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I am using a Stellaris EK-LM4F120XL for a project. The project includes an LCD module, servo motor, ultrasonic ranger and a numpad. The LCD, servo and ranger are rated at 5V operating voltage, and the Stellaris at approx. 3.3V. To have the project powered as a stand alone unit, I have a battery pack for 2x9V batteries. I will hook these in parallel to offer a longer running time. Then they will flow into a voltage-regulator, LM317, to provide 5V to the peripherals..

Then I will use this 5V source thru a non-inverting hex-buffer, CD4050BE, to then power the microcontroller.. I understand that no matter what voltage is input - the difference between Vcc and GND is output. Although, does that mean for a buffer you need to have a voltage-in, and a desired voltage-out - but also need to supply a Vcc voltage to maintain the desired output?

Therefore, if I have my 5V source from my regulator, do I really need to create circuitry using a resistor-divider or transistor voltage-drop, to get my 3.3V - then input this into the hex-buffer..? What a waste of current, no?

Why not utilise another voltage-regulator with voltage-divider resistors like I used to get my 5V..?

What about a single-channel linear-regulator - LP2950-N, of 3.3V variation..? Or LM2936-3.3...? Why use buffers at all then.. instead of regulators..?

I have diodes/resistors/PNP BJTs/opAmps on hand.. I am using transistors, BC548, to operate my 'loads' that are in parallel with diodes already.. And will ground everything to the same point - negative terminal of the powerpack..

To clear it up - I need 5V for all peripherals, and approx. 3.3V for the uController..

Cheers :-)

(I have circuit diagrams but cant add pics..?)

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  • \$\begingroup\$ Do you have an URL to your image? \$\endgroup\$ – jippie Oct 19 '14 at 5:26
  • \$\begingroup\$ "Then I will use this 5V source thru a non-inverting hex-buffer, CD4050BE, to then power the microcontroller." What. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 19 '14 at 5:32
  • \$\begingroup\$ *Than.. That was the initial idea: to run the output 5V from the voltage-regulator, into a buffer to get my 3.3V... \$\endgroup\$ – Rogan Josh Oct 19 '14 at 5:34
  • \$\begingroup\$ That... makes no sense. If you need to regulate voltage then you use a voltage regulator. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 19 '14 at 5:38
  • \$\begingroup\$ s21.postimg.org/lg7t8ymp3/CIRCUIT.jpg \$\endgroup\$ – Rogan Josh Oct 19 '14 at 5:51
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The CD4050 is a digital logic buffer that can be used to shift signal levels from high to low voltage. It is not a voltage regulator - however you don't need one anyway because the EK-LM4F120XL already has an on-board 3.3V regulator. But perhaps what you are really worried about is the signal levels between your 3.3V MCU and 5V peripherals.

The LM4F120 has 5V tolerant I/O, so you needn't worry about blowing up its inputs. However its outputs can only go up 3.3V, which may not be high enough for devices that expect to see 5V. A standard CMOS logic gate powered by 5V could require a logic high level of least 3.5V for reliable operation (at 3.3V it might still work, but with greater propagation delay and lower noise immunity). Unfortunately that means that the CD4050 is not suitable for level shifting from 3.3V to 5V.

Most servos work fine with a 3.3V input signal, however the LCD display and ranger (and keypad?) might have problems. Check their data sheets to find out what logic levels they require. To level shift outputs from 3.3V to 5V You can use a buffer with TTL compatible inputs, eg. 74HCT4050 (note the 'T' which indicates TTL). If you are using a bidirectional bus such I2C then you may need special level translators that are designed to work in both directions at once.

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  • \$\begingroup\$ Are you saying the digital buffer chip treats Vcc as high and anything below a certain threshold as a low, then will output 0V? I tested the chip with a DC 1.5V Vcc and input DC 9V - and the output was as expected DC 1.5V.. My goal wasnt to have the MCU power any device, as I want to use the external DC battery-supply.. So in your opinion I am safe to use my voltage-regulator to drop my DC voltage from 9V to 5V to power all my devices - including the MCU? That is using the LM317 regulator and a divider as in my circuit diagram, excluding any buffers.. \$\endgroup\$ – Rogan Josh Oct 19 '14 at 21:34
  • \$\begingroup\$ Digital logic gates switch between low and high at a particular threshold voltage, but for reliable operation a wider range is specified. For TTL it is 0.8V low and 2.0V high, for standard CMOS it is 0.3*Vcc and 0.7*Vcc (actual voltages depend on Vcc). The EK-LM4F120XL can safely be powered from 5V through the USB port or +VBUS pad, which is dropped down to 3.3V by the on-board regulator. If you were using a bare LM4F120 MCU then you would need to provide your own 3.3V regulated supply. \$\endgroup\$ – Bruce Abbott Oct 19 '14 at 22:59
  • \$\begingroup\$ i2c bi-directional level shifting can be done with just two FETs - one for data and one for clock. It's good for sub-MHz operation. (i2c is normally 400KHz.) \$\endgroup\$ – fadedbee Dec 12 '16 at 12:45

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