Injection Transformers - Bode Plots - Application - Page 1
Forum section: electronics / testgear

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 02:43:34 am
content: Hi group,There has been lots of discussion on the forum about the construction of injection transformers for making Bode plots of power supply control loops. There has been very little discussion about the correct application of these transformers.ReferencesA discussion about various types of injection transformer:https://electronicprojectsforfun.wordpress.com/injection-transformers/The main thread on this forum:https://www.eevblog.com/forum/blog/eevblog-1104-omicron-labs-bode-100-teardown/ApproachI built a copy of the linear power supply described on the 'Electronics Projects for Fun' website.Here is the schematic:I designed a PCB with a generous amount of test points to make it easier to hook up the injection transformer.Injection Transformer SchematicFor the first tests I am using an injection transformer built around a Wurth 744 229 Common Mode choke.($2.46 USD Qty 1 Digikey)SchematicConstructionTestingThe first test with Injection Transformer connected to TP3 and TP4, location A in the schematic.The second test with the injection transformer connected TP6 and TP5, location B in the schematic.This I believe is the better of the two results. It shows a classical single pole, 20dB per decade slope and 90 degrees of phase shift.LTspiceA simple LTspice model was made to explore this:LTspice resultsWith the injection transformer located at the top of the voltage divider, the measurement gives the correct result which is the measurement V(a)/V(b)To be continued ..Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 02:58:03 am
content: Hi group,Let us explore the waveforms in the time domain, transient analysis:LTspice modelThe AC sources in the AC model have been replaced with sine wave voltages with 0.1V peak, 0.2Vp-p, 1kHz signals.ResultsThe signals at nodes A and B are visible and very measurable.The voltage at node D is very small.It is clear that placing the injection transformer at the 'top' of the voltage divider is the best location for measuring Bode plots.Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 03:32:05 am
content: Hi,Here is a model that matches the TL431 circuit:AC analysisAgain the model shows a nice single pole slope.Regards,Jay_Diddy_B

author: ivaylo
author_rank: Frequent Contributor
timestamp: April 18, 2020, 04:51:50 am
content: Good work!

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 18, 2020, 05:47:46 am
content: i think we can just use ac coupled signal generator to the feedback line, no?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 10:20:16 am
content: Quote from: Mechatrommer on April 18, 2020, 05:47:46 amCan you post a schematic to clarify the question?Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 10:50:38 am
content: Hi group,A lot of effort was placed on maximizing the low frequency bandwidth of the Omicron Injection transformer. I will explore the impact of this.ModelThis is a model showing the 50source impedance and a lower resistance load.I have chosen the 6500uH as the magnetizing inductance. This is the value for the transformer that I used.I get the result:If the magnetizing inductance is increased to 68mHI get the this result:In fact the low frequency -3dB point is given by:LF (-3db) = 1 / (2 x Pi x L/R)where L = magnetizing inductanceR = the parallel resistance of the source (50) and the load 25The calculated value is= 1 / (2 x Pi x 6500E-6H / 16.6) = 406HzSo the LF -3dB is easily calculated.Impact of the -3dB point on the Bode plotThe model presented earlier has been expanded to include the injection transformer model:The value of the magnetizing inductance is stepped from 6500uH to 68mH.This is the result:The legend has been included to show that the magnetizing inductance (or the LF BW) of the injection transformer does not impact the Bode plot.To be continued …Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 11:05:03 am
content: Hi,Construction of 68mH injection transformer.This common mode choke was used in the construction:It was mounted on a small board with the protection and AC coupling circuits:Power Supply PCBHere is a picture of the power supply that is being used for the tests:Bode Plot with the 68mH cm chokeThat is pretty much a textbook single pole slope.Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 12:20:14 pm
content: Hi,If the results from 6500uH and 68mH common mode chokes are compared:There is very little difference between the two results. This is consistent with the result predicted by modelling.In the high frequencies there is no difference. This the important area. This is where the phase margin is measured.At the lower frequencies there is more noise with the 6500uH magnetizing inductance.Time DomainTo determine what is happening here we can look at the signals in the time domain with an oscilloscope. The signals from the injection transformer were connected directly to the scope inputs (no 10x probes).At 50kHz the signal both signals are a reasonable size. This is in the 'noise free' part of the Bode plot.At 220kHz, close to the 0dB point for this power supply, the signals are equal in amplitude.At 10kHz the loop gain is higher, so the signal from TP5, is decreasing in amplitude.At 1kHz the loop gain is even higher and the signal on TP is very small.The loop gain is around 46dB so the signal on TP5 is 200x smaller than the signal on TP7.If the vertical setting on the scope is changed to 1mV /div and reduce the bandwidth:the waveform can be seen.The signal on TP7 is 40mV p-p (15mV RMS)The signal at TP5 is 15mV / 200 = 75uVGo down to 100Hz and the signal on TP5 is around 7.5uV RMS, if you assume a single pole slope.ObservationThe LF -3dB of the injection transformer does not directly impact the loop gain measurement.The LF -3dB reduces the signal size, and makes a difficult measurement harder because of signal to noise ratio, SNR issues.The low frequency performance of the power supply being measured is of little interest. The interesting part is where the gain is 0dB and the phase margin is measured.If the power supply has a low bandwidth control loop, the gain is lower and the signals are bigger at low frequencies. SNR is less of an issue.Regards,Jay_Diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 18, 2020, 02:01:00 pm
content: Quote from: Jay_Diddy_B on April 18, 2020, 10:20:16 ammaybe like this?

author: Sighound36
author_rank: Frequent Contributor
timestamp: April 18, 2020, 02:48:39 pm
content: I can see that, can you remodel the spice please JBD?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 04:39:36 pm
content: Quote from: Mechatrommer on April 18, 2020, 02:01:00 pmHi,If I model that:The results are:The result 80dB is the gain of the amplifier E1, not the gain of the control loop.If you could measure power supply loop gain without an injection device, it would be done that wayRegards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 05:00:32 pm
content: Hi,If I model this:The results are:This is the correct answer for the loop gain.80db from the amplifier-14dB (x0.2) from the divider R1, R2for a total of66dBRegards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 05:19:59 pm
content: Hi,What to do if you can't access the top of the divider, or it is a high voltage power supply?Consider these models:Both of these give the same and correct response:This requires that a network with the Thevenin equivalent impedance is placed between the injector and the amplifier. This will keep the loop gain correct.Regards,Jay_Diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 18, 2020, 07:58:23 pm
content: i got something from varying Rs, but not sure what i'm seeing and not sure how TL431 can do some regulation that way, maybe i can model some crappy psu that i did before and see how the respond when i have time. i thought the idea for injection is to simulate varying load sensed into the feedback loop, and then the reaction on the output is compared to the injected input signal. any peaking respond indicates instability, correct? or no?

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 18, 2020, 09:58:36 pm
content: Hi Mechatrommer,I believe that you have a model that it is very similar to the test circuit found on many of the xx431 datasheets.The reason that I say it is the same, is that Q1 in you schematic has a voltage gain of 1. So you are measuring the open loop voltage gain of the TL431. What we are trying to measure is the closed loop frequency response of the power supply.LTspice measurement of the TL431 Open Loop GainThese are the results:Note:I had to adjust the values C2 and G1 to get the model to match the Diodes Inc datasheet.All the xx431 are slightly different. Using the part like this is like using an op-amp without a feedback network. When other components are added in a typical application they determine the transfer function, not the part itself.Regards,Jay_Diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 18, 2020, 11:51:27 pm
content: Quote from: Jay_Diddy_B on April 18, 2020, 09:58:36 pmi tried to model your circuit in OP and inject a signal (current?) at TP4... (i adjusted R2 value there (try and error)obviously i'm clueless in this subject. i cant find a reason why the need to break the loop and apply injection using transformer. all materials that i read will go to this stage right away without explaining any basic about it. maybe i should google more. why simply injecting a signal at feedback node (without the need to break it) will make any difference and how to determine if its an open or close loop characterization

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 19, 2020, 12:17:32 am
content: Mechatrommer and the group,Consider this circuit:When the circuit is in equilibrium the voltage on the non-inverting and inverting inputs will be the same. Both of them will be 2.5VBy extension the top of the divider voltage will also be in equilibrium. Because R1=R2 the output voltage will be 5V.The circuit serves to regulate the voltage at the top of the divider at 5V. Therefore we have a regulated power supply.In this case the closed loop performance is determined by product of the gains in the control loop.Q1 has a voltage gain of 1The output divider has a gain of 0.5The op-amp u1 has a gain of whatever the chip designer gave it.The total loop gain in this circuit is 0.5x the open loop gain of the op-amp.To measure this loop gain, a disturbance is introduced. The a floating supply, implemented with an injection transformer, is added to the top of the feedback divider.The op-amp will still try and hold the top of R1, point 'b', constant. The voltagebetweennodes 'a' and 'b' is always equal to the disturbance.If you measure node a to ground and node b to ground, the ratio of these voltages will change depending on the loop gain. If the gain is high the majority of the voltage will appear on node 'a' and remainder on node b. If the gain is 1, 0dB, the magnitude of the signals on nodes a and b will be the same (with respect to ground). If the loop gain is low the majority of the signal will appear on node b.The measurement V(a)/V(b) is the control loop gain.Does this help?Regards,Jay_diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 19, 2020, 03:01:20 am
content: Quote from: Jay_Diddy_B on April 19, 2020, 12:17:32 amyes it helps a bit thanks. from what i understand, we introduce extra voltage in the loop Vab = Vfg (function generator). its still hard to imagine... few questions playing in mind..1) if the intention is to add extra voltage, why cant we simply use a floating function generator instead of transformer? the FG's gnd pin (floating) connected to point b, and FG's signal pin connected to point a...2) since Vab is tied to transformer/FG's output voltage swing, any change to point a will affect point b vice versa. if a has to swing larger, so will b, vice versa. and also this makes V(a)/V(b) term will highly dependent on its common mode. if they are further away from ground, the ratio will be closer to 1, as they move nearer to gnd, the ratio will converge to infinity, this doesnt make sense.still many more questions to the puzzle.. i suspect there is math formulation behind this, for example, injection at point B (TP? and TP5) in OP will affect something in the formula, similarly injection at A (TP3, TP4) will affect something else in the formula.

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 19, 2020, 03:27:56 am
content: playing some more with crappy psu... break and connect a FG inside the loop, then i get something else...i think you mean V(AG) / V(AB) right? see attached(its indeed V(a)/V(b) in app note). is the purpose of using transformer so to provide isolation to FG? usually the non-floating type FG in the market right? if we have truly floating FG, preferably with low impedance output, i think we can just directly connect as in the attached picture, correct? the sim looks stable using THS3091, is that what i'm seeing?

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 19, 2020, 03:36:36 am
content: adding load Rl, plot for Rl = 10 - 1K ohm (100 cases) in Gain plot, from bottom line to top (10 - 1K ohm) it seems like a stable PSU, that "crappy" simplistic setup

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 19, 2020, 02:09:59 pm
content: Hi Mechatrommer and the group,Regular readers of the EEVBlog will know that I use LTspice exclusively to illustrate various concepts and model circuits.Today I am going to break with tradition and use TINA.I am going to create a model in TINA where the answer is already known, that way we can tell if we are using the tool correctly.TINA SCHEMATICThe components R1, R2 and C1 define the loop gain in this circuit.There is a pole at 1/(2 x Pi x R x C)Where R = R1 in parallel with R2 (The Thevenin equivalent)R= 7.5kF = 1/(2 x Pi x 22E-9 x 7.5E3) = 964 HzIf we consider the loop gain the divider R1 and R2 provides 0.5x amplification (-6dB)so the loop should have a dominant pole and a 0dB point at 964 / 2 =482 HzSet the range of frequencies for AC AnalysisSet the minimum and maximum frequencies to cover the range we are interested in.Run the simulationThis shows the signals on the test points with respect to ground.Create a User Defined FunctionThe desired function is VF1(s) / VF2(s) to display the Bode plot of the closed loop.Display only the desired curveSelect only the user defined function created earlier.Bode PlotThe result is a single pole slope that crosses 0dB at 480Hz. This is the expected result.ConfirmationThe values of R1 are R2 Changed. This changes the gain to 0.25 (-12dB) and maintains the Thevenin resistance of 7.5kThe 0dB frequency has moved to 240Hz. This is the expected to result.Back to LTspice ...Regards,Jay_Diddy_B

author: schmitt trigger
author_rank: Super Contributor
timestamp: April 22, 2020, 07:43:42 pm
content: First of all, thanks X1000 for such a comprehensive and useful explanation.I've learned more about the proper use of injection transformers in 15 minutes than what had been explained to me previously.However, I think there are a couple of typos. In the explanation below you talk aboutTP5andTP7, but in your original schematic,TP7is ground. See attached image.And what I believe isTP7, is labeled asTP?Quote from: Jay_Diddy_B on April 18, 2020, 12:20:14 pm

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 22, 2020, 10:04:14 pm
content: Quote from: schmitt trigger on April 22, 2020, 07:43:42 pmThank you for your kind words.You are absolutely right, I got TP? and TP7 mixed up.(I should stop using the 'young peoples font')Regards,Jay_Diddy_B

author: fenugrec
author_rank: Frequent Contributor
timestamp: December 26, 2021, 04:42:56 pm
content: Hi Jay_Diddy_B,thanks for taking the time to demonstrate all this.I'm trying to duplicate some of the results here with an HP 4195A VNA, and it just struck me that the 3577 can be set to either 50R or Hi-Z inputs... So for instance when using that injector PCB you showed (with 100n + 1k AC-coupling), would you set your 3577 to High-Z ? If my math is right with a 50R input you would get massive attenuation (1000/50 = about -26dB ?) and a ~ 1.5kHz high-pass, which may explain the weird results I was getting (the 4195 is 50ohm only)...