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I am trying to find a simple hardware solution to implement a 2-bit timer that gives out 2-bits continuously repeating data 00, 01, 10, 11, 00, 01.... whose time period is approx 1ms. The accuracy and jitter in the period is not an important design parameter. The simplicity in design is important. Simplicity means without any programmable logic. The voltage levels can be any.

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  • \$\begingroup\$ You mentioned 555, timer, counter and digital logic in your tags. Going from those alone seems like a straightforward way to solve this. Just get a 555 to generate a clock for a random counter chip. Do you just want something simpler than two chips and 3 or 4 passives? \$\endgroup\$ – Richard the Spacecat Oct 10 at 13:13
  • \$\begingroup\$ Alternatively, use the CD4060B, a counter with a built in oscillator. Usable with RC. I found it by looking for "counter with oscillator" on your favourite search engine. It seems to be quite cheap too. \$\endgroup\$ – Richard the Spacecat Oct 10 at 13:17
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Use a 74xx4060 14-bit binary counter IC, such as the 5 V 74HCT4060. This is a cheap and readily available part and uses a resistor and a capacitor to control the increment rate of a 14-bit counter.

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Its 10 output counter bits are different stages of a 14-bit ripple counter. So you can choose a relatively high oscillator frequency that requires a small capacitor and high resistor, then use a divider stage that gives you your 1 kHz 2-bit value.

enter image description here

It uses an external resistor and capacitor to set its internal oscillator frequency.

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If the accuracy of an RC oscillator is insufficient, you can use a crystal instead.

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As said by @RichardtheSpacecat, a 555 to generate your clock signal.

For the counter, two D flip-flops would be sufficient: (assume the flip-flops output are Q_not instead of Q)

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ One can still have a variant of the 4060B for a lower cost than both a 555 and another chip (perhaps a 7474?). Having less parts may lead to higher reliability, which the OP might be after. \$\endgroup\$ – Richard the Spacecat Oct 10 at 13:57
  • \$\begingroup\$ I did not know what was the 4060B ! \$\endgroup\$ – Wheatley Oct 10 at 14:03
  • \$\begingroup\$ Shouldn't those be the ¬Q outputs that you use there, not the Q outputs? \$\endgroup\$ – Hearth Oct 10 at 14:37
  • \$\begingroup\$ @Hearth exactly, I wrote it in the text but was not able to find a proper symbol in the schematic editor. \$\endgroup\$ – Wheatley Oct 10 at 15:17

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