Center_Tapped_Transformer

Transcription

1 V1 is 1.0 Vp-p at 10 Khz William R. Robinson Jr. p1of 17

2 Purpose The Bspice transformer model only provides for a center tapped transformer. In addition, in the study Transformers.doc 1 I discovered that the Bspice model for a transformer, which has the tap on the primary side, should not be simply flipped to provide a tap on the secondary side due to poor frequency fidelity when compared to a real transformer. Biege Bag provides a paper that suggests a method of using two transformer models to model a tapped transformer. This study looks at the fidelity of using that model. I formed the following conclusions Using the two transformer model for center tapped secondaries has a poor frequency response. At least in this case it is closer to the measured response than the flipped model from transformer.doc Theory and Design I choose to study the transformer in the following configurations Two loads one on each side of the secondary (Rload1 and Rload) One load on ½ of the secondary (Rload1 is removed) Gain (ideal transformer) Two loads one on each side of the secondary Ns 3 o Vout Vin * Np Np = number of turns on primary Ns = number of turns on (both halves of) the secondary Ns/Np is also known as the turns ratio One load on ½ of the secondary Ns 4 o Vout Vout * Ns Ns = number of turns on a tapped part of the secondary Ns = number of turns on all of the secondary o In this case Ns = ½ Ns Vout = ½ Vout across all of the secondary Input Impedance (ideal transformer) Assuming the ideal transformer is lossless then the power out of the transformer must equal the power out of the transformer Two loads one on each side of the secondary o Vout * Iout Vin * Iin o Substituting the equation above for Vout Ns Vin * * Iout Vin * Iin Np o Dividing both sides by Vin William R. Robinson Jr. pof 17

3 Ns * Iout Iin Np o Substituting Rload/Vout for Iout Ns Vout * Iin Np Rload o Substituting for Vout from the top equation Ns Vin Ns * Iin Np Rload Np o Rearranging Vin Np Rload * Iin Ns o Rin = Vin/Iin therefore Np Rin Rload * Ns One load on ½ of the secondary o Replacing Ns with Ns Np Rin Rload * Ns The ideal transformer has a flat frequency response however for the model o RP, LB form a high pass circuit R 4 F cutofflower = L * RP F cutofflower = (* LB) o XP, (Rs + Rin) form a high pass filter R 4 F cutoffupper = L Np Rload * Ns F cutoffupper = ( * LA) William R. Robinson Jr. p3of 17

4 Calculated Gain (ideal transformer) Two loads one on each side of the secondary o I was unable to find a data sheet for the transformer I used in the real circuit, based on using it forward and backwards I determined that its turns ration was 14.4:1 o Gain = 14.4 One load on ½ of the secondary o Vout = ½ Vout across all of the secondary o Gain = 7. Input Impedance (ideal transformer) Two loads one on each side of the secondary Np Rin Rload * o Ns o When Rload = ( * 500) Rin = 1K * (1/14.4) Rin = 4.8 ohms One load on ½ of the secondary o When Rload = 500 Rin = 500 * (7./1) Rin = 9.6 ohms See the real section for the measured values used in the calculations below One load on ½ of the secondary * RP o F cutofflower = (* LB) *0.8 F cutofflower = (*3.6mH ) F cutofflower = 35.3 hz Np Rload * Ns o F cutoffupper = (* LA) 1 500* 7. F cutoffupper = (*11u ) F cutofflower = 69.8 Khz Two loads one on each side of the secondary William R. Robinson Jr. p4of 17

5 Vout is just * Vout for one load on ½ of the secondary o so frequency response should be the same F cutofflower = 35.3 hz F cutofflower = 69.8 Khz William R. Robinson Jr. p5of 17

6 Simulation (B Spice) See the real section for the measured values used in the calculations below Gain Two loads one on each side of the secondary Tapped Transformer-Small Signal AC-1-Graph vm(vout) vm(vout1) k 100k 1000k 1.000M 100M 1000M 1.000G 100 o Gain = 11./5.6 The gain is significantly lower than ideal transformer but calculations did not account for internal resistance Calculations did not account for internal resistance(s) The drop across the secondary resistance RS o V drop_secondary = Vout*RS/(RS+Rload) o V drop_secondary = 14.4 * 110/( ) o V drop_secondary = 1.43 V quite significant The drop across the two RPs in the model o Iout = 14.4V/( ) = 14.4 ma o In = 1/turns ratio * Iout = 14.4 ma/(1/14.4) = 07 ma o V drop_primary = * RP In = * 0.8 * 07mA = 0.331V So V drop_secondary and V drop_primary drops about 1.8V As discussed in Transfomer.doc 1 : In a real transformer power loss in the primary and secondary are distributed across the inductances but in the model RP and Rs in series leading to higher voltage loss than in the real transformer If the Resistances are removed then the model would draw more current than the real transformer When significant currents and/or large internal resistances are involved the BSpice model can show significantly lower output than a real transformer will. William R. Robinson Jr. p6of 17

7 One load on ½ of the secondary Tapped Transformer-Small Signal AC-10-Graph vm(vout) m -5000m k 100k 1000k 1.000M 100M 1000M 1.000G 100 o Gain = 5.6 Gain is lower than ideal transformer but calculations did not account for internal resistance Calculations did not account for internal resistance(s) The drop across the secondary resistance RS o V drop_secondary = Vout*RS/(RS+Rload) o V drop_secondary = 7. * 55/( ) o V drop_secondary = V The drop across the two RPs in the model o Iout = 7.V/500 = 14.4 ma o In = 1/turns ratio * Iout = 14.4mA/(1/7.) = 104 ma o V drop_primary = * RP In = * 0.8 * 104mA = 0.165V So V drop_secondary and V drop_primary drops about 0.88V As discussed in Transfomer.doc 1 : In a real transformer power loss in the primary and secondary are distributed across the inductances but in the model RP and Rs in series leading to higher voltage loss than in the real transformer If the Resistances are removed then the model would draw more current than the real transformer When significant currents and/or large internal resistances are involved the BSpice model can show significantly lower output than a real transformer will. Input Impedance (ideal transformer) Two loads one on each side of the secondary o The input impedance is calculated below William R. Robinson Jr. p7of 17

8 Tapped Transformer-Small Signal AC-6-Graph vm(vsense) m 1.000m m 100m 9.000m 8.000m 7.000m 6.000m 5.000m 4.000m 3.000m.000m 1.000m m k 100k 1000k 1.000M 100M 1000M 1.000G 100 Iin = Vsense/ Rsense Iin = 1.6 mv/1 ohm Iin = 160 ma Rintotal = (Vin Vsense)/Iin Rin = (1V- 1.6 mv)/ 160mA Rin = 5.5 ohms Rin (refected) = Rintotal RP* Rin = 5.5 *0.8 Rin = 3.65 ohms One load on ½ of the secondary o The input impedance is calculated below vm(vsense) Tapped Transformer-Small Signal AC-5-Graph m 1.000m m 100m 9.000m 8.000m 7.000m 6.000m 5.000m 4.000m 3.000m.000m 1.000m m k 100k 1000k 1.000M 100M 1000M 1.000G 100 Iin = Vsense/ Rsense Iin = 815uV/1 ohm Iin = 81.5 ma Rin = (Vin Vsense)/Iin Rin = (1V- 815 uv)/81.5ma Rin = 1.3 ohms Rin (refected) = Rintotal RP* Rin = 1.3 *0.8 Rin = 10.6 ohms William R. Robinson Jr. p8of 17

9 Two loads one on each side of the secondary o F cutofflower = 40 hz o F cutoffupper = 91 Khz One load on ½ of the secondary o F cutofflower = 41 hz o F cutoffupper = 89 Khz William R. Robinson Jr. p9of 17

10 Real Circuit Center_Tapped_Transformer I was unable to find a data sheet for the transformer I used in the real circuit, based on using it forward and backwards as a non-tapped transformer.i determined that its turns ration was 14.4:1 o For 1./ the secondary this would be 7.:1 I measured the following parameters for the transformer and used these values for the model (this is the same transformer used in transformer.doc 1 o ½ Primary resistance = 0.8 o Secondary resistance = 55 ohms each side o ½ Primary inductance = 55 or 500 mh o Because the two halves mutually induct, the total inductance is 4 times as large as the inductance of ½ of the turns o Secondary inductance = 3.6 mh LA was determined by the method discussed in reference 5 o Measurements for Rin with only ½ of the secondary produced nearly identical results up through 800Khz so the same value of LA is used in theses simulations o Zin vs Impedance ohms All of secondary shorted 1/ of secondary shorted Khz o The work for this can be seen in Figure Transformers.doc 1 LA = 1/L = 10.6uH With source Vin conveniently set up to be at 1.0V Gain (ideal transformer) William R. Robinson Jr. p10of 17

11 Two loads one on each side of the secondary Measured Gain vs, Two 500 ohm loads one on each side Gain 3 open 1K Khz o For Frequencies below 0KHz (audio range is 0 0 Khz) o Gain = 10.6 (in audio range) One load on ½ of the secondary Measured Gain vs, One 500 ohm load on 1/ of secondary Gain open 500 ohms Frequecy Khz o For Frequencies below about 40 Khz William R. Robinson Jr. p11of 17

12 o Gain = 6.5 (in audio range) Input Impedance Two loads one on each side of the secondary o The input impedance is calculated below Vsense over 1 ohm resistor was measured at 150 mv Iin = Vsense/ Rsense Iin = 150mV/1 ohm Iin = 150 ma Rin = (Vin Vsense)/Iin Rin = (1V- 150mV)150mA Rin = 5.7 ohms Rin (refected) = Rintotal RP* Rin = 5.7 *0.8 Rin = 4.1 ohms One load on ½ of the secondary o The input impedance is calculated below Vsense over 1 ohm resistor was measured at 100 mv Iin = Vsense/ Rsense Iin = 100mV/1 ohm Iin = 100 ma Rin = (Vin Vsense)/Iin Rin = (1V- 100mV)100mA Rin = 9 ohms Rin (refected) = Rintotal RP* Rin = 5.7 *0.8 Rin = 7.4 ohms Two loads one on each side of the secondary o F cutofflower = <100 hz o F cutoffupper = 55 Khz One load on ½ of the secondary o F cutofflower = <100 hz o F cutoffupper = 100 Khz William R. Robinson Jr. p1of 17

13 Comparison Center_Tapped_Transformer Two loads one on each side of the secondary o and Gain response of the model is not precise. This is supported also by reference 5 you should always be aware that these seemingly simple structures are in fact very complex elctromagnetic devices. Linear circuit models are a very crude approximation to the real component, and most of the elements of the circuit models have strong nonlinearities in them. For this reason, you can only expect very limited results in trying to run circuit simulators on power supplies. You should always make extended frequency response measurements on transformers when you are developing components. This will show increase in resistance with frequency, and change in leakage inductance, allowing you to properly specify the test conditions for a tightly-controlled part. The changes in the winding resistance and leakage inductance will be strongly dependent upon the physical winding layouts of the transformer, and great care should be taken to control this as tightly as possible during design and manufacturing. Comparision Measured vs Simulated, Two (500 ohm) Loads one on each side of secondary Gain K Simulated 1K Khz Real-Measured Simulation Calculated Gain Rin ohms F cutofflower hz < F cutoffupper Khz William R. Robinson Jr. p13of 17

14 Center_Tapped_Transformer One load on ½ of the secondary o and Gain response of the model is not precise. Note measured has addition peak at about 60 Khz This is supported also by reference 5 you should always be aware that these seemingly simple structures are in fact very complex elctromagnetic devices. Linear circuit models are a very crude approximation to the real component, and most of the elements of the circuit models have strong nonlinearities in them. For this reason, you can only expect very limited results in trying to run circuit simulators on power supplies. You should always make extended frequency response measurements on transformers when you are developing components. This will show increase in resistance with frequency, and change in leakage inductance, allowing you to properly specify the test conditions for a tightly-controlled part. The changes in the winding resistance and leakage inductance will be strongly dependent upon the physical winding layouts of the transformer, and great care should be taken to control this as tightly as possible during design and manufacturing. Comparision Measured vs Simulated, one (500 ohm) load on 1/ of secondary Gain ohms Simulated 500 ohms Khz Real-Measured Simulation Calculated Gain Rin ohm F cutofflower hz < F cutoffupper Khz > William R. Robinson Jr. p14of 17

15 Comparison of two transformer model to flipped transformer.doc model. It is interesting to note how two (500 ohm) loads one on each side of the two secondary s compares one 1K load in transformer.doc Two loads one on each side of the secondary Real-Measured Simulation Calculated Gain Rin ohms F cutofflower hz < F cutoffupper Khz K load from transformer.doc Real-Measured Simulation Calculated Gain Rin ohm F cutofflower hz < F cutoffupper Khz Correctly so the real circuits match but the Gain and F cutoffupper for do not correlate well F cutoffupper o The single transformer model seems reasonable for simulation and Calculated, however the two transformer to emulate a center tap does not. Cutoff Np Rload * Ns o F cutoffupper = (* LA) For transformer.doc 1 1K * 14.4 F cutoffupper = (*11uH ) For Tapped transformer.doc * 7. F cutoffupper = (*11u ) Cutting Rload in two and dividing by ½ the turns ration squared cannot possibly give the same results unless LA was also ½ as large. Several data points of Zin verses frequency with the secondary shorted do not support this Futhermore if it is halfed in simulation we still do not get the same frequency respone William R. Robinson Jr. p15of 17

16 o Using the two transformer model for center tapped secondaries has a poor frequency response. o At least in this case it is closer to the measured response than the flipped model from transformer.doc Two loads one on each side of the secondary vm(vout) vm(vout1) Tapped Transformer-Small Signal AC-1-Graph k 100k 1000k 1.000M 100M 1000M 1.000G 100 Gain vs frequency for flipped model vm(vout1) Transformer Step Up by flip of Step Down-Small Signal AC--Graph k 100k 1000k 1.000M 100M 1000M 1.000G 100 o The advantage of the two transformer model is that it can handle noncenter tappeed transformers for proper gain and Zin William R. Robinson Jr. p16of 17

17 References 1. Robinson, William AJ4MC, Transformer (another of these short papers), online, accessed Morehouse, Harvy, Magnetics Transformer Modeling online, accessed UNKNOWN, The ARRL Handbook For Radio Communications, (ARRL 010) P.6 4. UNKNOWN, Winding Configurations, online, accessed Dr. Ray Ridley, (Power Systems Design Europe January/February 007), High Power Transformer Measurement and Modeling, online, accessed 010. William R. Robinson Jr. p17of 17