Discuss About Stepper Motors in the UK Electrical Forum area at ElectriciansForums.net

The problem is the very high power over the driving Transistor. That's the key for everything to work.
I did the Heavy math there. So you can do the easy stuff and make it work. Hahahaha.
----
Someone from the other forum, but also form UK, because I asked, told me this:
Notice that they can never get more than +5V, because they are being driven by the 4017 at 5V, regardless of how high your motor supply is. So their emitters only go to ~4.3V, and all the rest of the voltage is dropped across the transistor, causing them to heat up.
And my answer was this:
This is the theory I dont know at all ! This is the first time Im reading it ! Formidable.
So... a BJT transistor is voltage dependent? I never knew. I was having the impression that is behaving like a mechanical switch, also in the sense of letting whatever current and accepting whatever voltage on its C-E to drive higher voltage components/loads. Also in any Transistor datasheet is specified a maximum voltage over C-E, I was sure I got it right. Very curious !
For example:
1666612353502.png

Or maybe you know this thing in a context I dont know.
Try to understand me here- I really dont know what your logic is and Im trying to get it.
It has sense for the case you described. But also is weird from the datasheet point of view. I cant say I get it.
----
I want some light over this information here. Like I said, I believe what it said, but Im confused.
Try to explain.
Thank you.
 
The problem is the very high power over the driving Transistor. That's the key for everything to work.
I did the Heavy math there. So you can do the easy stuff and make it work. Hahahaha.
----
Someone from the other forum, but also form UK, because I asked, told me this:

And my answer was this:
This is the theory I dont know at all ! This is the first time Im reading it ! Formidable.
So... a BJT transistor is voltage dependent? I never knew. I was having the impression that is behaving like a mechanical switch, also in the sense of letting whatever current and accepting whatever voltage on its C-E to drive higher voltage components/loads. Also in any Transistor datasheet is specified a maximum voltage over C-E, I was sure I got it right. Very curious !
For example:
View attachment 102981
Or maybe you know this thing in a context I dont know.
Try to understand me here- I really dont know what your logic is and Im trying to get it.
It has sense for the case you described. But also is weird from the datasheet point of view. I cant say I get it.
----
I want some light over this information here. Like I said, I believe what it said, but Im confused.
Try to explain.
Thank you.
You know that a diode has a peak inverse voltage rating. Well a transistor, npn or pnp also has diode like junctions to form the collector, base and emitter on semiconductor substrate. Being diode like these junctions have ratings for reverse/inverse voltage to prevent so called avalanche breakdown and damage. Two reverse voltage ratings are specified Vcbo which refers to the cb junction with emitter unconnected ( Ie = 0) and Vceo which refers to the ce junction with the base unconnected (Ib = 0). In the forward biased direction the usual voltages associated with the space charge region apply between transistor junctions for base emitter and base collector which are 0.7V for silicon and 0.2V for germanium.
Have a look at these links:

Transistor terminal voltages - https://www.physics-and-radio-electronics.com/electronic-devices-and-circuits/transistors/bipolarjunctiontransistor/transistorterminalvoltages.html

What is Avalanche Breakdown? | Electrical4U - https://www.electrical4u.com/avalanche-breakdown/
 
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I know all that basics. Your explanation didnt help. And is ok, dont worry.
Lets concentrate over the bigger problem.
- How to make a more powerful transistorized driver for a 1.7A motor coil? The answer might be using mosfets. Now is a question of how.
 
So you want me to design a driver for the transistors I select from those you have to pass 1.7A in wave mode to the 2 phase bipolar stepper 17HS4401 using the switching waveform produced by your circuit (CD4017)?
 
yes
Design from scratch or copy a circuit that is already made and tested in time.
Hopefully one that is not overheating anything, not the motors, not the Transistors or anything else. To run cool.
 
Good morning. I may be a few days starting because I am overdue to deliver on a promise to my wife to install an outside light to cover the far end of the garden so that we can see our Jack Russell when she goes out at night. I am accompanying my wife on an outpatient appointment and this weekend we are away. Thus, next week I will begin.

I have in mind using optocouplers like the ones I mentioned earlier to make the transition between the 5V logic waveforms and the high side transistors. MOSFETs and perhaps some changes to your waveform generator. I have found the NEMA motor and will buy one.
 
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I made this movie last night and is completely unrelated to what we are doing here. Im braging with it and show it off. You can actually see on the table this project we are working right now with the H-bridge and the motors, but I push them away to make room for the new one. Is good to watch a technical video while you are away with your troubles. Especially when you are waiting for something and starts to be boring. See how much I care for you?
 

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