Frequency voltage divider - Page 1
Forum section: electronics / beginners

author: fxr
author_rank: Contributor
timestamp: July 01, 2020, 07:31:38 pm
content: I need electronics guru advice)I have a crystal oscillator with an output signal level of 3.3V, but I need get 1.2VCan a resistive divider be used to reduce the clock signal voltage?Can this method degrade the quality of the clock signal?Maybe there is a better way to make this work?

author: TimFox
author_rank: Super Contributor
timestamp: July 01, 2020, 07:39:10 pm
content: A resistive divider can certainly be used to reduce the voltage from 3.3 to 1.2 V.You need to specify the problem:1. Output drive capability.2. Input impedance of load.3. Frequency (determines need for compensation capacitors and allowable impedance level).4. Degradation: is this a sinusoidal signal? If not, are the rise and fall times critical?5. Are there DC considerations on the drive and load voltage waveforms?

author: fxr
author_rank: Contributor
timestamp: July 01, 2020, 09:17:32 pm
content: How many questions, you are really professional.1. 15pF2. I not found this params in datasheet3. 24.0MHz4. This a sinusoidal signal5. I don't knowYou know a method for calculate the voltage divider?

author: TimFox
author_rank: Super Contributor
timestamp: July 01, 2020, 10:00:10 pm
content: You still need to know how much current the source can provide, or you are willing to draw. Below, R1 is from the source to load and R2 is from load input to ground.If you know that, then the total resistance (R1 + R2) of the two resistors will be the source voltage (3.3 V) divided by that current.To divide the voltage, the two resistors will be in this ratio:R1 / R2 = (3.3 - 1.2) / (1.2).Now, if the input resistance of the load is not huge compared to the calculated R2, you need to increase the resistance so that the actual resistor in parallel with the input resistance equals the desired R2.Furthermore, if the load capacitance C2 (part of the input impedance of the load that you don't know yet) is not tiny compared with the parallel combination of R1 and R2, you may need to add a capacitor in parallel with R1.(R1 x C1) = (R2 x C2)where R2 includes the input impedance of your load. Depending on what you're doing, the input impedance could be anything from 50 ohms up to 100 k, so you need to specify it, at least approximately. If your load device is AC coupled, so that the DC doesn't matter, you could use two capacitors instead of the two resistors, but that would depend on the unknown input capacitance of the load.These are really a minimal set of questions that must be asked to determine the design, and we haven't even discussed accuracy. Are you sure the oscillator has a sinusoidal output? That voltage sounds like a TTL-level square wave.

author: fxr
author_rank: Contributor
timestamp: July 01, 2020, 10:38:03 pm
content: Thanks for basic calculations.I attached screenshot from datasheet.You right, this oscillator have TTL-level square wave.

author: TimFox
author_rank: Super Contributor
timestamp: July 01, 2020, 11:27:25 pm
content: Are you running it into a lower-voltage CMOS device or something else? This affects the DC conditions for the divider. Unfortunately, the spec sheet assumes a straight CMOS load at the same supply voltage, which has almost no DC conductance, so it did not specify the source current in the "1" state.

author: gcewing
author_rank: Regular Contributor
timestamp: July 02, 2020, 12:28:26 am
content: Do you have access to an oscilloscope? If so, you can do it by trial and error.Assuming the oscillator output is a typical logic gate output, it should be capable of supplying at least a few mA, and if the load is a CMOS gate its input resistance will be high enough to ignore.So start with something like 560 ohms and 1k, which should give you about the right ratio.First check that the oscillator output hasn't been loaded down too much; if it's noticeably less than it is when unloaded, increase the resistors while keeping the same ratio.Then check that you're getting something close to the required output voltage; adjust the resistor ratio as needed.Finally, have a look at the output waveform. If it's rounded off too much, add some capacitance across the top resistor until it looks square again.If you don't have a scope, you can probably make the voltage measurements with a multimeter set to AC. You won't be able to see the waveform shape, but if the circuit works, it's probably good enough.

author: fxr
author_rank: Contributor
timestamp: July 02, 2020, 12:17:23 pm
content: Thanks you very much for answers!TimFoxAre you running it into a lower-voltage CMOS device or something else?I'll use the controller USB2412 with CMOS logic.gcewingDo you have access to an oscilloscope?Yes, I have oscilloscope for tuning the schematic.Is the capacitor installed correctly?What the capacity should I use for signal tuning?I attached the test schematic, please let me know if I can make it better.

author: TimFox
author_rank: Super Contributor
timestamp: July 02, 2020, 02:18:12 pm
content: What is the Cin spec for your lower-voltage CMOS device? I'm not sure that your source can supply the 2mA current required in the high state: you might want to increase both by x10 or so, and play with the capacitors.If you can't find these specifications, you need to follow gcewing's advice and do trial and error with an oscilloscope, with a low-capacitance x10 probe.

author: fxr
author_rank: Contributor
timestamp: July 02, 2020, 07:27:08 pm
content: Quote from: TimFox on July 02, 2020, 02:18:12 pmHow I can see in datasheet Cin is 6pFDid I place the capacitor correctly on schematic?

author: TimFox
author_rank: Super Contributor
timestamp: July 02, 2020, 07:40:11 pm
content: That looks like the datasheet for the oscillator. What are the data for what you are driving? Since we are dealing with logic levels, one of which is near ground, we need a DC-coupled divider (i.e., resistors), but we may need to add capacitors. The problem is how much current can the oscillator output source in the high state?When interfacing a logic source (in this case higher voltage) to a logic load (lower voltage), we need to ensure that the voltage at the load is within spec levels and that the current from the source is within spec levels when inserting a divider between them.

author: fxr
author_rank: Contributor
timestamp: July 02, 2020, 10:03:58 pm
content: Quote from: TimFox on July 02, 2020, 07:40:11 pmNo, this is a receiver device datasheet .Quote from: TimFox on July 02, 2020, 07:40:11 pmWhere I need add capacitors (parallel, series...)?For make a clear, datasheets for oscillator and signal receiver attached.

author: TimFox
author_rank: Super Contributor
timestamp: July 02, 2020, 10:11:09 pm
content: According to the datasheet on the USB2412, youdon't need a divider. See page 17 for the specification on the external clock input. 1.4 V is theminimumhigh-state voltage on the external clock input, but it can accept voltage up to the "VDD33" supply voltage, nominally 3.3 V.

author: fxr
author_rank: Contributor
timestamp: July 03, 2020, 11:25:29 pm
content: Quote from: TimFox on July 02, 2020, 10:11:09 pmThanks you very much, this is helped for me!But I don't understand, why in the datasheet at page 21 indicate voltage level 1.2V for the clock...?The external clock is recommended to conform to the signaling level designated in the JESD76-2specification on 1.2 V CMOS Logic.

author: TimFox
author_rank: Super Contributor
timestamp: July 04, 2020, 02:20:52 am
content: I don't know why that statement is there, but (without a crystal connected) Pin 24 is the external clock input, and its specification only says that the logic levels on that pin are (<+0.5 V) for low and (>+1.4V) for high, so 1.2 V pk-pk would suffice, but it is not necessary to limit it to that level. Have you tried a direct connection from your source to the load?The operating range of voltage Pin 24 is from -0.3 V to VDD33 (3.3 V), see page 16, and the maximum range (higher on that page) is -0.5 to +4 without damaging the device.Since both devices are CMOS, the usual assumption is that the DC current is negligible and that only capacitor-charging current is relevant to the interface.

author: fxr
author_rank: Contributor
timestamp: July 04, 2020, 07:48:09 pm
content: Quote from: TimFox on July 04, 2020, 02:20:52 amNo, I not have a physical oscillator.I want to try connect the functional generator for test clocking at 3.3V.

author: TimFox
author_rank: Super Contributor
timestamp: July 04, 2020, 09:33:13 pm
content: From the spec, you just need the external oscillator to go from below the logic 0 level (but not below the operating level) to above the logic 1 level (but not above VDD).It is much easier to answer your questions if you state what this "functional generator" is, and what its properties are.Specifically, you don't want to use a sine wave symmetrical around ground into the CMOS binary input. If that's what you have, then capacitor couple it to a DC voltage divider to establish the proper center voltage level. With CMOS inputs, you don't need to use small resistances, and the capacitor should be much larger than the input capacitance and large enough to pass your frequency into the resistors.Earlier, when I asked you for the external source, you posted data for a CMOS oscillator that runs on 1.2 V. Its output would be just enough peak-to-peak voltage to drive the load you posted (3.3 V CMOS part), but the DC levels are not quite right. Since your original post said 3.3 V signal, I assumed you had a 3.3 V CMOS oscillator part.If you mean a function generator that can drive a 50 ohm load, the appropriate voltage divider would probably load down a CMOS source, and might even damage it. However, if the function generator square-wave output can be adjusted, keeping the lower voltage set to zero, you can verify (carefully) how much voltage is required to clock the load.Please specify exactly what you are connecting: the "voltage divider" you asked for, as a black box, will affect the driver and the load will affect the voltage divider. One cannot choose the interface without knowing what is being interfaced.

author: fxr
author_rank: Contributor
timestamp: July 05, 2020, 03:12:30 am
content: I just make clear, I have a USB controller USB2412 mouned on pcb and I want to order oscillator for him.But before I will get this oscillator, I want to make a experiments with function generator.I have a Siglent SDG2042X 40MHz Function / Arbitrary Waveform GeneratorThis function generator can give sine and square wave, also can drive load 50 ohm.

author: TimFox
author_rank: Super Contributor
timestamp: July 05, 2020, 02:58:03 pm
content: Then program your generator to make square waves with 0 or a few tenths of a volt at the low end and +3.3 down to +1.2 V at the high end and see what happens.The oscillator whose data you posted does not have quite enough output (due to its 1.2 V supply) to meet the specification for the clock input of the USB2412. There are plenty of oscillator devices out there with 3.3 V supply.Why did you want a voltage divider in the first place?

author: fxr
author_rank: Contributor
timestamp: July 05, 2020, 11:08:02 pm
content: Quote from: TimFox on July 05, 2020, 02:58:03 pmIt was my mistake, I thought that the required voltage level for clocking the USB2412 is 1.2V.I wanted to apply a voltage divider so as not to burn the chip. Because mostly crystal oscillators working on 3.3VWhat do you think, what minimal level of current power should a generator have for clocking USB2412?

author: TimFox
author_rank: Super Contributor
timestamp: July 06, 2020, 12:52:40 am
content: It’s a CMOS input, so power is not the variable. I told you the voltage levels from the data sheet.