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A beginner style question. I'm attempting to perform some logic level conversions on 3.3V signals to 5V using the 74HC4050D powered at 3.3V and 74HC541 powered at 5V. Is it OK to directly connect the outputs of the 74HC4050 to the inputs of the 74HC541 ? or do I need pull up/Pull down resistors from the output of the 74HC4050D to the input of the 74HC541 for each pin?

What should I be looking for in terms of the datasheets when determining these things (in my case I don't really aim to source or sink current to other devices.I just aim to do the voltage conversion. Is it only leakage current that would be output in this instance? If one didn't add a pull up/down resistors or a series resistor would the only risk be that of an unkown logic state if VCC were to change abruptly or signal noise was present?

PS I'm aware of that the maximum current into these devices Icc is about 50ma and a max output of a particular pin Io is around 4-5ma.


Links for you reference:
74HC4050D
74HC541

stevenvh
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Peter H
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  • [74HC4050 datasheet](http://www.nxp.com/documents/data_sheet/74HC4050_CNV.pdf) and [74HCT541 datasheet] – Peter H Jul 02 '12 at 13:34

3 Answers3

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Whether a 3.3 V signal can reliably drive a 5 V logic input depends on the minimum guaranteed logic high level of the 5 V logic input. Whether that is sufficiently below 3.3 V for reliable operation varies with logic families and manufacturers, so consulting the datasheet is essential. For example, Microchip PICs with Schmitt trigger inputs usually require 80% Vdd for guaranteed logic high. That would be 4.0 V, which is too high to be driven from 3.3 V logic.

A easy hack is to use logic with "T" in its name, like 74HCT as apposed to 74HC. The T stands for TTL, and means the logic input levels are compatible with the old TTL logic. This had a finite guaranteed logic high threshold sufficiently below 3.3 V so that driving one of these from 3.3 V logic is fine, although its a good idea to check the specs in any one instant. So if you need a logic gate in there anyway, make it a HCT type and you get the 3.3 V to 5 V conversion for free.

Olin Lathrop
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    @stevenvh: I can never remember. It doesn't help that another local EE I know has the last name "Schmidt". It all depends on how the German was transliterated into American when they got off the boat ;-) – Olin Lathrop Jul 02 '12 at 14:28
  • I remember because it's the only Schmitt I know. I don't even know if it's a common name in neighboring Germany. Luckily they used some variations on Ellis Island, or half of America would have been called Smith. – stevenvh Jul 02 '12 at 14:31
  • @stevenvh: "Schmidt" is certainly not unusual in germany. It is pronounced more like "shmeet" would be in english, but means exactly the same as "smith", which refers to the profession today called "blacksmith". Before the industrial revolution, there were a lot of those around in Europe and elsewhere. There are plenty of Smith, Schmidt, Schmitt and even some Smythe here in the US. – Olin Lathrop Jul 02 '12 at 14:39
  • No, I know that "Schmidt" must be one of the most occurring names, I meant "Schmitt". BTW, "De Smedt" and variants is the most occurring name in the Flanders' part of Belgium, too. – stevenvh Jul 02 '12 at 14:50
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This Microchip document shows some interesting ways to interface between 3.3 V and 5 V logic.

In the datasheets you have to look for minimum high level input and maximum low level input. The former is of 0.6 Vcc or 0.7 Vcc, so 3.3 V may be just enough, or not. Also keep in mind that 5 V is probably not exactly 5 V, but may vary by 5 %, and the same for 3.3 V. Worst case you don't even meet the 0.6 Vcc level.

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Well, the NXP datasheet doesn't say; NXP has a HCMOS family specification for all its HCMOS DC characteristics not mentioned in the datasheet. For input high a minimum of 3.15 V is indicated for 4.5 V power supply. Linearly extrapolating to 5 V that becomes 3.5 V, that's the 0.7 Vcc I mentioned. Low level input should be less than 1.35 V at 4.5 V supply, or, again linearly extrapolated 1.5 V at 5 V supply.

stevenvh
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  • Helloo again Steve. Thanks for your reference document. I guess my question is without resistors connected how would one know whether it's safe to connect the outputs of the 74HC4050D to the inputs of the 74HC541 as this chip can do the logic conversion? Cheers, Pete – Peter H Jul 02 '12 at 13:59
  • I may be misreading the 4050 datasheet, but I think you are :-). It's meant to accept input from 4000 series CMOS, which can work at voltages up to 15 V. But its output is a 3.3 V HCMOS level. – stevenvh Jul 02 '12 at 14:05
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Good beginner questions;

3.3V signals to 5V using the 74HC4050D @ 3.3V and 74HC541 @ 5V.

Is it OK to directly connect ...the 74HC4050 to the inputs of the 74HC541?

OK yes, wise? no. First off the noise immunity is imbalanced so that potential risk from ingress noise is imbalanced. Pull-up/down resistors assist high level current and balance the noise immunity when the "Thevanin equivalent circuit" equals the transition voltage. such that the driver moves away from that termination threshold with equal power. ( an approximation of optimum immunity in noisy world)

Consider consider 3.3 is above 50% of Vcc=5V the noise immunity is asym metric, so pull-up/down balances this around the 50% threshold for best immunity from asymmetric driver current. as previously stated, HC threshold = 50% of Vcc, and HCT threshold = TTL threshold . By the way relatively few EE designers know that TTL input threshold is= "two Vbe diode drops" = 1.30V so no shifting is needed.

The 4050 chip is a down-voltage converter hex buffer. So unless you are trying to do a closed-loop or loopback test? ... you may not use a 4050 to level-shift up.

The 4050 input translator ir related to the output Vcc by having a threshold slightly more than Vcc so that a translator range of 2:1 to 3:1 is possible. e.g. 5V to 2V logic or 15V to 5V logic, but not 15V:2V ;)

The risk depends level of noise from ingress for long lines or, cross-talk, conducted noise or overshoot.

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Tony Stewart EE75
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    Could you elaborate on this please Tony "... you may not use a 4050 to level-shift up." If I replaced the 74HC541 with a second 74HC4050D and attempted the logic conversion are you saying that signal output would not be level shifted from 3.3V to 5V? Or just that your not recommending it because of the noise immunity and fact that its a CMOS device and NOT TTL compatable? – Peter H Jul 03 '12 at 03:14
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    The 4050 chip is a down-voltage converter and cannot be used at all for up converting. YOu may use a (74LVC4245) Bidirectional 3.3v to 5volt transceiver is used to switch between the two logic families. Bus "B" is the TTL +5volts side, Bus "A" is the +3.3volt side. http://www.nxp.com/documents/data_sheet/74LVC4245A.pdf A pull-up/down terminator for 3 or 3.3 to 5V is also acceptable. – Tony Stewart EE75 Jul 05 '12 at 20:29