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I recently bought a lot of TTL components like 8-bit registers, bus transceivers, etc. to build an ALU and 2 registers. My issue is that when I build a register, the voltage drops to 4.2, sometimes even below 4V even though I put resistors in every output in series with the LEDs. I built 2 8-bit registers with bus transceivers using 74LS377N for registers and 74LS245N for bus transceivers. These two registers are built on different breadboards. And I connected the power to the bottom breadboard and wired the bottom breadboard to the other breadboard to provide power to it. The problem is that voltage drops significantly across these 2 breadboards.

Then I tore it all apart and individually tested the TTL components to see what's gonna happen. I powered up the breadboard and connected the two power lines on the sides. I put an 8-bit register and then only connected the power and the ground pins, leaving the rest disconnected. Then I saw the voltage drop below 4.5V from 5.1V. And I had like 6 of these registers so I tested them all. Some of them dropped to 4.9V and one of them didn't drop at all. But mostly they dropped below 4.5V. What am I missing? Or is there something I don't know about or I'm doing wrong? I'm kinda new to this so every bit of knowledge is appreciated :)

EDIT: Later, I went on to try this on other logic gates like 7400, 7408, etc., and saw that they behave like this too. Most of them drop the voltage, and a couple of them keeps it the same. I feel like it's about the logic gates I bought. But I have like tens of them.

EDIT: The alu I'm building is the one that Ben Eater built in his video, here is a screenshot.

Screenshot of breadboard from Ben Eater video

I built the exact same circuit but the LEDs that are at the bottom, which are the ones that are farthest away from the power supply, barely get any current. And I have checked everything like 10 times over the past 2 weeks. AND YES I have connected everything as Ben Eater did, except he was using some LEDs with built-in resistors so therefore I added 220 ohm resistors in series with the LEDs. But besides that, everything is the same.

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    \$\begingroup\$ Perhaps the voltages mentioned are power supply voltages? Or are they voltages measured at gate outputs? Its not clear. Be aware that TTL logic "0" pulls to ground easily. TTL logic "1" pulls to Vcc weakly. \$\endgroup\$
    – glen_geek
    Dec 5, 2021 at 15:10
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    \$\begingroup\$ Trailblazer17 - Hi, Rather than using words, I recommend you to create a small circuit which shows the problems you have, supply its schematic, some photos, and start with the exact voltage measurements of concern on that (showing whatever "voltage drops" worry you), then readers can focus on those specific concerns. At the moment, with no photos, no schematics or exact measurement points on those schematics, this question is inviting readers to give useful, but general advice which may not address your specific problem(s). So, please, fully describe a specific circuit. Thanks. \$\endgroup\$
    – SamGibson
    Dec 5, 2021 at 15:25
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    \$\begingroup\$ Re you recent edit - we really need to see what you built, not what you're trying (perhaps unsuccessfully ) to copy. \$\endgroup\$
    – brhans
    Dec 5, 2021 at 15:31
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    \$\begingroup\$ Please describe your +5V DC supply. You really need a carefully-regulated +5V supply for TTL. Two breadboards of TTL will require significant current. PeterSmith's answer addresses some of TTL's output pull-up, pull-down quirks. \$\endgroup\$
    – glen_geek
    Dec 5, 2021 at 15:32
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    \$\begingroup\$ I'm sorry but i tore it all apart because I've literally been trying every possible solution i got for the past 2 weeks and i got no changes. \$\endgroup\$ Dec 5, 2021 at 15:32

3 Answers 3

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TTL requires a +5V DC supply having less than 5% variation (from Texas Instruments 74LS00 data sheet):

from Texas Instruments 74LS00 data sheet

  • Measure your 5V supply voltage with no connection to breadboards, nor to anything else other than your voltmeter.
  • Then measure your 5V supply voltage when connected to TTL chips on breadboards.

Both measurements should lie within the range of 4.75 to 5.25V.
You might make a further check by measuring at each TTL chip: measure from its Vcc pin to its GND pin...this DC voltage should also be within the range of 4.75V to 5.25V.

If your measurements are outside this range, then you need a stiffer, better regulated +5V supply. It is also possible that voltage drop along wires connecting chips to the supply have too much resistance or are too long.

Andrew Morton mentions that OP's photo is Ben Eater's version of the ALU. I'm amazed that there are no Vcc-to-GND bypass capacitors! Wow, I'm surprised that such construction works reliably. Not only should a multimeter show Vcc voltage to be near +5V, but an oscilloscope should show no short-term variations of that +5V. Bypass capacitors are meant to smooth any variations. You'll often see a 0.1uf capacitor connected from Vcc-to-GND at each TTL chip:
Zenith CPU printed circuit board using LSTTL chips

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    \$\begingroup\$ @Trailblazer17 A "stiff" voltage supply means one that has good load regulation, i.e. when the load current changes quickly, the voltage does not change very much. If the wires are "too long" then they have quite a bit of inductance, which is bad for keeping the voltage at the ends of the wires constant when the current changes, which is why decoupling capacitors near to the chips are used: they act as a local supply of energy without the inductance in the way. \$\endgroup\$ Dec 5, 2021 at 17:18
  • \$\begingroup\$ Your reply taught me really useful knowledge thanks for that but i have some questions. I had a little bit of knowledge about bypass capacitors but not much. Where should I exactly put them? Near Vcc or does it matter? And should I put them on every chip I have? btw I use the power outlets in my room as my power supply like Ben Eater and i adjusted my phone charging cable too. I made ground and power wires wrap around each other i think that helps with impedance. Also, could it be about the adapter? \$\endgroup\$ Dec 5, 2021 at 18:40
  • \$\begingroup\$ The picture added shows TTL chips similar to yours, with many bypass capacitors - the designers are being VERY careful to ensure Vcc is unvarying. Need you be so careful? Perhaps not. I'd add at least ONE capacitor to every Vcc and GND rail of your breadboard (assuming your breadboards have similar distribution rails as your BenEater photo). Yes, I suspect your 5V power supply is not steady enough - do the voltage measurements outlined. TTL is fussy about its +5V and its GND. \$\endgroup\$
    – glen_geek
    Dec 5, 2021 at 19:34
  • \$\begingroup\$ I connected 1 uF capacitors to every breadboard rail. It raised the voltage from about 3.3V to 4V but its still not near 4.75V and the voltage drops as I add more breadboard. The bottom breadboard(closest to the power supply)rails read 5V, the next one reads 4.90V, the next one 4.30V, Then the last one reads about 4V. So voltage drops as I add more TTL chips even with capacitors. I reckon its about the power supply so I'll try some stuff out. \$\endgroup\$ Dec 5, 2021 at 19:51
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    \$\begingroup\$ @Trailblazer17 You may want to increase the gauge of wire you're using for the power connections; most breadboards will accept as large as 22 gauge. Trying to squeeze 20 gauge into them will probably damage them. Use multiple wires in parallel if needed. This may also, however, be a problem with using cheap breadboards (or ones that have been previously damaged by using too large wire!), with high contact resistance. I don't think the decoupling capacitors will fix your problem if you're seeing that low a DC voltage, though they're definitely a good idea. \$\endgroup\$
    – Hearth
    Dec 5, 2021 at 20:09
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The guaranteed high output of a TTL gate is greater than or equal to 2.4V.

That is usually at some specified current. With no load 4.5V is inside the requirements and perfectly reasonable.

TTL open inputs look like a high but any noise coupling can cause it to be interpreted as a low.

The guaranteed low output is less than or equal to 0.4V. At no load it is still unlikely to be very close to zero.

Do not leave CMOS inputs floating as it is possible to burn out the input stage.

Note that TTL high outputs cannot drive much current so putting LEDs on the output to indicate a high is not a good idea.

Indicating a low is common.

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  • \$\begingroup\$ When I'm measuring the voltage of an output, where should I connect the ground of the multimeter? \$\endgroup\$ Dec 5, 2021 at 15:25
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    \$\begingroup\$ @Trailblazer17 A multimeter has a high input impedance when measuring voltages, so connect the ground to the ground of the circuit, as that is what you want to measure the potential difference to. \$\endgroup\$ Dec 5, 2021 at 16:12
  • \$\begingroup\$ no it reads different values when i measure at different points. When I connect the ground to the ground wire of the farthest breadboard, it reads a smaller value. I think I should connect it to the ground of the charging cable. \$\endgroup\$ Dec 5, 2021 at 18:24
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By definition, Ground is wherever you define 0V, normally at the Vee power source.

TTL has asymmetrical drive currents and for noise immunity asymmetrical voltage thresholds, unlike CMOS which is designed to be symmetrical with some process variances for tolerance.

from datasheets:

SN74LS245 (Hex TTL BUffer)

IOH = –3 mA for VOH= 2.4 min., 3.4 typ.

  • Thus ZOH the high level resistance is (5V-2.4V)/3mA = 867 ohms MIN. (5V-3.4)/3mA= 533 ohms TYP.

IOL = 24 mA for VOL = 0.5V max.

  • Thus ZOL , the low level switch R is 500 mV/24 mA ~ 21 Ohms MIN

74LS377N

IOH= -400 uA for VOH= 2.7 Vmin 3.5 max.

  • Thus ZOH = (5-2.7V)/0.4 mA = 5.75 Kohms
    IOL = 8 mA for VOL = 0.35V typ , 0.5V max
  • Thus ZOL = 350 mV / 8mA= 43.75 Ohms MAX. and 40 Ohms typ

SN74HC245 CMOS

enter image description here

(Image from: Texas Instruments 74HC245 datasheet)

IOH = –6 mA for VOH= 3.84 min.at Vdd=5V-10%.

  • Thus ZOH the high level resistance is (4.5V-3.84V)/6mA = 110 ohms This is reduced for higher Vdd and rises for lower Vdd, unlike TTL.

IOL = 6 mA for VOL = 0.33V max.

  • Thus ZOL , the low level switch R is 330 mV/6 mA ~ 55 Ohms MIN There is no typ. but expect this to reduce for 5V.

Conclusion

For TTL high current loads e.g. LED's ~ 10 to 20mA, Use the low "0" level to drive LEDs with Anode to Vcc and series R which has about 1/25th of the resistance for ZOL/ZOH typ..

Also for HEX buffers

If the output is not used for logic to another stage, do not worry about the VIL VOL threshold but rather Pd and temp rise. from V*I=P * Rth ['C/W]

For HC family CMOS use 50 Ohms +/25% tolerance @ 5V is generally true.

For 3.6 V 74ALCxxx CMOS Logic at 3.3V, by design, this reduces to about 22~25 ohms +/-25% in general.

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  • \$\begingroup\$ thanks for taking your time to reply. \$\endgroup\$ Dec 5, 2021 at 15:57
  • \$\begingroup\$ This ought to be taught in school, but it wasn't. \$\endgroup\$ Dec 5, 2021 at 15:59
  • \$\begingroup\$ That's not considered sexy anymore. Everyone wants to learn VHDL/Verilog, FPGA/ASIC design, and C++ or Python. \$\endgroup\$
    – SteveSh
    Dec 5, 2021 at 16:13

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