ETU diagnostic tool

[exMech]

Well, he is trying to impart his wisdom to everyone. As I said before, I tried to do the electronics thing and didn't quite get the hang of it. I've been going through a bunch of online videos, and I get the idea of how a diode works; transistors is kinda like two diodes pointing at (or away from) each other. I'll have to dig up the picture... nevermind. One extra browser tab and a quick google comes up with this wikipedia article for the transistor diode model.

I am just trying to share what I know because I enjoy helping others. But to be honest, I'm not sure I'm doing any good here. Not sure if people are reading it and, if they are, if they are learning anything or are even interested in learning it. It's good to go see what others have to say on the subject but if you have to do this then it might mean I'm not presenting the information clearly enough for you to understand. All this information is available on the web but my goal was to present the information here in relation to the project at hand so people didn't have to search the web to try to figure out what is going on. I've got to admit, it's getting a little frustrating. I'm starting to think it's time just to wrap this up and move on to other projects. If it's helping anyone then I don't mind but it takes a lot of time to do these posts and if they aren't doing anyone any good then I have to wonder why I should keep doing them. If I present something and you don't understand, just ask. I have many ways I can present the information and one of them is bound to ring a bell.

**********

Okay, so the pin is being brought low. Now that you bring it up, I remember reading about that on one of the tutorial sites, but obviously it didn't take. I did a quick check on what a "pull up" resistor was, and curiously enough, it lowers the current/voltage coming in, but still keeps it above zero. It doesn't "pull" anything "up".

You are mixing a few things up here. As far as the transistor goes, the resistor connected to the base isn't acting like a pull up or pull down resistor. I didn't state that it was so in this aspect, the concept of a pull up resistor doesn't apply. In the case of the transistor, the resistor is acting like a current limiting resistor. It's still just a resistor but it's the manner it is used that makes it a current limiting resistor. We need to limit the current in the base of the transistor to protect the transistor and to also protect the microcontroller pin. It is the pin on the microcontroller that pulls the base of the transistor up (high) or down (low) according to the program running on the chip.

pull updn.JPG

This shows a resistor in each of the three different configurations discussed here. A pull up resistor DOES pull a line high. In the first drawing, if we didn't have the pull up resistor on the uC pin then the pin is said to be floating. We don't know the state of a floating pin is nor what voltage we may read on it. If we read the pin, we could read anything. We use a pull up or pull down resistor so we have the pin at a definitive state. With a pull up resistor, when we read the pin the uC will read the pin as a high level.

The middle pic shows the resistor being used as a pull down resistor. A pull down resistor DOES pull the line low. It works just like a pull up resistor except it pulls the pin low. When the uC reads the pin, it will read a low pin.

The resistor network on the tdm board (R22) is used to pull the uC pins high. The uC will read the pin high until the transistor in the optocoupler pulls the pin low. This gives the two definitive states, high and low. If you tie the input pin to 5 V without the resistor then you create a dead short and you can burn out the uC pin and you also don't have the means to toggle the input. So yes, pull up and pull down resistors do pull the pin high and low respectively and are used primarily on input pins.

The third image shows the resistor used at the base of the transistor. Here it is neither a pull up nor a pull down resistor. It is used as a current limiting resistor and works as described above.

I think you are over thinking the transistor operation so lets take another look at it.

DIODES.JPG

Here we have 5 diodes. Current flows from left to right (anode to cathode) when the diode is forward biased. Let's assume these are silicone diodes and so the anode needs to be 0.7 V higher (more positive) than the cathode voltage in order for the diode to conduct.

In the top diode, we have 5 V on both sides of the diode. Since we need a difference of voltage potential for current to flow, current will not flow through the diode.

In the second diode, we have 5 V on the anode and 4.5 V on the cathode. Even though we have a higher voltage on the anode than the cathode, we only have 0.5 V difference and so current will not flow. Remember, we said we need 0.7 V difference for the current to flow in the diode.

In the third diode we have 5 V on the anode and 4 V on the cathode. We now have a positive voltage difference of 1 V which is greater than the 0.7 V difference needed for the diode to conduct. We will now have current flow in the diode.

In the fourth diode we have 5 V on the anode and 0 V on the cathode. Again the differential voltage is greater than 0.7 V so the diode will conduct i.e current will flow. This would be like having 5 V on the anode and having the cathode pulled low by a uC.

In the fifth diode we have 4 V on the anode and 5 V on the cathode. We have a voltage difference of 1 V, however, the cathode voltage is higher than the anode voltage so the diode will not conduct.

************************************************** *******
Now let's see how this all relates to transistors. In order to have current flow in the collector of a BJT or in other words; in order to have current flow through our load, we need to have current flowing in the base of the transistor. I will show an NPN transistor and a PNP transistor below. For simplification purposes, I will show the collector for reference only but I don't show it connected to the transistor. Since we need current to flow in the base to turn the transistor on, let's see what it takes to get current flowing in the base.
************************************************** ********

NPN SYMBOL.JPG


Here is the NPN transistor symbol for reference purposes. Note the diode placement and direction.



NPN.JPG

Here I show the NPN bipolar transistor with only the emitter and base leads connected. Notice that the diode placement is exactly as it is in the NPN symbol from above. The collector is shown for reference only. Until we have current flow in the base, no current will flow in the collector or the load and so we will focus this discussion on the base current requirements.

In a common emitter configuration discussed previously, the emitter will be connected to ground (0 V) and the base will be connected to our microcontroller which we will say is a 5 V microcontroller. If the base voltage is pulled low by the uC by setting the pin low, we will have 0 V on the base and 0 V on the emitter. Since there is 0 V differential voltage from base to emitter, the diode will not conduct and so the transistor will be off.

Assuming this is a silicone diode as before, we need 0.7 V differential voltage for the diode to conduct. Now let's pull the base voltage to 5 V by setting the uC pin high. We now have 5 V at the base and 0 V at the emitter and so now the diode will conduct because we have a differential voltage greater than 0.7 V. The transistor will turn on and the load will energize. We didn't need the full 5 V to turn it on, just anything above the 0.7 V difference.

Note the direction of the current flow. Current will flow from the uC to the base of the transistor, through the diode and out the emitter to ground. From this you should be able to see why it takes a high signal from the uC to turn an NPN transistor on and a low signal to turn it off. We don't have a switch to turn the voltage on and off so we do it by creating a differential voltage (ON) or having a condition of no differential voltage (OFF).

You can also see here why we need the current limiting resistor on the base of the transistor. Once the diode starts to conduct, the uC pin will be connected directly to ground and a short will be created which will burn out our uC pin and possibly our transistor. The resistor has nothing to do with whether the transistor base is pulled high or low, but simply limits the current flow from the uC to ground.


PNP SYMBOL.JPG

Here is the PNP transistor symbol for reference purposes. Note the diode placement and direction.


PNP.JPG

Here I show the PNP bipolar transistor with only the emitter and base leads connected. As before, notice that the diode placement is exactly as it is in the PNP symbol from above. The collector is shown for reference only. Until we have current flow in the base, no current will flow in the collector or the load.

This time the emitter will be connected to the positive rail (5 V for this example) and the base will be connected to our microcontroller which we will say is a 5 V microcontroller. If the base voltage is pulled high by the uC by setting the pin high, we will have 5 V on the base and 5 V on the emitter. Since there is 0 V differential voltage from base to emitter, the diode will not conduct and so the transistor will be off.

Assuming this is a silicone diode as before, we need 0.7 V differential voltage for the diode to conduct. Now let's pull the base voltage to 0 V by setting the uC pin low. We now have 0 V at the base and 5 V at the emitter and so now the diode will conduct. It is the change in the way the diode is connected that has changed what voltage is needed to turn the transistor on or off. The transistor will turn on and the load will energize. Note the direction of the current flow and why it is the opposite of an NPN transistor. Current will flow from the positive 5 V supply rail to the emitter and through the diode. It will then leave the diode and out the base of the transistor to the uC where it will be sunk to ground. From this you should be able to see why it takes a low signal from the uC to turn a PNP transistor on and a high signal to turn it off. Again, we don't have a switch to turn the voltage on and off to the base of the transistor so we do it by creating a differential voltage (ON) or having a condition of no differential voltage (OFF).

The same explanation of the current limiting resistor given for the NPN transistor holds true for the PNP transistor shown here.

Electrical and electronics are no different. There are certain distinctions that have one referred to as electrical and the other electronic but in both cases you have a voltage potential difference which causes current to flow in a circuit to perform some function or work. So don't sell yourselves short by saying you understand electrical but not electronics. If you apply what you already know from the electrical field to your study of electronics, you will find you know more than you think.

I hope this clears things up. If not, just say so. I'd be happy to answer questions.

[Mok]

Well, don't I feel stupid. Even more than I already do.

Remember how I said the transistors on this thing were 2N4403s? Well, I said that after I had removed one from the source board and looked at one or two others. Thought all of them were the same. You all know what "assume" does, right?

I had run down to the center yesterday and grabbed the box of bad ones, thinking I might be putting something together to try and diagnose them. Started cleaning a few of them up with the Q-tips and alcohol. Was looking at one of the boards with a different layout when something caught my eye... namely, one of the transistors had a different part number. All of the boards had them. Went back to the reference board, and guess what... same thing.

Turns out Q2 and Q6 are *not* in fact 4403 PNP transistors, but 5210 NPN transistors.

So, a couple of quick edits to the diagram, which I'll get done shortly. Probably won't affect a whole lot, as it turns out, because those two sit in the "score" section along with Q1 (which is still a 4403 PNP), which we don't use anyway.

[Mok]

The updated "score" section, as promised.

[exMech]

A couple comments on your drawing.

1) You show the emitter of Q2 going to the coil of K1. While this is technically true, K1 has absolutely no impact on this circuit and so tying this to K1 creates confusion to the person reading the schematic. It also leaves the reader of the schematic wondering why it's tied to K1 and so they have to search down the schematic with K1 and try to figure out why they are tied together.

Remember the purpose of the schematic is to show the function of the circuits. The connection points of the components is left for the pcb layout. If you look at the pcb, you will notice that the place where the emitter of Q2 ties into the coil of K1 is simply a ground. It is no different than if it tied into any other ground on the board and the fact that it ties in here makes no difference but mentioning K1 adds confusion to the person reading the schematic. Since ground is the end of a circuit, change this to a ground symbol and make no mention of the K1 relay.

2) As above and for the same reasons you don't want to show R25 being connected to the emitter of Q3. Again, saying this leaves the reader confused as to why it's tied to Q3 and they have to track down the schematic having Q3 on it to try to figure this out. Remember that on these boards, emitters of PNP transistors are tied to +5 volts. The fact that it picks up +5 volts at Q3 is irrelevant and Q3 has nothing to do with this circuit. So as above, Instead of saying this is tied to the emitter of Q3 simply say it is tied to +5 volts.

With these changes made and a slight re-arranging of the components, it is easier to see the functionality of the circuit.

score_1.JPG

Here we have the components arranged so the positive voltage is at the top and the ground is at the bottom as we discussed previously.

When the uC pin 16 is low, this puts the base of Q6 low. When the base of an NPN is low, the transistor is off. This is because we have 0 V difference between the emitter and the base and as discussed in my last post, we need a voltage difference between the emitter and base to turn it on. This transistor will not affect the rest of the circuit at this point since it is off.

With Q6 off, 5 V will be seen at the base of Q2. When the base of an NPN is high, the transistor is turned on. This is because we now have a voltage difference between the emitter and the base. With Q2 on, this allows current to flow from 35 V, through R8 and R7 to ground. This sets up R8 and R7 as a voltage divider. Since some voltage will be dropped across R8, the emitter and base of Q1 will be at different levels with the emitter being higher in voltage than the base. As discussed in my last post, when the emitter of a PNP is higher than it's base by the required voltage, the transistor will turn on. That means Q1 is on and current can flow to the score connector.

When the uC pin 16 is high, this puts the base of Q6 high. When the base of an NPN is high, the transistor is on. With Q6 on, this will put the junction of R25 and R9 at 0 V by sinking it to ground through Q6. R25 prevents a dead short scenario. Because the junction side of R9 is now 0 volts, this will put the base of Q2 at 0 V. When the base of an NPN is low (at 0 V) the transistor will be off. So turning Q6 on turns Q2 off.

With Q2 off, no current can flow through the voltage divider setup by R8 and R7. Ohms law tells us that V = I * R where V is the voltage, I is the current and R is the resistance. Since I = 0 the voltage drop across R8 is 0 so with no current flowing through R8 it will not drop any voltage. This means the voltage at the emitter and base of Q1 will be the same. As discussed in my last post, with no voltage difference between the emitter and base of a PNP transistor the transistor will be off. Since Q1 is off, no current can flow to the score terminal.

As you said, you don't use this and I don't know anybody that does but it will be nice to have a complete schematic that includes everything. A failure in one of these components can lead to an issue with the power supply which can affect other circuits, so while we can ignore them for the most part, we can't forget about them completely. I will add this to my schematic.

[Mok]

With these changes made and a slight re-arranging of the components, it is easier to see the functionality of the circuit.

Oooooh, I like the layout.

As you said, you don't use this and I don't know anybody that does but it will be nice to have a complete schematic that includes everything.

I was trying to do that from Day One. I was told not to worry about it.

In any event, I got to playing with one of the bad boards I brought home from the center. Plugged in the power supply, and as the tag I put on it says, it goes into lockout the instant it gets power. Near as I can tell, the voltage regulator is good, as I'm getting 5VDC across C5 and on pin 24 of the rom/controller. Your initial diagnostic procedure said to check the voltage between terminals 14 and 16; it did not yield the 24Vrms that you were suggesting. Much more than that, it seems.

I'll sit down and look over these circuit diagrams some more; I wouldn't mind figuring this one out on my own.

[Mok]

As you said, you don't use this and I don't know anybody that does

...and I had to read through that more than once before I realized you were talking about the score connector and not the software we're using for the diagrams. Oh, well.

Makes me wonder what the deal is with the other unused connectors -- RS, Display and Sensor.

[exMech]

"Your initial diagnostic procedure said to check the voltage between terminals 14 and 16; it did not yield the 24Vrms that you were suggesting. Much more than that, it seems."

Actually, if you refer back to post #44 I said you should get around 35 V. This is DC voltage. If you look at the diagram for the power supply, you will see that the voltage at TS-16 comes from the rectified and filtered supply voltage. Without anything connected to TS-16, no current will flow through R15 and so there will be no voltage drop across this resistor and the full rectified and filtered voltage of about 35 VDC will be present at this terminal.

For reference from post #44:

3) Next, measure your voltages. Measure for 5 volts across C5. Measure the voltage between TS-14 and TS-16. With nothing hooked up, this should be between 30 and 35 V give or take some either direction.

"Makes me wonder what the deal is with the other unused connectors -- RS, Display and Sensor."

My main computer is having issues and I haven't got it back up yet. All my documents are on that computer (I do have backups) but from memory, and don't quote me on this:

RS is to trigger the rake drop solenoid.

Display - I don't know much about this one but based on the diagram, it appears data only is sent to the unit and other than that the unit is fully self contained.

Sensor - You can use infrared ball sensors with the board. When installed, the board will use the ball sensor as the trigger instead of the rake up switch. This is used in conjunction with the RS to drop the rake after the ball sensor is triggered.

Score - not sure what this does.

As you requested, I'll let you work on troubleshooting your board but just as a hint, with power only to TS-13 and TS-14 measure your voltages on the outputs of the optocouplers. This is what the microcontroller will see. I'll let you consider what the values should be and why the uC is doing what it is. Let me know if you get stuck.

[Mok]

Actually, if you refer back to post #44 I said you should get around 35 V. This is DC voltage. If you look at the diagram for the power supply, you will see that the voltage at TS-16 comes from the rectified and filtered supply voltage.

I don't know about "filtered", as I have the diagram up on my other monitor; Terminal 16 ties into the AC input line just past D15, which is taking the voltage from Terminal 13 and spitting out a half wave rectified signal.

The output from the regulator links directly to pin 9 on the controller (which happens to be the reset pin, oddly enough) after going through C6. Obviously, it goes more places than just there; I'll have to relabel that connection.

Without anything connected to TS-16, no current will flow through R15 and so there will be no voltage drop across this resistor and the full rectified and filtered voltage of about 35 VDC will be present at this terminal.

I'll check it again, but I thought it was higher than that.

My main computer is having issues and I haven't got it back up yet. All my documents are on that computer (I do have backups)

If the hard drive is intact, you should be good. If the file system hasn't been corrupted, worst case scenario, you could pull the drive (based on the premise that it's a SATA drive and not M.2/mSATA/NVME) and stick it into a docking station to hook into another system via USB.

but from memory, and don't quote me on this: RS is to trigger the rake drop solenoid.

Interesting. I've never seen such a thing. Must be long before my time. The install manual for the board lists the connector as "future optional output".

Sensor - You can use infrared ball sensors with the board. When installed, the board will use the ball sensor as the trigger instead of the rake up switch. This is used in conjunction with the RS to drop the rake after the ball sensor is triggered.

So, a light ball detect? Like what Vector does with its ball detect, except it apparently triggers the machine through the camera board ("distribution board", if you're a Brunswick employee) and ties into the reset button.

Score - not sure what this does.

Let me know if you get stuck.

Will do.

[Mok]

Okay. Not having anything else to do today, I sat down with the board in question. Here is what I have discovered.

Pins 10, 12, 14, and 16 of the optocoupler package should have 5VDC input for the transistor portion. This was verified.
Pins 9, 11, 13 and 15 have no voltage differential to be measured, as they tie into the ground rail.

Pins 1, 3, 5 and 7 are the voltage input pins (anodes) for the LED portion of each optocoupler. 1, 3 and 5 had 0 volts, while 7 had 5.
Pins 2, 4, 6 and 8 are the cathode pins for the LED portion of each optocoupler. 2, 4 and 6 had 0 volts, as expected. Given the voltage drop across the LED on opto 4, pin 8 had approximately 4 volts. This is curious, as this means the LED for opto 4 should be lit; therefore, voltage should be measured on pin 13.

Back to the input voltage; 5VDC comes into the R22 resistor pack on pin 1. For opto 2, it exits R22 at pin 3, and branches to rom/controller pin 23 and opto pin 14. 5V was measured on each.
For opto 3, 5VDC exits R22 at pin 4 and branches to rom/controller pin 1 and opto pin 12. 5V was measured on each.
For opto 4, 5VDC exits R22 at pin 6 and branches to rom/controller pin 21 and opto pin 10. 5V was measured on each.

Just to make sure, I measured input voltage to the rom/controller; it should be getting 5V in on pin 24 directly from the regulator via the top of C7 -- which it was -- but also on pin 9 after going through C6.

This is not happening.

Measurements are sporadic; I either get low voltage, or no voltage at all. Just for grins, I hooked up the component tester across the leads, and it claimed it was two diodes. Doing the same for two other capacitors worked just fine, thankyouverymuch.

I'm getting the idea that the controller isn't even starting up because it isn't getting the 5V it's wanting on the reset pin.

[exMech]

"I don't know about "filtered", as I have the diagram up on my other monitor; Terminal 16 ties into the AC input line just past D15, which is taking the voltage from Terminal 13 and spitting out a half wave rectified signal."

I'm not sure why you have "filtered" in quotes. I urge you to go back to post 35 where I discuss the power supply. As you do, think of the voltages at the following points and what the wave forms at each would be. Cathode (output) of D15, input to R11, input to R15, emitter of Q1, input of R8, and positive terminal of C2. I can go through this again but that post should help clear up your confusion. Consider the picture below following the trace from the output of D15 as a little hint as to what each of these voltages and wave forms should be.

image_12261.jpg



"The output from the regulator links directly to pin 9 on the controller (which happens to be the reset pin, oddly enough) after going through C6. Obviously, it goes more places than just there; I'll have to relabel that connection."

To understand why C6 is needed, I'll refer you to the datasheet for the microcontroller. .


reset.jpg


This says a lot but we only need to concentrate on the second sentence:

"An internal diffused resistor to VSS permits a power-on RESET using only an external capacitor to VCC."

It's common practice to tie the reset pin of a uC either high or low through a capacitor. To understand why, let's think about starting your car for a minute. You don't shift the car into gear as you are turning the ignition key to start it. If you ever tried to start a car with a manual transmission while the clutch is out and the car is in gear. The car will lunge forward and have a difficult time starting. The same thing is true for the uC. If we try to shift it into gear while the power supply and components of the uC are still powering up, we'll end up having trouble starting the uC properly. So we need to have a delay to allow the power supply to stabilize as well as the resonator and internal registers. Once these are all stabilized then we can shift the uC into gear and let it go.

As the second sentence states, this pin is tied to VSS (ground) internally through a resistor. When we add the external capacitor and tie that to VCC (positive voltage), we create a basic RC circuit. An RC circuit is a timing circuit and we have discussed them before. Because the resistor is internal to the uC, we can only change the value of the capacitor to determine how long it takes for our uC to shift into gear. When power is first applied, 5 volts will be applied to the capacitor + terminal. Because the capacitor will initially act like a short this puts the pin at 5 V preventing it from starting while it is powering up. As the capacitor starts to charge up, the voltage at the pin will start to drop. After a given amount of time, the voltage at the pin will be low enough to allow the uC to exit the reset condition and the program will start to run. Additionally, it also takes time for a capacitor to discharge so this prevents some glitch from accidentally resetting the uC.

So this is the reason that the reset pin is tied to VCC through the capacitor C6.

"I'll check it again, but I thought it was higher than that."

It's possible that the voltage may be higher than 35 V. That value will depend on the individual transformer and the loading on it. I have some that will have a peak voltage of 40 V with no load so the filtered voltage will be higher than 35 V. The more you load it the lower the voltage will be so a transformer with a higher VA rating will have a higher no load voltage. As you may recall, I did say give or take a few volts. But it shouldn't be more than approximately 1.414 times the RMS voltage of the transformer read by your meter.

"If the hard drive is intact, you should be good. If the file system hasn't been corrupted, worst case scenario, you could pull the drive (based on the premise that it's a SATA drive and not M.2/mSATA/NVME) and stick it into a docking station to hook into another system via USB."

Ah yes, yearning for the good ole days when you could easily work on your own stuff. Unfortunately, with today's tablets, that's not quite so straight forward. I have repaired my share of laptops and desktops but tablets basically require you to destroy the device to fix it. It is this reason that I have lost confidence in AMF and Brunswick as they try to keep people from working on the equipment they own. But we digress and is a discussion for another thread.

"So, a light ball detect? Like what Vector does with its ball detect, except it apparently triggers the machine through the camera board ("distribution board", if you're a Brunswick employee) and ties into the reset button."

I'll try to find the information on the RS connector. In the meantime, the attached pdf is for a different version of the board but describes the sensor connector in more detail.

I think I've addressed all your comments. As always, if you get stuck on something, just ask.

Merry Christmas to all and to all a goodnight.

[exMech]

I started my previous post before you posted your latest post, so I didn't see it before writing mine. And since I wrote mine while watching the ole TV, it took me a while to write it and so your's posted first so this is just the response to that post.

Your readings on pins 9, 11, 13 and 15 are correct. They should always be 0 V.

Your readings on pins 10, 12, 14 and 16 are also correct. With no rake up, no reset button press, no rake down, and no 2nd ball light, these all should be high.

Pins 1, 3, and 5 supply voltage to the led's when the switches are closed so with the switches all open, there should be 0 volts on these pins.

Pin 7 get's it's voltage directly from the regulator so you should always get 5 volts here.

Yes, with no voltage on the anodes of the led's you got what you expect at the cathodes of 0 V.

"Given the voltage drop across the LED on opto 4, pin 8 had approximately 4 volts. This is curious, as this means the LED for opto 4 should be lit; therefore, voltage should be measured on pin 13."

While you are reading 4 V on pin 8, as you can tell by the output of the corresponding transistor this led is off. With the rake down switch open, there is no path to ground so this led is in fact off. When you try to read the voltage on the cathode of the led, your meter is likely providing a path to ground through the meter. This is a high impedance resistance so the current through the led is very small and why it isn't lighting the led. You can verify this by measuring the voltage across R18. If there is no voltage drop across this resistor, then there is no current flowing in the circuit which means the led is off. So when you measure this with the circuit conducting and the led on, you should expect to see a voltage drop across the resistor. The 1 V drop you are seeing across the led is just due to leakage current and not sufficient to light the led.

The reason you see this voltage on pin 8 and not pins 6, 4 and 2 is because they have no power when the switches are open whereas pin 7/8 are supplied directly by the regulator.

Pins 7/8 have corresponding pins 9/10 not pin 13 so I'll take this as a typo. Irregardless pin 9 is tied to ground and so will never have a voltage across it. Pin 10 on the other hand is the pin I believe you are referencing in the last part of this statement. The above explanation should clear up why the led is off and you are measuring 5 volts on pin 10 even though you have a voltage drop across the led in opto 4.

"Just to make sure, I measured input voltage to the rom/controller; it should be getting 5V in on pin 24 directly from the regulator via the top of C7 -- which it was -- but also on pin 9 after going through C6.

This is not happening."

"I'm getting the idea that the controller isn't even starting up because it isn't getting the 5V it's wanting on the reset pin."

I explain this in my previous post which got posted after your post. That post should explain it but your thinking is backwards. The reset pin starts out high and goes low after C6 charges up and when NOT in reset mode. So when the chip is running, the reset pin should be low which is what you are reading. If you're watching this pin on a scope, you would see the transition from high to low but this happens fairly quickly so you won't likely see this using a standard meter.

"Measurements are sporadic; I either get low voltage, or no voltage at all. Just for grins, I hooked up the component tester across the leads, and it claimed it was two diodes. Doing the same for two other capacitors worked just fine, thankyouverymuch."

I'm not sure what component tester you are using to test the capacitor but you have to be careful relying on these testers as they are known to have issues. Based on your voltage readings, it looks like your cap and uC are good.

Next I would start closing switches and seeing what happens at the optocoupler outputs. It doesn't matter if the uC is in lockout or not even working. This won't affect the optocoupler outputs.

[Mok]

Measurements are sporadic; I either get low voltage, or no voltage at all. Just for grins, I hooked up the component tester across the leads, and it claimed it was two diodes. Doing the same for two other capacitors worked just fine, thankyouverymuch.

I'm getting the idea that the controller isn't even starting up because it isn't getting the 5V it's wanting on the reset pin.

Well, so much for that idea; after trying the same trick on several other boards, it's doing the same thing, so it's obviously picking up other components. Yes, I had tried it without removing the capacitor first. So, that's obviously not it. (Why can't this be a simple fix??)

I explain this in my previous post which got posted after your post.

...which you wrote before I got around to checking the other boards. I really shouldn't have this bulging gut on the front of me; I get plenty of exercise, what with my flying off the handle, jumping to conclusions...

I explain this in my previous post which got posted after your post. That post should explain it but your thinking is backwards.

Godfscking dammit. I'm always getting this backwards. Can I NOT get any of this right?!?

I'm not sure what component tester you are using to test the capacitor

See this post. Not the most comprehensive tool out there, but it works well enough. Anyway, let me sit down and work through your replies.

[exMech]

While working on another project I got frustrated with limitations of Fritzing so I fired up KiCad. I redrew the schematic and made some changes to the layout that I think will make things clearer. I still have work to do on it but take a look and see what you think.

[Wookie8662]

I prefer this version alot more.
Everything is viewable at a glance, on one page.

[Mok]

Without anything connected to TS-16, no current will flow through R15 and so there will be no voltage drop across this resistor and the full rectified and filtered voltage of about 35VDC will be present at this terminal.

I'll check it again, but I thought it was higher than that.

I did check it again, and it wasn't. About DC26Vrms. (I don't have anything that shows peak. Meters show RMS, so that's what you're getting from me.)

If the hard drive is intact, you should be good.

Ah yes, yearning for the good ole days when you could easily work on your own stuff. Unfortunately, with today's
tablets, that's not quite so straightforward.

Ah, you didn't mention that it was a tablet. If it were a standard PC, my previous statement would still apply.

In the meantime, the attached pdf is for a different version of the board

I have that in hardcopy.

Pins 7/8 have corresponding pins 9/10 not 13 so I'll take this as a typo.

Correct. Should have been pin 9. Then again, I shouldn't be picking up voltage there at all, as it's tied into the ground rail.

The above picture shows the same signal as in the second picture above. Notice now that the voltage no longer drops to 0 volts with the camel humps. This is the result of placing a filtering capacitor parallel to the load.

Okay, I was under the impression this was done with a *smoothing* capacitor. Teh Googlez tells me the function of a *filtering* capacitor is to remove unwanted frequencies from a signal.

*looks at diagram* *googles some more*

So "filtering" and "smoothing" are more or less the same in this context. To-may-to, to-mah-to... Also, I was thinking for some reason they were placed in series with the signal path rather than in parallel. *That* was why I had "filtered" in quotes, because I was thinking it hadn't been smoothed, or "filtered" yet. With the capacitor in parallel, rather than in series, I'm kinda wondering how that actually works; and PLEASE don't say "you don't need to know" because that just angers me.

re: 5V on pin 9 reset going away after startup. Just for grins I put the probe on pin 9 and terminal 14, and fired it up. The needle flickered a bit, then stopped. I'm guessing it's because the meter just didn't have time to register the 5V before the chip finished booting and brought the pin low. (It's an old Radio Shack analog. Radio Shack. That alone should tell you how old it is.)

So. Voltages and waveforms. Unless I miss my guess, one is getting half wave rectified DC24Vrms on the output of D15 before it hits the top of R11 and C2, which is the filtering/smoothing capacitor. Then again, since C2 doesn't need to be in series to do its thing, is that smoothed voltage migrating all the way back to the output of D15? And since the top of R11 is directly connected not only to the top of C2, but the bottom of R15, the emitter of Q1, and R8, those would also see the same DC24Vrms.

Or do I have that backwards again somehow?

*starts poking around with the meter*

No scope, so I can't see the waveform, but I'm getting about 26ish volts pretty much everywhere you suggested.

While working on another project I got frustrated with the limitations of Fritzing so I fired up KiCad.

Yeah, I was only using Fritzing because I had it installed, and that only because that's what Kanga suggested. I could give KiCad a go, I suppose.

I prefer this version a lot more.

It's free and open source. Download it and start to playing around.

There's also Eagle, which the EEVBlog guy was using, but it's paid software from Autodesk and requires a subscription.

In the meantime, just because, I made a couple of tweaks to the power section drawing. Yes, it's in Fritzing. Deal.