I am building a homebrew 8-bit computer on breadoards and PCBs with TTL logic IC's (LS, HCT and ALS families).

This is a small portion of the circuit (3 modules): enter image description here

I'm using an ordinary phone charger to supply the required 5V, but some IC's are getting only 4V or less, even though it's working OK that's not good at all.

The total current required is pretty low, around 200mA, very within the power supply range. I've tried with a few different power supplies with no success. I've done some research and found out that cell chargers are regulated, so it wouldn't be a problem.

The computer has 6 modules, the module closer to power supply reads 5V on its IC's, and the more it gets away from power, the voltage gets weaker.

I tried to make a power rail directly connected to each module, with same results.

I could put the modules schematics here but it would make the question very long, so I will provide the links to them:

1- Clock & Sequencer

2- Program Counter & Memory

3- Instruction Register

4- Control Unit

5- Registers

6- ALU

Modules 1 and 3 are already in PCBs, the others still in breadboards.

Link to the project here.

Are there any obvious project mistake here? Any help would be appreciated.

Edit: Could it be poor quality ICs from China the cause?

  • \$\begingroup\$ Are you measuring the VDD--GND voltage WHILE the computer is clocking along? Is each IC given a local 0.1uF bypass capacitor, located right at the IC? \$\endgroup\$ – analogsystemsrf Nov 23 '18 at 15:10
  • \$\begingroup\$ I measured the voltage with and without auto clock enabled (there is a manual clock mode). \$\endgroup\$ – André Baptista Nov 23 '18 at 15:58
  • \$\begingroup\$ There is 100nF bypass capacitors only on the PCBs, on the breadboards don't. could it be the reason of the problem? \$\endgroup\$ – André Baptista Nov 23 '18 at 16:00
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    \$\begingroup\$ All chips need decoupling, but as Captain says, double up your wiring to far loads or use thicker wires. Use your voltmeter and measure the voltage drop across your power lines. There will be a voltage drop. If your power supply was not large enough to supply load all voltages would be down. If you want to use a power rail, make VCC and ground larger. \$\endgroup\$ – StainlessSteelRat Nov 23 '18 at 17:25
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    \$\begingroup\$ This is almost certainly a design flaw, not an IC quality flaw. Especially at slow clocks cmos chips would probably consume less than 74xx or 74LS. Your biggest issue is with wiring however: if you are going to build something complex with breadboards you need to attach them to a carrier, route your wires flat not up in the air, and consider staking down the bundles for example with short wires stuck in the ground row or even wire tire points beside the modules. Breadboarding a computer from discrete components is technically absurd - you are purposely choosing to make things difficult. \$\endgroup\$ – Chris Stratton Dec 18 '18 at 16:19

the more it gets away from power, the voltage gets weaker

This usually means that the wires are too thin (ie their resistance is too high) for the current. 200mA is however not much, so this seems strange. Try measuring the voltage drop over the power supply wires directly and/or try using thicker wires (or two wires per link) to see if this is really your problem.


As I supposed initially, the problem were the cheap chinese IC's from Ebay. I bought all again from a reliable store and replaced them (luckily I always use IC sockets, so replace is pretty easy), and now it works like a charm:


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    \$\begingroup\$ How much current does it draw now, and what voltage do you get on the furthest module? \$\endgroup\$ – Bruce Abbott Jan 20 '19 at 19:30
  • \$\begingroup\$ I haven't tested all modules together yet. But this module pictured (ALU) wasn't working before and the tension on VCC pins was below 4V. All 9 IC's replaced and now it reads 4.9V and it works perfectly \$\endgroup\$ – André Baptista Jan 21 '19 at 12:17

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