DIP to SOP - Solved Fake Chinese IC

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
I designed a board and breadboarded using CMOS DIP (ripple counter). Everything worked great. So my PCBs arrive and I put them together to discover the SOP package does not behave the same.

Basically the issue is that I'm using a diode as an AND gate on output pin of a ripple counter. It works great in DIP package but with SOP it seems like it can't take the full current to give it a good low on 01 or 10 output... Obviously without the diode gate everything works great (without the decoding I want). I'm going to put the SOIC on a DIP adapter board and test some more.

I think the solution may be a MOSFET gate so it grounds the output of first output to ground when the second input is low (P Channel)... Ideas comments?

I wish I could draw a picture but logic outputs get messy for me to draw.
 

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
It's a simplified version.



This works using DIP... does not with SOP16... using a TI CD4060BM ripple counter. The output are limited by the chip to about 15mA so no resistor needed. Feeding it 5V. My guess is that the Diode is not able to sink full current from output as SOIC - so I get weak flashes where I should not.

Even tried replacing the diode on the breadboard using the chip diode from the board using header pins... works fine.

The Q outputs are labeled... So the low on Q13 should keep LED 1 from turning on for the first half of it's cycle.

Datasheet shows pins are same for DIP and SOIC - confirmed as it runs fine otherwise

testing123.jpg
 

kubeek

Joined Sep 20, 2005
5,625
I don´t see where you get that the output current is limited to 15mA. Use resistors in series with the leds. Which datasheet are you actually using?
What is the supply voltage?
 

kubeek

Joined Sep 20, 2005
5,625
Let me ask again, what is your supply voltage? Where do you see that it is safe to let the output run into a short circuit? Why don´t you try adding the resistors and see what happens?
Do you realize that when Q10 is high and Q13 is low you are shorting the output?
 

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
It can sink current... I can try with the resistor.. but like I said it works great using DIP... this SOIC is acting differently. Actually might help the diode... You're talking about a current limiting resistors for the LED so with about 5V output on white LED about 120 ohm in series with LED?

Why does it work with DIP???

I think the solution to the issue will be to change the AND gate to MOSFET going to ground instead of the pin.
 

djsfantasi

Joined Apr 11, 2010
5,702
What are the specific part #s for the DIP and SOP ICs? There may be subtle differences between them. That’s not obvious given just the one datasheet. Since you’re experiencing a difference, in order to compare we need to know the explicit parts.

Also, what diode are you using as an AND gate? D1? I’m not sure, but the diode AND gates I am familiar with require two diodes. That would prevent the existing short between Q10 and Q13.

Adding resistors to the LEDs must be also placed to minimize the effect of that short. I’d recommend a placement, but I don’t fully understand your diode AND gate.
 

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
The part numbers:
I've tried two different SOP chips (one was generic) I've order 2 more from other manufacturers to rule this out.
DIP Package: Texas Instrument CD4060BE
SOP16 - Texas Instrument CD4060BM

The Diode is not a factor.. I replaced the 1N4005 (what I had at time) with the shottky chip diode (MBR0520) 0.5A@20V with 0.26 drop and DIP design works fine.

I ordered some SOP20 - DIP adapter boards to play around on a breadboard. It almost seems like the DIP is doing something it's not suppose to now that I've gone through the logic again.

I have probed with O'Scope but the O'Scope ground is affecting the logic outputs... So getting limited view.

I believe I'm using the output (LED) as the other part of AND gate...

I'll keep at it... I'm sure I'm missing something.
 
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Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
I know that if I use a p channel mosfet to gate q13 I should be able to bring it to board ground but it’s annoying that it works with DIP but not as SOP. Grrrr
 
Last edited:

mvas

Joined Jun 19, 2017
538
I know that if I use a p channel mosfet to gate q12 I should be able to bring it to board ground but it’s annoying that it works with DIP but not as SOP. Grrrr
Your circuit, in message #3, violates "common sense" electrical engineering design rules.
Actually, you are lucky that the DIP circuit works at all.
You must energize your LED's with the proper Current Limiting Resistors.
You have decided to ignore common sense, therefore you will be "annoyed" by your poorly designed circuit.
 

kubeek

Joined Sep 20, 2005
5,625
I still have no idea what you are trying to do, or what the last drawing means. Your circuit is using outputs Q1, Q2 and Q3, yet that last drawing shows some diodes drawn between Q1 through Q4.
With three outputs you have 8 distinct states S0 through S7. In which states do you want which LED to be on? Can you please make a table with that so that we know what you want to achieve?
 

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
I still have no idea what you are trying to do, or what the last drawing means. Your circuit is using outputs Q1, Q2 and Q3, yet that last drawing shows some diodes drawn between Q1 through Q4.
With three outputs you have 8 distinct states S0 through S7. In which states do you want which LED to be on? Can you please make a table with that so that we know what you want to achieve?
I've updated the diagram. I want the LED to be on for the green circles. Funny or not, I've read that the chip limits the output current and tested it to be true. I'm wondering if it can only sink so much current as SOP.
 

kubeek

Joined Sep 20, 2005
5,625
Ok, now it makes sense. Connect Q10 to resistor say 1kohm, then that to D1, then to Q12.
Connect top end of D3 to the juncion between resistor and D1, bottom end of D3 to Q13.
 

Thread Starter

Wolframore

Joined Jan 21, 2019
1,321
I've tried to draw this out... I don't get it... what's D3?

BTW this is from old Elektor magazine article for a CD4060 circuit:
The CMOS output stages deliver constant current to the outputs. The current is highly dependent on supply voltage. This inherent current limiting allows the LEDs to be directly connected to these outputs without the need for additional current limiting resistors.
 
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kubeek

Joined Sep 20, 2005
5,625
Now I see where you come from with that. That is a misuse of the physical characteristic of the device, I bet it worked like that only on the really old devices.

The current limiting is not a designed in feature, it is a characterisation of the device from times when it was first made, so that you know how much the output typically sags under load. In no way is it meant to be relied upon and used as current limiting, since modern manufacturing process will increase the short circuit current significantly, as you most likely just witnessed.

See in the datasheet that the Output low current IOL min is a minimum, meaning that if the output is held at 0.5V when active low, the sink current will be at least 3.4mA, typically 6.8 mA. Nowhere does it gurantee that it will not be 50mA or 500mA. So you know it will limit at some number, but can´t know when, and should not rely on such a number.
 

kubeek

Joined Sep 20, 2005
5,625
My guess is that for the DIP version they still use the old silicon die that works like you want it to. On the smd device they used a new smaller die with much modern process, and the parameters of the transistors, especially of the outputs are much better.
 
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