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  • pechkata
    replied
    Originally posted by Detectorist#1 View Post
    Hi pechkata,

    The right picture is wrong! Read carefully my post #108 - " If gating pulses from uP are negative( +5V -no gating pulse and 0V - pulse) - emitters have to be connected to +5V and 10K resistors have to be connected from collectors to minus 5V supply. The control signals have to be taken from the collectors." In this case, the bases of the transistors have to be connected via 10K to 47K resistors to the control signals of uP. Voltage converters will be inverting in this case.
    Sorry, Vlado, but I didn't understand anything from this google translator...
    If he had drawn it...​

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  • Carl-NC
    replied
    Here are the two ways -- common-emitter & common-base -- side-by-side. You can see they are exactly the same circuit and work exactly the same way but the CE version inverts and the CB version does not. Eduardo, use the CB version.

    Click image for larger version

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  • Detectorist#1
    replied
    Hi pechkata,

    The right picture is wrong! Read carefully my post #108 - " If gating pulses from uP are negative( +5V -no gating pulse and 0V - pulse) - emitters have to be connected to +5V and 10K resistors have to be connected from collectors to minus 5V supply. The control signals have to be taken from the collectors." In this case, the bases of the transistors have to be connected via 10K to 47K resistors to the control signals of uP. Voltage converters will be inverting in this case.

    Leave a comment:


  • pechkata
    replied
    That's how I understand it...

    Click image for larger version  Name:	viber_изображение_2025-02-22_14-32-33-479.jpg Views:	0 Size:	273.5 KB ID:	434430 Click image for larger version  Name:	viber_изображение_2025-02-22_14-32-33-638.jpg Views:	0 Size:	305.1 KB ID:	434431

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  • Detectorist#1
    replied
    Hi GeoMax,
    Yes! This is right connection.

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  • Detectorist#1
    replied
    Hi pechkata,
    The connections of BC807 in these level shifters are totally wrong. If gating pulses from uP are negative( +5V -no gating pulse and 0V - pulse) - emitters have to be connected to +5V and 10K resistors have to be connected from collectors to minus 5V supply. The control signals have to be taken from the collectors. If gating pulses from uP are positive ( 0V - no gating pulse and +5V - pulse), the emitters of BC807 have to be connected to corresponding outputs of the uP and the bases have to be connected to GND via 10K resistor. The output signals have to be taken from the collectors and the collectors have to be connected to minus 5V supply via 10K resistors.

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  • GeoMax
    replied
    Originally posted by Detectorist#1 View Post
    Hi GeoMax,

    No. CD4066B have to be supplied from +5V and -5V. And the collector of BC557 have to be connected via 10K resistor to minus 5V supply.
    Like this...

    Click image for larger version

Name:	Non-inverting level transistor circuit for MCU.png
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  • pechkata
    replied
    Is this the switch you are talking about...?


    Click image for larger version  Name:	PFD-DIGITALNI.jpg Views:	0 Size:	798.3 KB ID:	434422

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  • Detectorist#1
    replied
    Hi GeoMax,

    No. CD4066B have to be supplied from +5V and -5V. And the collector of BC557 have to be connected via 10K resistor to minus 5V supply.

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  • GeoMax
    replied
    Hi detectorist#1

    does this small circuit below shows what you wrote about the non-inverting level transistor circuit in post #103 (above)?

    Click image for larger version

Name:	Non-inverting level transistor circuit for MCU.png
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  • Detectorist#1
    replied
    Hi eduardo1979,

    For controlling of 4066 no need to use inverting pulses from uP. Emitters of the PNP have to be connected to corresponding uP outputs and bases have to be connected to the Gnd via 10K-47K resistors. Thus, PNP will works as non-inverting level translators from 0/+5 volts to -5/+5 volts.

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  • eduardo1979
    replied
    Hi Carl and thanks again for answering my question.
    I need one last favor from you, as I have no experience with PIC microcontrollers. I have only worked with AVRs and I don't have any kind of compiler for the PIC. Could you post a HEX file on the forum with the four signals inverted? This would save me a lot of work, as I have opted to use a HEF4066 and the four level converters. I'm in no hurry, do it at your convenience.
    Thanks in advance.​

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  • Carl-NC
    replied
    Correct.

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  • eduardo1979
    replied
    Thank you very much Carl for responding and clarifying my question. I'm thinking of using a HEF4066 and its four corresponding level converters. I'm thinking that when using level converters, the square waves of the four signals coming out of the PIC would have to be inverted. Am I right?

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  • Carl-NC
    replied
    Originally posted by eduardo1979 View Post
    Could I put a 74HCT4066 instead, making the corresponding modifications on the PCB on the power pins, since they are different? I've also thought about using a CD4066 and adding 4 voltage converters with 4 PNP transistors and 8 resistors just like in the Arduino Nano PI. In your opinion, which of the two variants would be the best?
    ​
    Any 4066 will require the level converters since the clocks must be referenced to -5V. Once you do that, probably most any 4066 will work as long as it's rated to handle the 10V power supply. Whether one is better than the other often depends on the company who made it and what IC process they used. You have to test and compare.

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