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Hi guys, got a Morley Bad Horsie Steve Vai 1 rev.3 made in USA dating to 1998 up on the bench. Diagnostic shows LDR-1 to be malfunctioning. Morley wants $10 for a LDR from their e-bay store, but Digikey has LDRs starting @ 80 cents each. The problem: I can't tell from the schematic what the specs are for this part, it just says 'LDR-1 150-200.' Can anybody tell me what the specs of the part I need are? Voltage on pin 13 of U1 has to drop from 8v ON to 1v OFF across R1 in order for the device to function properly. Schematic below:
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"If you can get the smoke back in the amp, it will work." -
I don't have an answer about the specs, but a few years ago i had to fix one of these quickly for a guy on the road. What worked for me was I switched the two photocells. The one that wouldn't turn the pedal on worked fine for the wah section and the one from the wah section worked fine to turn on the pedal. -
My thought was to jump the LDR thereby turning the pedal on, but the guy doesn't need it right away. He brings me lots of work so I figured it's worth a shot to save him $9 on a part. The LDR only showed a drop from 8.35vdc-7.6vdc in the circuit even when I hit it with an ultra-brite multiple LED bicycle light. Nowhere near the 1vdc needed to turn the circuit on. Does the input pin13 of U1 draw current? If not I could just use ohm's law to figure out the bottom dcr value of the voltage divider. I can test the component to get a 'dark' resistance readingbut I don't really want to spend any more time on it, just gonna order the part from Morley at a 1000% mark-up. I'm in the wrong business...Originally posted by 52 Bill View Post"If you can get the smoke back in the amp, it will work."Comment
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No, that CMOS gate draws almost zero current. If its input pin failed it would most likely fail short to ground anyway. If you were in doubt, you could remove U1 and see if the voltage started moving. But you could also just remove the LDR and measure its resistance alone - light and dark.
But let us assume the part is indeed the problem.
I charge a dollar a minute in my shop. And if I had no photoresistors in the shop, I could look at that 80 cent item, but if it takes me 10 minutes to decide which one and work up an order, why bother? I might as well just buy the expensive part. If I were stocking my shelves it would be a different matter, but this is one repair.
The circuit is just a logic gate. The voltages are not critical, the resistance is not critical. If I had to guess, I'd think that 150-200 meant the on resistance. I'd be surprised if the voltage had to go as low as 1 volt to toggle this gate, but if so, so be it. R1 and LDR1 form a simple voltage divider. All the LDR needs to be is down under maybe 500 ohms lit, and something high when dark. I;d guess a reliable logic high would be 6 volts maybe. (They toggle near the middle, but the standard transition voltages are for reliable operation.) SO if dark is even just 10k, that puts the voltage divider output at 6v. 100k? That makes about 8.5v. And most photocells have a higher off resistance than that anyway. SO worrying about the dark resistance is probably moot, and the on resistance needs to be relatively low. SO if I were to try to invent a substitute, I;d probably look at the list of available ones and choose the one with lowest lit resistance, assuming all of them have more than enough high resistance.
And you can just short across the LDR to see if the circuit toggles like it should. Because even though I think it rare, the pin 13 could be shorted to the plus rail.
A long time ago I bought an assortment of photocells from I think Jameco, just a couple of these and a couple of those, etc. Other folks sell them too. And over the years that little drawer full has served me well. I have made up new trem bugs for Fenders and others. I have fixed Morley things. Every now and then, there they are to bail me out of some situation. If you plan to order an 80 cent one from somewhere, might as well get a few. This sircuit is on the low end of resistance, in a tube amp, it might matter moire that the off resistance be over 1 meg or something. But in general, a very few dollars might work out in your favor, and if you never need another one, you are not out very much.Education is what you're left with after you have forgotten what you have learned.Comment
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Thanx Enzo. The chip is fine. Circuit functions when I jump the resistor and toggle the gate. I haven't ordered the part yet so I will order an assortment from Digikey(unless you know a better supplier.) I'll just go with a high 'dark' resistance and as low of an 'light' resistance as I can find and get a handful. I imagine that these parts in the photo-wah pedals see a lot of abuse being right next to the opening on top of the pedal enclosure. I tried cleaning the lens of the photocell with rubbing alcohol and a q-tip and that made no difference. My question is, what are some of the standard values you have used in tube amp tremolo circuits? I've only ever serviced oscillator-type tremolo units(usually a bad cap or resistor drifted way out of spec, maybe a dirty tube pin contact) so I'm not familiar with the range of parts I may need for these applications. Could you give me a suggestion for a parts list to keep on hand? People really like it when you have their amp fixed and out the door within 24 hours. I've gotten more than a couple $20 tips for same-day or next day completion..Originally posted by Enzo View Post
One other question: I've gotten inquiries about modding pedals, but the pedal forums I've visited seem a little less tech oriented(schematics are few and 'sketchy') and more oriented towards the magic capacitor raving type of threads. Do you know of any serious pedal-mod forums where accurate schematics and technical discussion are the norm?"If you can get the smoke back in the amp, it will work."Comment
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I don't get into pedal stuff. I'll fix them, but when someone wants one to sound different I leave that to people willing to fight in that war. yes, magic caps, magic IC brands, etc. Not for me.
I don;t imagine any of the large parts houses have that much difference in price and availability. If you like dealing with Digikey, then I imagine you will do fine there. I usually shop at Mouser because I am used to it and they most always cover my needs. But when I am putting together a large stocking order of mutiple parts I always make a chart with pricing and availability from Mouser, Digikey, Allied. Photocells is one thing I do look at Jameco for, because Jameco, though a smaller parts house, is a little more hobby oriented than the big guys, and so they sometimes have some interesting parts. Oh and nothing against Newark, I just never seem to go there.
I have not looked at photocells lately, seems to me that when you get real high dark resistance, the on resistance will be higher too. SO it might be a matter of do I want one that goes 5k to 1 meg or one that goes 200 ohms to 10k. I made the numbers up to illustrate the point.Education is what you're left with after you have forgotten what you have learned.Comment
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If somebody brought me a detailed schematic of what they wanted done and a specific list of which magic capacitors and IC's they would like me to install I would do the work, with the caveat that though I guarantee MY work, I won't guarantee that it will sound any better or sound at all like they want it to when I'm done. Vague requests to make it sound 'more this' or 'less that' would be declined. The 'true bypass' mods are the only ones that seem remotely useful to my eyes.Originally posted by Enzo View Post"If you can get the smoke back in the amp, it will work."Comment
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Yes, that is exactly what I tell customers. If they want me to do a published modification they can show me, I am fine with doing the work, but "make this sound more like Green Day" is not what I do.Education is what you're left with after you have forgotten what you have learned.Comment
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Measuring the 'dark' resistance of stock unit shows 200k. Digikey has a part that is 200k 'dark' and 3-11k 'light' for $1.86ea. Morley says that they buy quantities of LDRs and sort the parts for specific applications so I assume that they are looking for a specific range of 'light' resistance depending on the location in the circuit of the part to be replaced. If the 'light' resistance is too high I could always increase value of R1 from 4.7k to whatever value it would have to be to see appropriate ON/OFF voltages on pin13 of U1."If you can get the smoke back in the amp, it will work."Comment
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Just for grins, go to jameco and search "photocell". Aside from the grab bags, they do list the lite and dark for each type. Nothing as low as 200 ohms, but some under 10k anyway. And like a lot of things, a part may be rated at some amount, like 10k, but that is the spec sheet maximum low. That doesn;t mean the part doesn't go a lot lower.Education is what you're left with after you have forgotten what you have learned.Comment
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Good call on the Jameco site. They sell a more suitable part. 3k max LIGHT dcr and 200k min DARK dcr @$1.15 each That's a better part and a better deal. I was just sitting down to place the digikey order. Unknown is the amount of light produced by the LED, as there is no spec on it. The other option for using a LDR that has a higher LIGHT resistance than spec would be to increase vdc on the triggering LED. Right now it's sitting at 1.15vdc and not overly bright.Originally posted by Enzo View Post"If you can get the smoke back in the amp, it will work."Comment
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You wouldn't increase the voltage on the LED, you'd increase the current through it.. An LED is a diode, after all, and it will have a characteristic voltage drop across it, that is what your 1.15 volts is. LED L1 has a 6.8k current limiting resistor. Make that 5.6k or 4.7k and the current through the LED will increase, so it gets brighter.
See if a simple part replacment fixes it before you try modifying things.Education is what you're left with after you have forgotten what you have learned.Comment
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Correct me if I'm wrong, but my understanding of the LED is that the current draw is constant. Just like an incandescent bulb or a tube filament. I have used the published current draw specs of LEDs to calculate the voltage drop across a resistor using ohm's law and the results on the bench usually are pretty close(within 10%) to the results from the calculation. Decreasing the resistive load on the anode will yield a proportionately lower voltage drop across the resistor and resulting higher voltage potential between anode and cathode in accordance with ohm's law.Originally posted by Enzo View Post
The rest of my thought experiment was just that. Just imagining how I might be able to fix one on the go without proper replacement parts. Strictly theoretical. I ordered a handful of the photocells from Jameco(thanks for the recommendation) and I can only assume that they will come in very handy. My intuition told me that the LDR was the weak link in that circuit, due to the physical location, and that it would be the most subject to wear/most prone to failure. My instincts proved correct. I would guess that a large percentage of photo-wah pedal malfunctions are caused by failure/wear of the photocells.
Please do correct me if my understanding of the LED is incorrect."If you can get the smoke back in the amp, it will work."Comment
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A light bulb has a constant current draw only if its supply voltage is constant, ignoring that its resistance changes with its temperature. But it is basically a resistor, so change the voltage across it, and you change the current through it - Ohm's Law.
LEDs don;t have a current draw spec per se. They have a current limit, a maximum current. Any more than that and they are not happy. If an LED says 20ma on the package, that doesn;t mean the LED draws 20ma no matter what. That means you don;t want a lot more than 20ma. 20ma would be its rated brightness spec.
A diode or LED is not a resistor, it does not respond linearly.
I take a target current, subtract the LED voltage drop from the supply, and I have the amount of voltage the limiting resistor needs to drop. R = V/I SO knowing the voltage drop I need and the current I need tells me the value of the resistor I need. reducing the resistor value increases the current through the circuit.
Take a battery, an LED, and a pot. Wire them in series. For sake of discussion, a 9v battery and a 10k pot (used as variable resistor). Turn the pot up and down and see the LED change brightness. Now break into that loop somewher and clip in your meter set for current. Use the low current range. Turn the pot up and down and see what the current does. Current is uniform through a series circuit, so wherever you make this measurement, the LED current will be that amount.Education is what you're left with after you have forgotten what you have learned.Comment
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So I guess my mistake was confusing the 'target current' with the current draw and assuming that it is constant. I see what you are saying though. I rarely measure current, I take the known value of dc resistance, read voltage and calculate with ohm's law to arrive at current. That's how I work most everything out on the bench. I use LED indicators in homebrews and usually power em off a dc filament supply. The ones I grab at the local radio shack typically have a range of 3.15-3.55v or somewhere around there. The higher current rating is equivalent to the brightness. I always assumed that the current rating was a constant and calculated how much resistance I needed to drop the voltage by using current as a constant and applying ohm's law. It works ok on the bench, but your explanation is more accurate. Theory is not my strong suit. I just know how to make stuff work.
Speaking of theory, I have a tremolo question but I'm gonna start a new thread for it. I'm interested to see what you have to say about my 'current' dilemma....
That was probably the worst pun ever.
Sorry"If you can get the smoke back in the amp, it will work."Comment
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