Strange voltage readings from function generator - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: March 07, 2014, 09:39:58 pm
content: I picked up the student manual for Art of Electronics recently. Figured I'd work my way through it. In doing one of the experiments I got really weird (to me at least) results.The circuit looks like this:My function generator is set to 1kHz at 10Vpp.My expectation was that the output would look like a sine wave with half the amplitude of the input. Instead I get a 1kHz sine wave that's still 10Vpp.Next I tried hooking the function generator up to my oscilloscope directly with a BNC cable, ie. no circuit. I got the same 1kHz sine wave except now it was showing as 200Vpp!So I'm a little confused. What's going on here?

author: SeanB
author_rank: Super Contributor
timestamp: March 07, 2014, 09:50:51 pm
content: Your input voltage display is confused. That explains the 200V, it thinks a 10x probe is connected still. What the FG output is is a voltage source and a terminating resistor of 50R in series. Thus into a 50R load there is 10V, but into a high impedance there is the voltage output without the loss across the internal termination resistor, thus the 20V from the FG. Try looking at the FG output with a 50R load across the scope input ( or turn on 50R termination if the scope offers this, and you will see the correct voltage. For the divider you will need to have either a direct 1M input, or a 10x probe, and no terminator on the scope input.

author: Legion
author_rank: Frequent Contributor
timestamp: March 07, 2014, 10:20:53 pm
content: I tried matching the input impedance on the scope. Set it to 50R. Now it's showing as 100Vpp, it's off by a factor of 10.

author: electronics man
author_rank: Frequent Contributor
timestamp: March 07, 2014, 10:30:11 pm
content: it still seems you have x10 atenuation on your scope.

author: Legion
author_rank: Frequent Contributor
timestamp: March 07, 2014, 11:47:23 pm
content: Quote from: electronics man on March 07, 2014, 10:30:11 pmYeah, I did. Thanks.If I hook my FG direct to the scope I get 1kHz 10Vpp. If I hook my scope up to the circuit and measure the Vout, I get 10Vpp. My expectation was that I'd see 5Vpp.Next I tried hooking up channel 1's probe above the first 10k and channel 2 at Vout. Channel 1 shows 20Vpp and channel 2 shows 10Vpp. So voltage division is happening. But why am I getting twice the amplitude out of my FG through the circuit? But the proper 10Vpp when hooked up direct via BNC?

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: March 07, 2014, 11:53:02 pm
content: Read up on why setting your scope to 50R made a difference, and why it made the difference that it did. Then consider what the resistances are in the different configurations that you're measuring. A circuit behaving one way may not continue behaving that way when something gets attached to it; you must always consider the circuit as a whole including your test gear, probes, etc.

author: w2aew
author_rank: Super Contributor
timestamp: March 08, 2014, 01:15:08 pm
content: Quote from: Legion on March 07, 2014, 11:47:23 pmWhen you hook the gen to your circuit, it is seeing a 20k load, not 50 ohms. So you're not getting the same voltage division from the output. Some function generators give you the ability to tell them what the load impedance is, so that the output voltage can be properly set. If you parallel your 20k load with 50 ohms, all will work properly.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 03:13:51 pm
content: Quote from: w2aew on March 08, 2014, 01:15:08 pmI'm surprised this is necessary. When I hook up a 9V battery I don't have to worry about the circuit I'm hooking it up to, I know it's going to give me 9V. I thought the same would be true of a FG.1.) If I'm reading your suggestion right, paralleling a 50R with the circuit will give the correct value. If that's the case, why not just have an option to switch in a parallel 50R on the FG? Why would you need to "tell" the FG the load impedance? Unless the parallel resistor value changes for each circuit.2.) If the parallel resistor value has to change depending on the circuit, does this make the amplitude setting on the FG useless? Or at least less useful?3.) 50R is small compared to 20k, so there shouldn't be much sag when the load is hooked up. At any rate, I should see a voltage sag, not a voltage doubling. How am I doubling the voltage?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 08, 2014, 03:34:22 pm
content: Some generators do have the option to switch what the reading is relative to, but the point is twofold: 1. always measure your sources -- do not trust the number on the display, assume it is for convenience only; 2. real function generators are designed for impedance matched loads, and are not ideal voltage sources (rather, they are usually good approximations to a 50 ohm Thevenin/Norton source). Use them accordingly.Because of #2, connecting a resistor in parallel would be silly; it's already there. To understand the underlying purpose in greater detail, you'll need to learn about transmission lines, impedance matching, reflection and termination. It's noteworthy that a FG's square/pulse output may not meet specified rise/fall time if unterminated (especially through a long cable).Tim

author: robrenz
author_rank: Super Contributor
timestamp: March 08, 2014, 03:35:42 pm
content: This postand thread may help you.You have to see the circuit of output impedance of your source and input impedance of your measuring instrument to grasp whats going on. This Pic may help showing different conditions. Remember the FG is outputting 2x what you set it for because it is anticipating a 50 ohm termination in the receiving instrument that will form a 2:1 voltage divider. Anything other than the 50 Ohms is going to give you something different than you have the FG set for.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 03:46:48 pm
content: Quote from: T3sl4co1l on March 08, 2014, 03:34:22 pmI'm aware of impedance matching, reflection, etc, but my knowledge level isn't high enough to know what to do about it. So the proper way to do it then is to hook up the scope to the circuit inputs and then adjust the FG amplitude until you get the Vpp you want on the scope. Ignore the voltage display on the FG.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 04:00:14 pm
content: Quote from: robrenz on March 08, 2014, 03:35:42 pmThanks for the link. I'm going to have to spend some time on AC circuits to understand terminators, etc.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 04:03:34 pm
content: Quote from: w2aew on March 08, 2014, 01:15:08 pmThanks for the advice. I put a 50R is parallel and I get the proper values. I'll keep this in mind going forward.

author: robrenz
author_rank: Super Contributor
timestamp: March 08, 2014, 05:11:06 pm
content: Quote from: Legion on March 08, 2014, 04:00:14 pmLook at the picture I posted andthink DC only, it is just a voltage divider comprised of the output resistor of the FG and the input resistor of the measuring instrument. For conceptual purposes you can forget all the AC stuff. You can also set your FG to output DC and test it that way and forget any AC analysis.It seemed mystical and tough to understand for me also until I pictured what I drew on the picture in the above post and thought DC only.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 07:26:50 pm
content: Quote from: robrenz on March 08, 2014, 05:11:06 pmI think I understand what's going on in your picture. It looks like we're using the 50R resistor to act as a voltage divider with the 50R output impedance of the FG. This divides the output voltage by 2.The part that confuses me though is why the FG puts out twice the Vpp. In your picture it says "FG set at 1V actually puts out 2V". From some of the other posts in this thread it looks like this doubling of Vpp is caused by a reflection of the signal. If that's the case, then that's the part I don't understand, the physics of why putting a 50R terminator prevents a reflection of the AC signal.

author: robrenz
author_rank: Super Contributor
timestamp: March 08, 2014, 08:04:35 pm
content: The FG outputs twice the setting because it is expecting to be connected to something with 50 ohm input impedance forming the 2:1 voltage divider. Has nothing to do with reflections which are not going to occur at DC. There is more to impedance matching and reflections with AC signals but you don't need to go there to understand the voltage divider principles involved at DC and low frequencies.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 08:13:27 pm
content: Quote from: robrenz on March 08, 2014, 08:04:35 pmOK, it's by design. How come you consider this a DC circuit? The only voltage source is the FG and it's operating at 1kHz with no DC offset. Is 1kHz a low enough frequency to be approximated by DC?

author: Hideki
author_rank: Frequent Contributor
timestamp: March 08, 2014, 08:18:52 pm
content: To avoid reflections at high frequencies/steep slopes, function generators _assume_ that you have terminated the cable at the other end with 50 ohms.This this is the normal usage and it will always divide the output in half at the receiving end. However, to avoid the annoyance of getting 0.5 volts at the receiver when the generator is outputting 1 volt, the way the generators work is that they will output _double_ of what the display is showing. So normal business is that the generator outputs 2 volts when you set it to 1 volt, and the two 50 ohms divide this in half, giving you... 1 volt in the end.If you break the assumption and don't terminate it with 50 ohms, the generator won't automatically know about it and will still output 2 volts through its 50 ohm output impedance.The solution to this is usually found in the generator itself, where you can tell it how the signal will be terminated. 50 ohms or "high impendance" (Hi-Z). When set to high impendance it will no longer double the output voltage and actually output 1 volt when set to 1 volt.

author: robrenz
author_rank: Super Contributor
timestamp: March 08, 2014, 08:24:19 pm
content: Quote from: Legion on March 08, 2014, 08:13:27 pmIMO 1kHz might as well be DC. (I am no expert on this) Others may be able to comment on what frequency range you need to start considering AC aspects of impedance matching. My point was to get the basic principle look at it from a DC perspective. That makes it obvious why you need a 50 Ohm terminator when connecting to a high impedance instrument.

author: Legion
author_rank: Frequent Contributor
timestamp: March 08, 2014, 08:32:33 pm
content: Quote from: Hideki on March 08, 2014, 08:18:52 pmCould I just get a T-shape BNC splitter, put a 50R BNC terminator on one end and the FG output on the other?

author: robrenz
author_rank: Super Contributor
timestamp: March 08, 2014, 08:37:00 pm
content: Quote from: Legion on March 08, 2014, 08:32:33 pmYes. That will work the same as a 50 ohm inline terminator. They both are just adding a 50 Ohm resistor between the signal line and ground.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2014, 12:52:15 am
content: Quote from: robrenz on March 08, 2014, 08:04:35 pmActually... the counterintuitive part arises because you are implicitly used to reflections, and don't know it. It's not that the effect is counterintuitive, it's that your intuition is conditioned upon a different experience.If you run the analysis, you see that DC into an open circuit is 100% reflected, at all times, regardless of the length or impedance of transmission line attached -- the difference is, reflection and mismatch (typically a gross mismatch, going out of ones' way to make good voltage sources and all!) is standard design practice.From a conventional perspective, the reason the FG's output is "double" voltage is because theystart with thatand add the resistor after.However, from a transmission perspective, an alkaline cell, for example, is, say, 0.2 ohms and 11 watts. It is constantly delivering that power at all times, however, by leaving it connected to a high impedance termination (an open circuit), all the power is immediately reflected back, the superposition of which (reflected and incident waves) causes the appearance of 1.5V as measured on the terminals.(Shorting it causes an inverse reflection, resulting in zero terminal volts, 7.5A, and all 11W being dissipated in the cell.... it heats up. A matched load of 0.2 ohms draws 5.5W, i.e., 0.75V and 3.75A, and the cell still gets warm, though not as bad as the shorted case.)Another example: a switching supply normally connects an inductor to a voltage source, then to a load capacitor, alternately (a crude but sufficient description of one type of converter). In transmission terms, the inductor has impedance, and carries reactive power; when the load is light, the reactive power is reflected -- recycled -- through the source. When the load is heavy, the source delivers power into the reactance, and then into the load. (Taking a specific example, this is most apparent in bridge type converters, where the series inductor carries a triangular current waveform; this current doesn't deliver any output power, it's just reactive power cycled into the power supply rails. Under light load, the inverter is loaded only by this ripple; lots of power is reflected, so overall power consumption is small. Under heavy load, little is reflected.)Tim