use a 4049 gate to invert the counter LOW, or you could use another transistor as suggested elsewhere - here's chapter and verse:

The transistor used is the same type, BC447/448 or BC332.  Any NPN should work.V1 can be either 12V from the counter or whatever LOGIC supply voltage the Big easy is using (NOT the motor supply!)  V2 MUST be the Big Easy logic voltage.R1 and R2 can be anything around 10K - 47K, since the transistors are either on or off it shouldn't matter much, and I doubt the Big Easy board needs much drive.How it works... when the counter is not on the preset count the open collector pin 4 is just that - open - so current will flow through R1 into the base of our interface transistor, switching it ON.  This will bring the enable input of the Big Easy board close to ground (low) so it will drive the motor.When the counter reaches the preset count it switches the open collector transistor ON, shorting the base drive to our interface transistor, which will switch off.  R2 is now a pull-up resistor which pulls the Enable input of the Big Easy to a HIGH level, hence stopping it driving the motor.

So reference the circuit that you've drawn above (the one with the addition of the NPN transistor), and given that the voltage at V1 is 5V and that the voltage at V2 is also 5V  (5V is the logic level for both VM147 and BigEasyMotorDriver):
 What value for Resistor R1 would you recommend and what value for Resistor R2  would you recommend? Both the same or different?
Please forgive the dumb questions but it's not often I get an expert on the line.
Michelle

Hi MichelleSorry, being in Australia means that you have to wait a day for replies....You could use the Big Easy 5v line for both V1 and V2., or the VM147 5V line for V1 and the Big Easy 5V line for V2, it doesn't really matter.And since we are only interested in logic levels here, and the transistors are going to be either fully on or fully off, you could use pretty much any resistor from 1k to 100K for R1 and R2 and it would probably work.  Maybe take the middle ground and go 10K.  They will draw 1/2 mA approx with  a 5V supply which is well within the transistor capabilities.  I could not find the input requirements of the Big Easy but you can be sure it would not be much, there is a reference to the chip used somewhere and we can look that up if you do get problems.You could use a higher value for R1 if you want, since it is only driving the base of our interface transistor which has a gain of at least 100, so it won't need much base current.  But really I don't think it will make much if any difference, whatever you use.> Please forgive the dumb questions but it's not often I get an expert on the line.As someone once said, the only really dumb question is the one that isn't asked.  And flattery will get you everywhere   I asked plenty of dumb questions on my way to this exalted "expert" status, so you don't have to apologise, it's a pleasure to help.And I'll second Elizabeth, you are on the right track, these boards will do exactly what you want, from your earlier questions. (you could even make an egg timer

The nice thing about this is that you can use it for whatever you want, 
you're talking about Logic, so I'll first assume CMOS, the inputs of
which do not load a source much.