Equivalent Replacement for Obsolete Vintage IC?

[dg_diyer]

Hi, newbie here hoping someone may be able to help with this.

I am repairing a vintage binary clock that freezes which I have isolated to occurring in one of the integrated circuits. The IC, Texas Instruments SN7490AN, is listed as obsolete. I have seen some listings for the IC but would prefer to avoid a pricey part that may also be stale. Is there an equivalent replacement part for this?

Can the SN74LS90N, which isn’t obsolete, be used as a substitute for the SN7490AN? This data sheet implies they do the same thing but if so why do they have different part numbers?

https://www.ti.com/lit/ds/symlink/sn74ls90.pdf?HQS=dis-mous-null-mousermode-dsf-pf-null-wwe&ts=1687629047497&ref_url=https%253A%252F%252Fwww.mouser.com%252F

[ledtester]

The LS in the part number refers to a later variant of the original TTL technology which had improved performance in terms of power consumption and switching speed. The logical operation of the chip is the same. You can read more about the LS and other variants here:

https://en.wikipedia.org/wiki/Transistor%E2%80%93transistor_logic#Sub-types

I think the LS chip would work in your clock - it's at least worth trying out.

[edavid]

If this is a Graymark 540 clock, and it's one of the chips driving the LEDs (U4-U7), then no, a 74LS90 won't have enough drive current.

If it's U2 or U3, a 74LS90 would probably be OK.

You can get 7490s from Jameco:

https://www.jameco.com/z/7490-Major-Brands-IC-7490-Autosynchronus-Decade-And-Binary-Counter_50690.html

[dg_diyer]

edavid, thanks for your response. Yes, it's a Graymark 540 and the trouble spot is the U2 IC where it divides by 10. Fortunately I still have the manual from the kit so, as very much a novice at this, I have been able to diagnose the problem.

[edavid]

Good, a 74LS90 should work fine there.

Can I ask how you figured out that IC is bad?

If it's soldered in, and your clock has the older phenolic (brown) circuit board, you're going to have to be careful desoldering so as not to lift traces.  I would cut off all the pins from the body of the IC and then desolder them one by one.

I've attached the schematic in case anyone wants to follow along at home.

[TimFox]

The bipolar 74 devices are "current sinking", meaning that the NPN output transistor must sink the total of the low-state input currents of the driven devices.
The input currents for the high state are much lower, and are easier to drive.
When comparing 74LSxx to 74xx (ignoring speed), the important difference is the input (low state) current, and the output (low state) sinking current (while keeping the node voltage below the threshold for low state).
 7400:     low input current = 1.6 mA;  low output drive current = 16 mA; fan-out within series = 10 gates
74LS00:  low input current = 0.36 mA; low output drive current =  8 mA;  fan-out within series = 22 gates;  fan-out 74LS00 driving 7400 = 5 gates.

[wasedadoc]

The bipolar 74 devices are "current sinking", meaning that the NPN output transistor must sink the total of the low-state input currents of the driven devices.
The input currents for the high state are much lower, and are easier to drive.
When comparing 74LSxx to 74xx (ignoring speed), the important difference is the input (low state) current, and the output (low state) sinking current (while keeping the node voltage below the threshold for low state).
 7400:     low input current = 1.6 mA;  low output drive current = 16 mA; fan-out within series = 10 gates
74LS00:  low input current = 0.36 mA; low output drive current =  8 mA;  fan-out within series = 22 gates;  fan-out 74LS00 driving 7400 = 5 gates.

Note that the low input current is a maximum figure while the low output is a minimum.  So there is a good chance that in practice you can achieve slightly higher fanouts that those theoretical numbers.  Might be useful in a "one-off" build though not recommended for production.

[TimFox]

They also involve the actual output voltage at max sink current vs required input voltage to guarantee low input state.
If you follow the datasheet values, the result is guaranteed over the full production of the devices.

[floobydust]

U2 divide by 10 is not doing anything high current or unique to 7400 logic that I can see, so I think a 74LS90 would work fine.
If the clock gets stuck sometimes also check the Hold switch is not oxidized and needs cleaning, it's never pressed that often.
Or sometimes old IC sockets get oxidized and give bad connections. I have good success using PPE-based contact cleaners like DeOxit.

[wasedadoc]

They also involve the actual output voltage at max sink current vs required input voltage to guarantee low input state.
If you follow the datasheet values, the result is guaranteed over the full production of the devices.

Maybe I wasn't clear enough about what I meant by "production".  The specs are the guaranteed figures.  Many devices will be better than those figures and using a fanout larger by 1 than the theoretical numbers is likely to work in the majority of cases.  Could be worth trying if it avoids adding another IC in a home-build one-off but would certainly be frowned on if a commercial design where circuits are expected to function 100% even if all components are at the unfavourable extremes of tolerances.

[TimFox]

I agree as to what will almost certainly work for a "onesie" in ones own work.
As a rule, though, I try to be conservative even in my own amateur designs.

[floobydust]

The 74xx IC's alone will draw up 200-300mA for the clock board. Huge spread in current consumption specs. The LS IC's are at least 1/2 of that.

[wasedadoc]

The 74xx IC's alone will draw up 200-300mA for the clock board. Huge spread in current consumption specs. The LS IC's are at least 1/2 of that.

I think you mean "The LS ICs are at most 1/2 of that".

[MrAl]

Hello there,

Whether or not it works is not a matter of hopefulness, it is a matter of drive specifications for both the driver and the driven.

Since the driven is a 7400 series device, it should have a 7400 device to drive it.  What matters most here in this special case is just how many inputs are being driven.  If it is just one input being driven it might work, pending the investigation of the low and high state requirements of the driven.

There are writeups on this but i don't know where they are right now.  They also go by 'fanout' meaning if a drive chip has a fanout rating of 5 then it can drive 5 inputs of the same actual family.  Someone could check but i think the LS series is 1/2 of the regular series, so that if the driven inputs only total 2 and the fanout of the driver is 4 but of the lower LS series, it should still work.
Now if the driven inputs are 4 and the fanout of the driver with the LS series chip is only 4, then it's not a good idea to use that LS chip even if it seems to work at first.  That's because the chip specs are for a given temperature range so it may work one minute and not work the next.

It's not hard to find a 7490 chip.  If you use that you can be sure it will work, no question about it, unless something else is wrong too.

As someone mentioned, the copper clad may get ripped up for some pins, which is not really too unusual.  You can be careful but just in case there is a known fix.  Since the entire trace does not rip up, only near to the pin holes, you can get a AWG #26 gauge bare wire and make a small loop around the pin of the new device, then run the tail down the part of the trace that still remains.  You then solder both the pin again to the loop, and the tail to the copper that remained during removal of the old chip.  Of course be careful you do not create any solder bridges.  The AWG #26 wire is thin enough to go around a DIP pin without touching any other pins.  I dont think i would try this with AWG #22 though, although AWG #24 may work too.
If you cant get #26 wire sometimes you can pull a strand out of a piece of regular stranded wire.  It should be thin like AWG #26, but you can test it by wrapping it around one pin and make sure it is thin enough so that it does not touch any other pins, before and after soldering.

[floobydust]

The 74xx IC's alone will draw up 200-300mA for the clock board. Huge spread in current consumption specs. The LS IC's are at least 1/2 of that.

I think you mean "The LS ICs are at most 1/2 of that".

I went with SN7490A Icc typ. 29mA max 42mA, SN74LS90A Icc typ. 9mA max 15mA so it's max is 1/2 the typical.

Did anyone actually look at the schematic?
U2 divide by 10 input is not driven by another gate, it's got a discrete transistor Q2 from mains AC ref. which has no problem sinking 5mA.

[Terry Bites]

 if the last part of the number says it the same thing, then it must me... ER. oh maybe. 74 became 74LS, it got a faster output due to the addition of a shotky diode that stop it getting staurated "The Baker Clamp. Many new improved forms a have appeared: 74S,74AS 74F etc with no changes to anthing but speed and power.  Then there's CMOS, 74HC is just bog standard CMOS, just like bt better than 40000 series. To help you mix your old ttl with the new cmos, ther'es 74HCT, T for threshold. CMOS ic with 74LS thresholds. See sdyu001ab.pdf from TI