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Topic: 2.3" 16 segment common anode LEDs, HT16K33 and ULN2803 (Read 1 time) previous topic - next topic

MitchSF

From what I've learned here about this circuit, and that happens to be a lot, it looks like this can work.

The only question I have, is whether R3 is necessary, and if so, are the values of R3 and R8 appropriate?

If it does work, I'll probably recalculate the R values for a 9v supply, and use a smaller MOSFET. If no glaring errors are pointed out, I'll order a few parts and will breadboard next week.

Thanks again.

EDIT: With VCC at 7v maximum, I'm not sure that the 74HC07 will turn on the MOSFET. I was thinking that it could be connected to 12v. So maybe a 2N3904 as a switch switch driven by an inverting 74HC04 will work.


Can it work?

ROW pins pull high when active and high impedance when passive, the ULN2803 need input current to turn on, this looks acceptable.

COM pins pull low when active and is high impedance when passive, there is a non-inverting driver with 5V CMOS input and output. A floating input on a CMOS will often be staying in the previous state, i.e. low. The transistor input will be either 0V or 5V, but the transistor is supplied from 12V, this means it would be on all the time if it was a P-Channel transistor, but symbol shows N-Channel transistor wired to always pass current.

This is not going to work.

The problem is the two different supply voltages. A real mos driver chip could do it or a transistor more, but a transistor includes an inversion that you do not need, this has to be fixed with a transistor more or a inverter and you need a pullup on the COM outputs.

MitchSF

Thank you. Are these the changes you are describing?



Thank you. Are these the changes you are describing?

Exactly, you forgot to put 12V on the circuit.
It is a good schematic praxis to feed positive voltage from the top, i.e. turn you mosfet around and point S up.

And one more item, when using a BJT transistor you need a base resistor, you could use a small N-channel mosfet if you want to avoid that.

MitchSF

Changes made. Thanks again.

Is the NTR1P02T1G a good choice to substitute for the NDP6020P? That one seems like overkill in this circuit.



That improved the schematic, but move the gate resistor so it goes up from the gate to 12V, that way it is much faster to see it is a pullup.

What was the current? 0.6 to 0.7A? I would prefer a transistor with a bit more headroom than NTR1P02T1G, i.e. 2+A rating, but it is not really necessary.

MitchSF

Thanks again.

With all 16 segments lit, maximum current is 400mA. Hopefully this will be the final drawing.


I cannot see any problem with it now. I hope you agree with me that changing the transistor and resistor position makes it easier to read the schematic.

If you plan on making a PCB I would recommend making a full schematic and use a program that can coordinate PCB layout with schematic, this reduces the risk of errors.

MitchSF

Very good. Thanks again for the help.

I use DipTrace to design pc boards. A full schematic will be drawn. I'll post the results after the breadboard is completed with one or two displays.

PaulRB

is whether R3 is necessary, and if so, are the values of R3 and R8 appropriate?
R3 not neccessary. The inputs of 74hc07 are very high impedance, only a tiny current will flow.
R8? Did you mean R6? If so, yes, because without it, the 74hc07 output would be damaged by short circuit when it goes low.

EDIT: With VCC at 7v maximum, I'm not sure that the 74HC07 will turn on the MOSFET. I was thinking that it could be connected to 12v.
I said that 74hc07 has open-collector output, but did not explain what the benefit of that would be. The 74hc07 chip itself would be powered with 5V. Because it's outputs are open-collector, it won't be damaged if those outputs are exposed to 12V, unlike other chips (such as 74hc04 or ht16k33 or Arduino pins). Open collector outputs can only pull a signal low, they cannot source current to pull a signal high. That is why R6 is there. So to answer your question, the 74hc07 does not switch the MOSFET off. R6 does that. But the 74hc07 can turn the MOSFET on by pulling its gate low, overcoming R6 pulling it high.


I said that 74hc07 has open-collector output, but did not explain what the benefit of that would be. The 74hc07 chip itself would be powered with 5V. Because it's outputs are open-collector, it won't be damaged if those outputs are exposed to 12V

I do not agree with your, here is a schematic from the hc07 datasheet:



The diode sort of spoils your explanation.

PaulRB

ROW pins pull high when active and high impedance when passive...

COM pins pull low when active and is high impedance when passive...
Can you please tell us where this is explained in the data sheet?

These are the diagrams from the data sheet that show how the outputs are wired internally:

Looks like COM 0-3 are different from COM 4-7.

Exactly.


The two MOSFETs with a ring around is used while updated display, the other MOSFETs and resistors are for keyboard scanning.

PaulRB

I do not agree with your, here is a schematic from the hc07 datasheet:
The diode sort of spoils your explanation.
Yes, I see what you mean, I am also surprised by that. Will have to do some more research!

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