I have been playing with a uPC1237ha I have pulled out from an old working power amp trying to understand how the thing works. Here is the simple circuit I have set on the breadboard. As you can see I have created a bare minimum to just see if I get anything at the pin 6 when no input at pin 1 or 2.
The voltage as measured at pin 8 is 3.4V (by the spec). The voltage at pin 6 is 23.6V while I expect it to be smth close to 0 since pin 6 is the output of the relay driver and in the absence of input at pin 1 or 2 pins 8 to 6 are supposed to drive a relay. Is my IC broken or have I messed things up?
I never confirmed that DC or overload protection worked in the original amplifier I pulled it out from. All I know is that the amp was working OK in normal conditions. So there is a chance the IC is broken, however I can't see by the look of it that anything wrong with it (it is not visibly fried or anything like that).
The voltage as measured at pin 8 is 3.4V (by the spec). The voltage at pin 6 is 23.6V while I expect it to be smth close to 0 since pin 6 is the output of the relay driver and in the absence of input at pin 1 or 2 pins 8 to 6 are supposed to drive a relay. Is my IC broken or have I messed things up?
I never confirmed that DC or overload protection worked in the original amplifier I pulled it out from. All I know is that the amp was working OK in normal conditions. So there is a chance the IC is broken, however I can't see by the look of it that anything wrong with it (it is not visibly fried or anything like that).
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Hi reedcat,
No, I think the chip is fine.
You have to tie the unused inputs down to common, pin 4 should see just under a volt on it (AC power detect), but less than 10 VDC (abs max). If you don't tie the other inputs low, they could float high enough to make the IC think there is a problem. Ground pin 3 so it will automatically reset also.
I was going to upload the data sheet, but it is slightly too large. Darn! Follow this link to the data sheet : https://www.promelec.ru/pdf/upc1237ha.pdf There is a typical application type schematic on that page that tells you how to set it up, and even a test diagram earlier.
As a general rule, you can't leave inputs floating, you have to tie them high or low depending on the tripped state. There is a pin on a 555 timer that must be tied high or the chip won't function (and it isn't the power pin).
-Chris
No, I think the chip is fine.
You have to tie the unused inputs down to common, pin 4 should see just under a volt on it (AC power detect), but less than 10 VDC (abs max). If you don't tie the other inputs low, they could float high enough to make the IC think there is a problem. Ground pin 3 so it will automatically reset also.
I was going to upload the data sheet, but it is slightly too large. Darn! Follow this link to the data sheet : https://www.promelec.ru/pdf/upc1237ha.pdf There is a typical application type schematic on that page that tells you how to set it up, and even a test diagram earlier.
As a general rule, you can't leave inputs floating, you have to tie them high or low depending on the tripped state. There is a pin on a 555 timer that must be tied high or the chip won't function (and it isn't the power pin).
-Chris
Darn it! You are absolutely right! As soon as I had 2V at pin 4 and tied all other pins it worked like a clock. I was even able to test mute on power on. Thank you for your help. I spent all night yesterday banging my head at this. 🙂
I did look at the data sheet alright. It is the lack of experience that shows.
Here is one more question that hunts me. I hope you guys find it a piece of cake and relieve me from my misery.
Attached are the example circuit from the data sheet and my own drawing with uPC1237 power circuit and AC off detection circuit (going to pin 8 and pin 4 respectfully).
I am going to power uPC1237 with a separate low VA transformer (TR2) so that uPC1237 is not impacted by the slow start circuitry I use with the main PSU transformer (TR1). Originally I was going to connect a spare secondary of TR1 into the uPC1237 AC off detection circuit (AC3 / AC4 in my drawing) however I am thinking lately I want a short cut. Rather than send a wire from TR1 to the uPC1237 PCB why not use TR2 secondary directly and connect in the PCB itself? So my question is, can AC input in the upper part of my drawing be connected with AC input of the lower part - AC1 to AC3 and AC2 to AC4.
I suspect the answer is 'no' (since this will essentially short out D3) but I don't know what the right way to connect is. If AC3 and AC4 came from a different transformer (as shown in the data sheet example) then grounding one output of the transformer wouldn't be a problem, but if I use the same transformer to power uPC1237 and for AC off detection then I am thinking it will cause issues. What do you think guys?
P.S. I cannot feed DC power to pin 4 directly (via proper resistor) since I will be using a 470uF cap in the power circuit and it will delay AC off detection. The capacitor in AC off detection circuit is much smaller (data sheet recommends 4.7uF)
Here is one more question that hunts me. I hope you guys find it a piece of cake and relieve me from my misery.
Attached are the example circuit from the data sheet and my own drawing with uPC1237 power circuit and AC off detection circuit (going to pin 8 and pin 4 respectfully).
I am going to power uPC1237 with a separate low VA transformer (TR2) so that uPC1237 is not impacted by the slow start circuitry I use with the main PSU transformer (TR1). Originally I was going to connect a spare secondary of TR1 into the uPC1237 AC off detection circuit (AC3 / AC4 in my drawing) however I am thinking lately I want a short cut. Rather than send a wire from TR1 to the uPC1237 PCB why not use TR2 secondary directly and connect in the PCB itself? So my question is, can AC input in the upper part of my drawing be connected with AC input of the lower part - AC1 to AC3 and AC2 to AC4.
I suspect the answer is 'no' (since this will essentially short out D3) but I don't know what the right way to connect is. If AC3 and AC4 came from a different transformer (as shown in the data sheet example) then grounding one output of the transformer wouldn't be a problem, but if I use the same transformer to power uPC1237 and for AC off detection then I am thinking it will cause issues. What do you think guys?
P.S. I cannot feed DC power to pin 4 directly (via proper resistor) since I will be using a 470uF cap in the power circuit and it will delay AC off detection. The capacitor in AC off detection circuit is much smaller (data sheet recommends 4.7uF)
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Hi reedcat,
You can feed it from a supply with a large capacitor. You just put it's rectifier and small cap feeding pin 4, then take the next rectifier to the larger capacitor for the other power supply. That would work if the other supply was bipolar with a common ground. Of course, you could also feed it from the main power rectifier. It will detect as little as 0.8 Volts, and if the slow start delays it a little, who cares? Just shorten the time it takes to pull in.
The important thing is the common ground point.
-Chris
You can feed it from a supply with a large capacitor. You just put it's rectifier and small cap feeding pin 4, then take the next rectifier to the larger capacitor for the other power supply. That would work if the other supply was bipolar with a common ground. Of course, you could also feed it from the main power rectifier. It will detect as little as 0.8 Volts, and if the slow start delays it a little, who cares? Just shorten the time it takes to pull in.
The important thing is the common ground point.
-Chris
Let me make sure I understand you. Are you referring to connecting both pins to the small transformer (TR2) like the attached circuit shows? Or did I misunderstand you?
The main amplifier PSU is bipolar with common ground. I don't want to use it to power uPC1237 because I think slow start will interfere with uPC1237 ability to mute speakers on power on. If that is not the case then I will gladly use it - one less transformer, my chassis it already crammed.
Feed what? The uPC1237 itself or the AC off circuit?
I think I completely failed to understand the significance of this statement since I still don't understand how you suggest to power uPC1237 from the main PSU.
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Hi reedcat,
Your picture is correct. The reduced AC voltage will power the AC detect just fine. I guess what you really want is a "housekeeping transformer", but you don't need one.
Assume the uPc1237 is affected by the slow start. As far as it's concerned, power has not been applied yet and so it remains off. Soft start trips and only now can the chip work, so it begins a time out before activating the relay. This really has no effect beyond increasing the length of time before the speaker relay contacts close.
The uPc1237 will not connect the relay until it has powered up and gone through the timing period. In short, you are worrying about nothing. Try it.
-Chris
Your picture is correct. The reduced AC voltage will power the AC detect just fine. I guess what you really want is a "housekeeping transformer", but you don't need one.
Assume the uPc1237 is affected by the slow start. As far as it's concerned, power has not been applied yet and so it remains off. Soft start trips and only now can the chip work, so it begins a time out before activating the relay. This really has no effect beyond increasing the length of time before the speaker relay contacts close.
The uPc1237 will not connect the relay until it has powered up and gone through the timing period. In short, you are worrying about nothing. Try it.
-Chris
Alright, I think I understand now why slow start will not affect protection. I still don't understand however how I can power uPC1237 from main power supply.
Attached are two designs. One is for using the spare secondary - no questions there. The other one is my attempt at connecting w/o spare secondary. See if there are any issues there.
One difference of this design with the design in the data sheet is that in the data sheet the first out of the transformer goes in the pin 4 circuit while the second out of the transformer is grounded. In my case the second out goes into rectifier bridge and then grounded. Is that acceptable?
Attached are two designs. One is for using the spare secondary - no questions there. The other one is my attempt at connecting w/o spare secondary. See if there are any issues there.
One difference of this design with the design in the data sheet is that in the data sheet the first out of the transformer goes in the pin 4 circuit while the second out of the transformer is grounded. In my case the second out goes into rectifier bridge and then grounded. Is that acceptable?
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Hi reedcat,
Diagram #2 is correct, but you are introducing variables before you have a good understanding of the core question. See this attachment for the basic connection. Of course you need to add resistors between the diode and capacitor. I just wanted to show you the basic idea without any other things to clutter it up.
-Chris
Edit:
Sorry, something turned the sketch 90°
Diagram #2 is correct, but you are introducing variables before you have a good understanding of the core question. See this attachment for the basic connection. Of course you need to add resistors between the diode and capacitor. I just wanted to show you the basic idea without any other things to clutter it up.
-Chris
Edit:
Sorry, something turned the sketch 90°
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Alright, I get it now why the diodes should be connected in a series like that. The second diode will prevent larger capacitor discharging into the smaller one. Thank you Chris!
I posted the complete schematic of my uPC1237 circuit in this discussion. I would still like to use a diode bridge somehow since I use it to power the transistors of the overload detection circuit, plus my 12VAC would produce 18VDC, just right for my 18V rated relay. I don't know how to do it properly using a bridge. So the problem is not completely solved.
I posted the complete schematic of my uPC1237 circuit in this discussion. I would still like to use a diode bridge somehow since I use it to power the transistors of the overload detection circuit, plus my 12VAC would produce 18VDC, just right for my 18V rated relay. I don't know how to do it properly using a bridge. So the problem is not completely solved.
Hi reedcat,
Don't complicate things too quickly. Become comfortable with the basics before dressing it up with changes.
Hi thimios,
It is highly unlikely that these chips would be re-marked products. There are no low voltage equivalents or anything else that would be a functional equivalent that has the same pinout. This would be true of most specialised ICs, unlike op amps and STK power packs and the like. You might however run into a 4966 that was really a 4066 (much lower voltage but otherwise the same beast).
-Chris
Don't complicate things too quickly. Become comfortable with the basics before dressing it up with changes.
Hi thimios,
It is highly unlikely that these chips would be re-marked products. There are no low voltage equivalents or anything else that would be a functional equivalent that has the same pinout. This would be true of most specialised ICs, unlike op amps and STK power packs and the like. You might however run into a 4966 that was really a 4066 (much lower voltage but otherwise the same beast).
-Chris
Hi thimios,
Yes, of course. My point was that these kinds of parts are unique and there would be zero point in relabeling them in the first place. Only a complete idiot would spend the money on this enterprise!
-Chris
Yes, of course. My point was that these kinds of parts are unique and there would be zero point in relabeling them in the first place. Only a complete idiot would spend the money on this enterprise!
-Chris
the only risk in buying these unique parts is the "failed inspection and out the back door" variety that get sold by the criminals with no conscience.
I'd say they would have to be dumpster diving. All the semi companies today destroy the non-conforming product to prevent this very thing. It dilutes their reputation for quality, so that is a pretty strong motivator. Besides, they aren't stamped or labeled until they pass testing.
-Chris
-Chris
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