Construction Diary -- Part 1, The LED matrix/display

[Andrew Davie]

Very nice, Andrew. I see that there is a manual available for this scope at http://cgi.ebay.com/Tektronix-453A-4-Os ... dZViewItem if it does not come with the scope that you have purchased. There is also an interesting web page about the Model 453 scope at http://www.diyguitarist.com/TestEquipme ... nix453.htm .

It doesn't come with a manual, but I found a free online scan of it. I had also noticed that page with the review, one of the reasons I bid on the scope in question

[Roland]

BTW I sympathise with you entirely about the trauma of going to Dick Smiths, it also applies to Jaycar, Altronics, etc. They all seem to be moving to silly little gadgets and toys and it is becoming more and more difficult to put something together from scratch in this country (I wonder if that is happening overseas as well).

As far as the UK is concerned - about the only chain you can buy components over the counter is Maplins:

http://www.maplin.co.uk/

They do a pretty good web/telephone ordering service as well - though the real benefit to me is that I have a couple of fairly local branches I can use. They do stock superbright LEDs - but in tiny quantities (like 2 of each type) - so if you use them to build your array you need to mailorder. Variety is good - but prices are less so.

There are still independant electronics stores - one of the best I know of is NR Bardwell:

http://www.bardwells.co.uk/

fortunately they do internet ordering too (and probably phone) and have a huge range of surplus stock at good prices.

Another company I have had very good service (and prices from) is:

http://www.bowoodelectronics.co.uk/

After Andrew's success buying LEDs from Hong Kong - I tracked down the same Ebay seller (topbright88) and bought some too. Unfortunately they came back as 3mm rather than 5mm and as I'd have to return them at my expense I'll keep them anyway (must remember to try them out sometime soon too . I did buy some from a dealer in the UK too (via Ebay- mjcomponents). Took ages to arrive - but the price was good and they are a nice orange - much like the historical neon(?) tubes apparently used in the original mechanical systems.



Roland.

[DrZarkov]

In Germany there ist "Conrad" (www.conrad.de) with shops in many towns and a very good web-shop. I prefer a small local dealer here. No Webshop, the owner is an elder man and his wife. Of course they start selling "toys and gadgets", too, but they still have everything I need, for good prices.

[Andrew Davie]

The voltage across the transistor from collector to emitter (Vce) will vary between your supply voltage (when the input voltage to the base is 0) to about 0.2 when the input voltage is at it's max (actually this is only if it is designed to be fully on i.e. saturated at this level but since we want white (max input voltage) to be LEDs fully on this is a reasonable assumption).

The voltage at the emitter of the transistor will vary between 0 and 2V because at the base it varies between 0 and 2.7 and there is (about) .7 volts dropped across the base to emitter junction. This means there is a maximum voltage across the emitter (gamma) resistance of 2 volts.

After a bit of a break, I've returned to this, and wired up the boards and rectifiers all together again. Like before, the LED array lights up with no input signal. This time I measured the voltage across the leads of the transistor, and I got the following

from base to emitter, 0.53V
from base to collector, 0.48V
from collector to emitter, 0.26V

I don't know what this means, really. One would assume that there is only 0.5V or thereabouts running through my LED matrix. Obviously not, though -- because all the LEDs are on. In fact, I can measure the voltage... hang on

OK, so the LED array is connected to my stable, rectified 12V supply which actually puts out 11.8V DC. If I measure across the LED matrix, I get 6.94V. What is this really telling me... that my LED matrix is using (11.8 - 6.94)V = 4.86V. Divide by 3 that gives each LED using 1.62V which I consider unlikely, but then again the LEDs aren't shining full intensity.

SO, where to from here? Why are my LEDs lighting at all. With no input signal, I would expect the voltage from base to collector of the transistor to be... well, 0.

-- One thing that just occurred to me -- I have NOT recalculated/replaced the resistors in the gamma circuit -- I am still using the originals. I must do that next, as a matter of priority -- that is, replace the three with a single one. A bit of a bummer, as they're soldered in. --

[gary]

The voltage dropped across the LED matrix is the voltage that you measured (6.9V). The 11.8-6.94=4.86 is the voltage dropped across the rest of the circuit. Does your measurement include the balancing resistors? If so there is a voltage dropped across that as well. Measure (if you can) the voltage across one string of LEDS only (i.e. NOT including the resistor) and divide by 3 to get the actual voltage across each LED. Technically, if this is less than the forward voltage of the LED it should not light.

Please measure the voltage from base to ground, and from emitter to ground.

It would appear that the transistor is being biased into saturation so my guess is that the base voltage (and hence the base current) is not zero (which it should be if there is zero input to the circuit). If the base voltage is higher then we need to check the rest of the circuit (in particular measure the voltage to ground at pin 3 of the second op-amp).

BTW have you got those wire links in? (easy to overlook, especially the one under the IC socket).

Anyway, these tests may give us another clue.

[Andrew Davie]

The voltage dropped across the LED matrix is the voltage that you measured (6.9V). The 11.8-6.94=4.86 is the voltage dropped across the rest of the circuit. Does your measurement include the balancing resistors? If so there is a voltage dropped across that as well. Measure (if you can) the voltage across one string of LEDS only (i.e. NOT including the resistor) and divide by 3 to get the actual voltage across each LED. Technically, if this is less than the forward voltage of the LED it should not light.

Please measure the voltage from base to ground, and from emitter to ground.

It would appear that the transistor is being biased into saturation so my guess is that the base voltage (and hence the base current) is not zero (which it should be if there is zero input to the circuit). If the base voltage is higher then we need to check the rest of the circuit (in particular measure the voltage to ground at pin 3 of the second op-amp).

BTW have you got those wire links in? (easy to overlook, especially the one under the IC socket).

Anyway, these tests may give us another clue.

The voltage across a single string of LEDs is 6.91V, giving a per-LED voltage of 2.3V. This doesn't include the balancing resistor. With the resistor the voltage is 7.04V.

Since I don't really know which is the emitter and which the collector on the transistor -- I will go look that up now -- the measurements are for all transistor to ground possibilities.

middle pin of transistor (base) to ground = 1.23V
"left" pin of transistor (closest to LEDs) (emitter?) to ground = 0.72V
"right" pin of transistor (closest to gamma) (collector?) to ground = 0.69V

I note that shorting the base and right pin (collector?) causes the LEDs to brighten by about double. The LEDs, in case I hadn't mentioned before -- are shining, but only dull shining, not full-unbearable-brightness.

I did put those 'other side' wire links in, and have double-checked they're there and soldered OK.

Thanks for the help!

[gary]

Mapping your results to what the specs of the transistor BUT11:

left = base (LED pin side) 0.72V
middle = collector 1.23V
right = emitter 0.69V

(BTW are you actually using a BUT11? I find them difficult to obtain locally).

So... The voltage at the base looks perfect. i.e. this corresponds to black as stated in the doc for the circuit. This voltage corresponds to the cut-off voltage of the transistor and so there should be little current flowing through the transistor and the LEDs should be off (or very dim).

Unfortunately, I cannot understand the other voltages (collector and emitter). If the voltage drop across the LEDs + resistor = 7V then the voltage at the emitter of the transistor to ground *MUST* be 5V if your supply voltage is 12V. Can you re-measure these please? also measure your supply voltage in case loading is causing this to drop. Also, I can't see how the LEDs can be on since they seem to be well below their minimum Vf.

(I am assuming that you don't have another temporary limiting resistor in the circuit as I previously suggested).

Have you tried applying an NBTV signal yet? It would be interesting to see if you are getting modulation.

[Andrew Davie]

Mapping your results to what the specs of the transistor BUT11:

left = base (LED pin side) 0.72V
middle = collector 1.23V
right = emitter 0.69V

(BTW are you actually using a BUT11? I find them difficult to obtain locally).

So... The voltage at the base looks perfect. i.e. this corresponds to black as stated in the doc for the circuit. This voltage corresponds to the cut-off voltage of the transistor and so there should be little current flowing through the transistor and the LEDs should be off (or very dim).

Unfortunately, I cannot understand the other voltages (collector and emitter). If the voltage drop across the LEDs + resistor = 7V then the voltage at the emitter of the transistor to ground *MUST* be 5V if your supply voltage is 12V. Can you re-measure these please? also measure your supply voltage in case loading is causing this to drop. Also, I can't see how the LEDs can be on since they seem to be well below their minimum Vf.

(I am assuming that you don't have another temporary limiting resistor in the circuit as I previously suggested).

Have you tried applying an NBTV signal yet? It would be interesting to see if you are getting modulation.

The salesman at the electronics store substitued the transistor with an "equivalent" -- marked 414/BDX54C. I forgot that he had done this.

I have re-measured the voltages at the points listed, and get the same measurements for the transistor pins.

I include pictures of my rats nest to help make sense of things. There are a few things I'm uncomfortable with -- most particularly the fact that the negative of the 12V circuit rectifier is not connected to ANYTHING.

Having re-measured the voltage of that circuit (across the + to -) I get 11.81V. I'm very sure this is stable and doesn't change significantly -- if at all -- under load.

Since I re-measured the voltage across the LED matrix, 7V, and I have 11.81V in the +Ve of the power input, yet I'm getting 0.69V from emitter to "Ground" this says to me I've just not understood the way that "ground" operates, and this is the problem...? Perhaps the pin positions of the transistor is differently ordered... or it's not a suitable transistor at all...?

The LEDs, by the way, are most definitely ON! This can be seen by the bright bit at right of the ratsnest3 picture.

[gary]

Hmmm curiouser and curiouser.

The replacement transistor is a Darlington Transistor, whilst it's max ratings are fine for this application. and the pinouts are the same, the cutoff Vbe for this transistor is 1.4 not .7 as for the BUT11. This means, as I see it, that black level would now need to be offset to 1.4V. It's much higher gain may also have some ramifications.

None-the-less I can't for the life of me see how the LEDs are on with the measurements you are getting. It seems to be against the laws of physics.

The negs (I assume you mean ground or 0V here) should all be common although in your case it doesn't seem to make any difference because your bridge diodes are effectively in parallel and so you are getting a return through the diodes of the other bridge. Note that the second bridge is unnecessary as you could have taken your rectified voltage from the output of the other bridge.

[Andrew Davie]

The negs (I assume you mean ground or 0V here) should all be common although in your case it doesn't seem to make any difference because your bridge diodes are effectively in parallel and so you are getting a return through the diodes of the other bridge. Note that the second bridge is unnecessary as you could have taken your rectified voltage from the output of the other bridge.

OK, now the first thing I think I need to do is understand the difference between ground, and "the other terminal". I understand with an AC source, I have two places to "put my wires", one is a positive and the other is a negative. But in reality it doesn't matter and this labeling is misleading, because the voltage at each terminal swings through a full sine wave from full-positive to full-negative and back.

Now with a DC source, I think it does matter. If I have +12VDC then one of the terminals (red/positive) will have 12V across it, and presumably the other will have 0V. Here I'm getting confused, really. If I measure the voltage across them, it is 12V. This is how I'm measuring my rectified voltage. Do I assume/say that the 'negative' terminal is at 0V?

And this is where the question REALLY starts. Is this '0V' the same as 'ground'? That is, whenever I see GND on a circuit diagram, can I connect it to 0V on the power supply. On ANY power supply?

Because that's what I've gone and done. All the GND connections route to the 17V rectified DC's 'black/negative' terminal. Is this correct?

[Andrew Davie]

Well, my oscilloscope arrived today. It looked incredibly complex, but when I started to play with it, it actually all became fairly simple to use. Of course I'm only using the most basic of features -- but I think I've been able to test it enough to say that it's working rather well. Not bad for what must be a piece of 30 year old equipment.

I hooked up the club CD that I have and played track 2 (which is a simple white screen (I think)) and hooked up the oscilloscope 'probe' to the output of each channel. This allowed me to quickly identify which was the sound and which was the video (I didn't previously know for sure). The sound channel shows varying sine waves, increasing in frequency, etc... depending on the track.

The video channel has a fairly distinctive pattern (see picture). Using the various scaling and positioning knobs, I got to the point shown where I measure the duration of each 'burst' at 2.5ms (which comes to 400Hz which then becomes 400/32lines --> 12.5fps). So that's a good test to show that the oscilloscope is fairly well calibrated (at least in the time axis).

The video signal appears to have a peak-peak reading of something like 32mV, with the main signal line looking like it's hovering around 28mV. I don't understand these figures yet, but I'll have a think about it.

Overall, considering I've only had this thing for an hour or two, I'm really very very pleased with it. A good 'cheap' eBay purchase once again, and I'm sure this will be very useful as I learn more about how all of this stuff fits together.

One of the first things I did was measure the rectified DC output of my two AC->DC boards. Interestingly, there was no detctable ripple in either of them. The one with the transistor-thingy, I didn't expect ripple. The other (17V one) I did, but not a thing I could do with the vertical scale knob made that line wobble. I might have to come back to this one. Later note: I did come back to this, and with a different probe I did find significant ripple -- see http://www.taswegian.com/NBTV/forum/vie ... ?p=422#422

Now, though, I have no excuse to put off trying to diagnose what I've done wrong with the NBTV circuit setup.

[gary]

OK, now the first thing I think I need to do is understand the difference between ground, and "the other terminal". I understand with an AC source, I have two places to "put my wires", one is a positive and the other is a negative. But in reality it doesn't matter and this labeling is misleading, because the voltage at each terminal swings through a full sine wave from full-positive to full-negative and back.

Now with a DC source, I think it does matter. If I have +12VDC then one of the terminals (red/positive) will have 12V across it, and presumably the other will have 0V. Here I'm getting confused, really. If I measure the voltage across them, it is 12V. This is how I'm measuring my rectified voltage. Do I assume/say that the 'negative' terminal is at 0V?

None of the terminals have absolute values, they are relative to each other. What you are measuring is the "potential difference" between the two terminals. When you place your meter across these terminals what you are measuring is the potential difference between the two i.e. you can say that one terminal is, say +12v higher than the other. So you can use the convention that the lower terminal is 0V and the higher terminal is 12V. You could just as easily say that the lower is -6V and the higher +6V but your meter would still read 12V because that is the voltage (or potential difference) between the two terminals.

This applies to AC as well as DC. The term ground is just a convention derived from the common practice of attaching the lower value terminal to earth ("the ground"). In this case however the "ground" may not actually be connected to earth (although generally it is), it is just a common potential to which all other potentials are measured.

So we connect all the lower potential terminals together to get a common "ground" potential so that your 12V terminal is 12V relative to this common connection and your 17V is likewise 17V relative to this common connection and the difference between the 17V and 12V terminals will be 5V (I know that that sounds obvious but if they don't have a common ground it may NOT be the case - very bad). If this is not done then the potentials are "floating" with relation to each other and thus the difference between them is indeterminate.

And this is where the question REALLY starts. Is this '0V' the same as 'ground'? That is, whenever I see GND on a circuit diagram, can I connect it to 0V on the power supply. On ANY power supply?

Because that's what I've gone and done. All the GND connections route to the 17V rectified DC's 'black/negative' terminal. Is this correct?

Yep, with respect to your DC voltages, that is correct.

[Andrew Davie]

The video signal appears to have a peak-peak reading of something like 32mV, with the main signal line looking like it's hovering around 28mV. I don't understand these figures yet, but I'll have a think about it.

I note that these measurements were made with just the probe, no ground-connection from oscilloscope to circuit ground. I was trying to measure the DC voltage of the rectifying circuits and was getting very odd low results. This appeared to ... well, rectify, when I connected ground on the oscilloscope to 0V on the transformer (at least for the "17V" circuit. When I went to connect the probe to the 12V rectifing circuit, I got blue sparks and figured... hey, time for bed.

I guess I probably should be turning things off before making connections, just to be safe/sure.

[gary]

If you are measuring the DC output of the bridge rectifier you do NOT want to place the 'scope probe's ground (black lead) on the 0V of your transformer (if that is what you are actually doing). You want to put it on the ground point of the bridge rectifier.

[Andrew Davie]

If you are measuring the DC output of the bridge rectifier you do NOT want to place the 'scope probe's ground (black lead) on the 0V of your transformer (if that is what you are actually doing). You want to put it on the ground point of the bridge rectifier.

Not quite. I was connecting a separate wire from 0V on the transformer to the ground connection on the oscilloscope. I was connecting red and black of the oscilloscope lead to the + and -, so to speak, outputs of the rectifying circuit. I did (this morning) power things up with things connected this way, and after a few seconds noticed a burning smell so quickly switched everything off. I'll be more careful in future.

It's all black magic to me at the moment, though I know a lot more than I did when I started. Many thanks for the help/advice.