I have a few questions about transistors in a car amplifier! - Page 1
Forum section: electronics / beginners

author: RokasBum
author_rank: Newbie
timestamp: February 13, 2025, 03:36:25 pm
content: [ Attachment Invalid Or Does Not Exist ]Hello, fellow travelers of reality! Today I come to you with a question that has been bothering me for a little bit of time. I added the pictures with isolated traces of this IC controller chip of a car audio amplifier. As you can see there are two different sections; upper one with data communication with head unit and lower one with voltage sense from 12v battery and power on signal. Keep in mind everything is digital I think???On the upper left side of both sections, you can see the voltages as described by the manufacturer in on state of the amplifier. On the upper right section there is also an oscilloscope view of the data communication.I guess we can start with the lower section maybe it will be easier for me to understand what is going on. So, as far as I can tell the logic behind this is that the COM IC checks if there is battery power coming in (lower section, upper left side). It goes through a coil then a resistor then a diode and sooner or later ends up at the base of the transistor (Q005). The voltage of the battery opens (or doesn’t allow to open) the transistor and allows (or doesn’t) the current to flow. As long as 3.3 v is generator on the power supply section of the circuit, we’re going to have 3.3 volts. That 3.3 v goes inside of the chip at BATT-IN and continues all the way to the collector of the transistor? Battery either makes the connection or doesn’t. According to the manufacturer, the voltage is zero, therefore, the transistor is always opened making it open circuit and no current flow in the circuit or is it always closed and making straight connection to ground, but since voltage is potential difference and there’s nothing in the circuit the voltage is 0?The same thing is happening on the right side, I think. 12 volts come in from the top at Q004, the base Is connected to p-on rail. When head unit gives low voltage (below 1 volt) that means that ACC ignition key is turned? As long as that condition is met the current flows from emitter to base and collector. Collectors current turns on the next transistor Q006 and once again as before opens the circuit or closes it? I’m so confused by this.Now jumping back to the first section we see that when the amp is on, those three signals In the upper left show 3.3 volts. Judging by manufacturers data, the data rail is always on and at battery voltage. This leads me to believe that COM IC sends communication signal through its AUDIO-SO rail and receives data through AUDIO-SI and AUDIO-SIO-IN rails. What I don’t understand is when are the transistors opened and when are they closed? It seems like the COM IC sends its signal through Q002 transistor and receives communication through Q003 transistor. Do I understand that correctly? I think that when either head unit or COM IC send out 0 volts the other side picks it up? When everything is operational, it’s always at 12 volts?I’m pulling my hair out and none of this makes sense. Why is it that lower section has constant 0 volts and upper one is always operational? If lower section had no battery or had an open circuit it would believe that it was still operating? WOOOT! Thank you for your help!

author: RokasBum
author_rank: Newbie
timestamp: February 13, 2025, 03:37:19 pm
content: https://i.sstatic.net/zO2YeBY5.pngCan't seem to attach the image sorry ^^

author: PGPG
author_rank: Frequent Contributor
timestamp: February 13, 2025, 04:28:24 pm
content: First.I don't like when someone describing transistors uses terms open/closed.When switch is open then current can't go through it and when closed current can go through it. Do you understand transistor the same way or totally opposite? Hard to guess.I use always therms switched on (conducts current) and off (not conducts current).Second.3V3 and 12V looks like being simply supply voltages. So your sentences that they (voltages) come from transistor (I don't want to read it once more to search where but remember something like this) are wrong.Third.I don't know what is the device at this schematic.At BATT-IN through R021 there is 3V3.But Q005 is switched on so shorts this point to GND giving there 0V (all voltages are measured against GND - you seem to be not sure about it).Why Q005 is switched on? Because 12V supplies its base through resistor and Zener diode.So what is the function of Q005?The COM IC is probably supplied by 3V3. So when it is working 3V3 have of course to be.Q005 allows IC to know if 12V supply exists or not.Fourth.Try to understand the same way other transistors. Notice that some of them are not transistors but so called 'digital transistors' having integrated resistors in the same case.

author: RokasBum
author_rank: Newbie
timestamp: February 13, 2025, 06:45:48 pm
content: Quote from: PGPG on February 13, 2025, 04:28:24 pmYeah thanks for the clarification. I understand that the transistors can be open, half open and they can vary talking about how much current they let through. With these transistors it seems like they're either on or off, but I could be wrong.Quote from: PGPG on February 13, 2025, 04:28:24 pmLet me link the full schematic, maybe that will help clear everything up and make sure that I understand everything correctly and haven't skipped something:https://faq.out-club.ru/download/common/MITSUBISHI_CV-0MW7B51.pdfYes they are supply voltages, but they're also connected to the signal rails which is kind of confusing to me. Talking about the lower section of my schematic, it seems like both BATT-IN and POWER-IN have 0 volts because they are shorted to the ground through the transistors Q005 and Q006 and as long as there is power coming from battery and P-ON rail, the transistors will be closed. If there is no power coming from these rails, the BATT-IN and POWER-IN rails will read 3.3 volts since the Q005 and Q006 transistors will be open? Do I understand this correctly?Quote from: PGPG on February 13, 2025, 04:28:24 pmIsn't this redundant? If there is no 12 v, the IC would not turn on, why does it need to check if battery is there? And not only that, it does it through a transistor instead of checking 12v directly? I guess it's because the IC operates at lower voltages?Quote from: PGPG on February 13, 2025, 04:28:24 pmAre these digital transistors operated by AUDIO-SO for Q801 and DATA for Q001? When AUDIO-SO shorts to ground (IC chip draws current from Q801) it shorts Q002 to ground and that sends DATA signal to ground also? Is that how DATA rail knows about 0v spike from the IC? Alternatively when DATA shorts to its ground on the head unit it closes Q001 which activates Q003 and it shorts AUDIO-SI and AUDIO-SIO-IN to the ground and gives IC a spike to 0v instead of 3.3v? Do I understand this correctly?Thank you for your reply!

author: PGPG
author_rank: Frequent Contributor
timestamp: February 13, 2025, 08:35:58 pm
content: Experiment with inserting pictures to your post. I tried it once. It is horrible, but probably can be done.If remember well I have added 3 pictures (with first - big problem, with next - less) then after posting I sow that only pictures 2 and 3 were visible (who did that software ?!!!)I sow people writing that they post without pictures and then using Edit add them, what works better (if I remember well what I have read - didn't tried it myself).The problem is that I am very, very busy now. We are finishing something that so called 'dead line' had 2 weeks ago and we need at least also one week (live tells you that if you think you need a week than you need 3 or more).I have opened your whole pdf. No time to analyse it ever. It is your task.Quote from: RokasBum on February 13, 2025, 06:45:48 pmI also think so, but don't look into schematic to ensure myself (hurry, hurry - it is 21:30 and I'm still at work).Quote from: RokasBum on February 13, 2025, 06:45:48 pmYes.Quote from: RokasBum on February 13, 2025, 06:45:48 pmAre you sure?May be 12V comes form the other source so it can be not present even 3V3 is.If 3V3 comes form 12V than it is possible that device has some info to safe in Flash and starting to write wants to know if it will have power for enough time. So if 12V is 12V than there is enough charge in capacitors that even supply will be disconnected I will have enough time to finish just started flash write. Such write can need for example 50ms and if broked during writing you get unspecified values that can next time confuse you. If you know that each write process was correct than you can protect yourself against breaking supply at any time.When you do upgrade for some devices you always see "Not disconnect supply during upgrade". How can you ensure this if you not have UPS?This is stupid approach. For all our devices you can break supply during upgrade how many times you want and device will always end with old version or new and never with corrupted software.Our devices are supplied from 12V. How do you think: what for microcontroller in our devices gets info if supply is over 10V or not?Quote from: RokasBum on February 13, 2025, 06:45:48 pmSorry, will not be going into it. Just look at them as transistor with two resistors.But the fact that they were used suggests that they work as digital - only on or off. There is no rationale to use them in analog circuits.

author: wasedadoc
author_rank: Super Contributor
timestamp: February 14, 2025, 12:21:52 am
content: "open" and "closed" have opposite meanings in different contexts. Closing the contacts of a switch lets current through. Closing a farm gate stops cows passing through.

author: coromonadalix
author_rank: Super Contributor
timestamp: February 14, 2025, 05:36:13 pm
content: call or see the transistors like a variable valve ...