Hola AdamI did, but here is an updated one. Thanks for pointing it out...
Yes, sure; thanks a lot! But I'm having a lot of problems doing toner transfers. I tried it 3 times in my cyber-cafe's laser printer, and these where the results:Well if you want please send me the files and I can try printing them out for you and if you like them I can send them to you....
Yeah, sure. I attached another one with the background in white. Might still not be good enough -as I captured those images- so you might like to see it through the files I uploaded to Dr. Killjoy.Hola Adam
Could you please change the black background to white? I do not know why actually, but reading that schematic as is, is very hard for me. Age counts, maybe...
Gracias.
Yes, I already thought about that; but then dropped the idea because it would involve reprogramming the PIC and I wasn't absolutely sure about the final position of the PIC. It wont be too difficult, but it will take additional subroutines to reorder the bits of the data sent out to the LCD.Pretty sure you can simplify somewhat by swapping some pin numbers around on the MCU. Pins 1&2 for example...
HTH Steve.
Thanks, and sorry for the delay; I needed to check some things first.A few things that jump out... though I haven't critiqued heavily.
I now routed pins 1 and 2 of the PIC by re-routing another track, but I still have 5 left. I can't get those, so I will connect them with a wire outside the PCB. Four of them carry digital data to the LCD, so don't worry me much, but the last one carries a high frequency square wave (up to 20Mhz).It appears that you have 6 or 7 traces that have not been routed.
I had a 50Ω resistor at the output of the MAX, as the datasheet's schematics suggest, but took it off when I added the op-amp; which also has a 1k potentiometer between them, but it's output is directly connected to the BNC connector.The output of function generators typically have a 50 ohm resistor - 2W inside that protect the device from shorting and to help with 50 ohm transmission lines (50 ohm coax). You could get away with a 1W resistor if you're still using a 5V supply, but I wouldn't go any smaller. This will create 100mA short current... and 50mA operating currents... this can cause voltage dips on the power rails if too small of a trace is used... because of this - I offer up the next two suggestions:
I'm not sure what "high-side power plane" and "power fill" mean. Do you want me to make the +5V and the -5V wider or make them fill the empty spaces on the bottom side?You do not have a high-side power plane... typically you will see a plane on the top and bottom... usually one will be power and the other will be ground... I suggest doing a power fill on the top layer once all your routing is complete.
I used 1uF caps for the MAX038 because it's what its datasheet recommends for all configurations. I just noticed that if I use electrolytic caps I also need to use a 1nF ceramic; as the datasheet states: "bypass capacitors are 1μF ceramic or 1μF electrolytic in parallel with 1nF ceramic".1uF seems like pretty large decoupling capacitance for high-speed design... you probably want more on the order of 0.01uF ceramic as close to the part as possible (really, making the smallest loop possible), 0.1uF ceramic right next to those, and then some bulk caps on the order of one or two 4.7uF tantalum spread out on the card. I have taken this approach on my high speed designs and I have never had a problem. I usually leave out the 0.01uF capacitors if there is nothing over 1MHz going on with the chip I'm trying to decouple.
I just added the 50Ω resistor next to the BNC connector; but not sure what you mean by "design output traces properly for 50 Ω"; you mean wider?I'd also design my output stage amplifier traces properly for a 50 ohm transmission line or add the 50 ohm resistor and keep it as close to the BNC as possible... possibly doing a little bit of both.
'Good?... Enough!' for you is a compliment for me.With that said... with the exception of the first bullet, your design will probably be 'Good?... Enough!', as my boss likes to say, if you proceeded with the current design... and would make a good functioning function generator. These are just the other suggestions that us anal types like to make sure are designed correctly.
Good tindel! Short and to the point. Thanks for sharing.I'd personally get those last few traces on the board... much easier to do it once than every single time... don't be afraid to put a few short traces on the back of the board - but do it smart too... electrons are lazy... they always take the path of least resistance... and high frequency electrons are even more lazy... high frequency electrons tend to want to go thru large copper planes in the exact same path as they went to the load (I.E. a mirror of the traced path, see: https://www.sigcon.com/Pubs/news/8_08.htm for more info). I would get that 20MHz square wave placed with a good return path... you need that placed... even though it's a 20MHz square wave, it will have harmonics up to 200MHz pretty easily...
Thank you for the suggestion; I think I just got the 20Mhz line connected through the bottom plane. Been trying all configurations for more than a week; starting from the beginning over and over again dozens of times. I got the best results with the PIC just under the LCD's row of pins; but if I do it that way I would need a bigger board to place all the controls and the power supply in an accessible place, while leaving the central area under the display practically empty.Maybe try turning the PIC 90 deg clockwise and the other IC 90 degrees counterclockwise....
layout is an iterative process don't be afraid to just start over.
Yes, the top is the ground plane; all the tracks go on the bottom plane, to which I also added a ground fill as suggested by tindel.I think you have the ground plane on the top layer?
Now that I connected that 20Mhz line, and with the exception of those 4 jumper wires that I need to connect the PIC to the LCD data, I don't need to use tracks on both layer; so I just need to cover the ground plane completely while etching, and draw the tracks only on the bottom plane. I also won't need to align them, as I can first drill the pad's holes, and use those as a guide to use a wider drill bit for the space between each pin and the ground on the top plane; which makes is far easier.You'll have to figure out how to line-up the top and bottom layer... because of the copper voids in your planes... there are some youtubes that show how to do this.
I was just saying that I took that 'Good?... Enough!' as a compliment. You have definitely touch a lot of subjects that I never even heard about before; so please continue.Like I said - it's probably 'Good?... Enough!' if you want to build it and get something working... go for it! I was just offering suggestions for improvements and further learning, really.
Since I'm leaving the top plane entire as ground, I'm pretty sure there will be a good return path under every high frequency line. I checked it anyway; just in case.Yes, I think you understand the concept. You want a return path for all of your currents... usually at DC through ~1MHz this is simply the ground plane... for higher currents, the return path is usually the closest plane to the source path... think of all of your currents in loops... every time current moves it has to go back to the same place it came. The higher the frequency becomes the more likely it is to return via the same route that it came because it is the lowest impedance path because it is also the lowest inductance path.
I'll leave the 50 Ω resistor -since it's already well placed- and make that track as wide as possible.It looks like you put a 50 ohm resistor as close to the output as you reasonably could, so you will probably be okay. Do note that this will reduce your output voltage by half if you properly terminate your 50 ohm transmission line. If you want to get rid of the 50 ohm resistor, I'd suggest using the calculator to calculate the trace width so that you can use proper 50 ohm impedance transmission lines. This will have the benefit of getting the entire voltage on the output, but have the downside that a short may take out your op-amp - or more. You can calculate your trace width by using calipers to figure out your board thickness, subtract out your copper thickness (usually 1oz copper (o.0014")). FR4 relative permeability Er is 4.8, so that will get you in the ballpark... it doesn't have to be perfect, just close. Assuming your board thickness is about 64 mil - this would make the trace about 120 mil wide - pretty thick. usually you want to keep voids around the trace, just as thick as the trace itself... so this will get you around 120mil * 3, about 3/8" wide.
Glad you're back.Wow. A lot has happened while I wasn't looking.
I just connected the unused pins of the THS4222 and added 100nF bypass ceramic capacitors to it. Also changed all other bypass capacitors of ICs to ceramic.Some things I noticed:
You need to connect the unused pins of the THS4222. Connect pins 6 and 7 together and ground pin 5.
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All bypass caps other than the 10 uF bulk bypass capacitors should be ceramic types.
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You want the ceramic bypass capacitors on both the MAX038 and THS4222 power pins.
It's quite confusing in that area: the feedback resistor is R22, which is 4.7k; and the input resistor is R21, which is 1k. Those where the values I've been using in the breadboard with the AD822AN opamp. Should I use different values for the THS4222?The gain of the THS4222 seems a bit high and the feedback resistor is the wrong value. You must use a 2 K ohm feedb ack resistor -- any other value will cause either a reduction in bandwidth or peaking in the frequency response.
The input resistor (R21) should be more like 510 ohms.
I had to use long tracks for both supply rails -there was no other way- but I think the opamp input and output are fairly short; just marked them in this image for clarity:The traces going to and from the THS4222 look way to long to me.
How can I fix that issue without changing much the power supply; should I take the 50Ω resistor off, recalculate the regulator to account for the resistor, or maybe add transistors for better regulation?Your shunt regulated power supply will not work well when the output is driving an external 50 ohm termination resistor.
I have a 20k pot plus a 1k limiting resistor on the FADJ; which will allow me to select the recommended 12k on the datasheet. It's what I've been using in the breadboard, and worked fine up to now.The values on the Fadj and Iin pins of the MAX038 don't look quite right to me.
Thanks for all your help.... My laptop battery is dieing so more later...
Yes. The AD822 is a low frequency op-amp. The THS4222 is a very high frequency op-amp. The THS4222 data sheet shows a 2 K Ohm resistor for a gain of 5. that is what I would use. The value is not as critical as I was thinking in my earlier post since the THS4222 is a voltage, not current, feedback amplifier. Even with the 2K feedback resistor, just a few pF of capacitance will effect the bandwidth of the amplifier. Note that at a gain of 5, your amplifier only has a bandwidth of about 25 MHz. This is because the gain bandwidth product of the part is only 120 MHz. You may want to consider using both amplifiers of the part each with a voltage gain of about 2.2.It's quite confusing in that area: the feedback resistor is R22, which is 4.7k; and the input resistor is R21, which is 1k. Those where the values I've been using in the breadboard with the AD822AN opamp. Should I use different values for the THS4222?
I am not sure why you have 2 frequency setting pots...I have a 20k pot plus a 1k limiting resistor on the FADJ; which will allow me to select the recommended 12k on the datasheet. It's what I've been using in the breadboard, and worked fine up to now.
The datasheet recommends a 20k pot for the Iin, but I got better results using 1M and 100k pots with a 10k limiting resistor in the breadboard. Both worked well, but the 1M gives me too much range and little precision, while the 100k is a tiny bit short for the range caps I've used. The ideal would be a 200k pot, but I couldn't get that value, so I went for the 100k.
The right way of doing it is to supply positive and negative unregulated input voltages to positive and negative 5-volt regulators. As I have said, making an artificial ground will not work very well. The problem is not the 50 ohm resistor. Instead, the problem is the current into your external circuit under test. This load current can cause large current mismatches (as much as 100 mA) on the plus and minus power supplies. You might be able to limp along by putting emitter followers on the Zener diodes but that is not what I would do.How can I fix that issue without changing much the power supply; should I take the 50Ω resistor off, recalculate the regulator to account for the resistor, or maybe add transistors for better regulation?