ePanorama.net - Electronic components
Well there is nothing I like better than finding the right intuitive model for something. Intuition is so fast if you can keep out the bad intuition.
Theoretical transfomer models
Based on 
Questions about transformers are often easier to answer, if you consider a "t" equivalent circuit. You lose the notion of isolation with the "t" equivalent but you can get it back by pretending there is an ideal transfomer connected between the "t" and the load. You can also put the turns ratio in the ideal transformer if you want, so the all values are as seen by the primary.
Example of "t" equivalen circuit
Here is the "t" equivalent circuit for a 1:1 audio
isolation transformer (designed for a 300 ohm load):
Primary | Secondary
Side Lm Side
- R1,R2 = primary and secondary winding (copper) resistence. Typically about 50 ohms. Not necessarily equal.
- L1,L2 = primary and secondary leakage inductances. About 5 mH. Not necessarily equal.
- Lm = mutual inductance, about 2H.
I called Lm the mutual inductance and tha's probably not the best term, though I think that in a 1:1 the mutual inductance is about the same as the self inductance or shunt inductance or magnetizing inductance or what ever it is best called.
For simplification you can wind up combining both leakage inductances into a single inductance on either side of Lm.
Description of model operation
Well lets assume that there is 1.25Vrms at 1KHz on the primary and no load. The full 1.25V appears on the mutual inductance so that there is about 0.1 ma through the mutual inductance. This is the current that gives rise to the core flux. There is 0.1ma (.995 ma) through the leakage inductance and primary resistance too. In short through the primary winding lumped circuit.
Now, let there be a 300 ohm load. The voltage on the mutual inductance is decreased very little (you do not need to make complex analysis). Even if you put a short on the secondary, the mutual current is decreased only by a factor of about two.
This circuit above has neatly divided the current in two paths. In real transformer there is only 1 conduction path through each winding not two, but this model circuit behaves like the real one because of the cancellation effect.
Magnetic flux cancelation effect
But how much current goes through the primary winding ?
The answer is 0.1ma + 4.2ma. Why doesn't this current add to the flux in the core? Because the current in the secondary cancels it's effect. Energy goes into the load, not into the ferrite, because two magnetic fields bucking each other do cancel. That is fundamentaly what is meant by electromagnetic equations being linear. Of course in the near fields of the windings this is not true as can be easily seen by just drawing a closed curve around the circumferance of the coils wire at a single place. The directed integral of the B-field around the curve has to be proportional to the current inside. But in the bulk of the core the fields do cancel. You can think of it as bucking if you like, but a hall effect probe inserted in the center will read a very low field due to the almost complete cancelation. The integral of the stored energy in the magnetic field integral( B dot H ) over all space will be much less than the intgeral for currents in only one winding or the other but not both simultaneously.
There is an unavoidable magnetizing-flux present in any transformer, and the primary magnetizing-flux current. Of course, this current is through the primary inductance, and is +90 degrees WRT the voltage, and doesn't directly consume any power. However, this current does cause losses in the primary winding resistance. The magnitude of the flux is set by the voltage and frequency across the primary, and not by the load current (if any).
Remember the basic AC transformer formula, V = k f N Ac Bm, which tells us how much flux is present for any voltage and frequency ? This is the formula used to find Bmax, so we can be sure the transformer core isn't too close to saturation, which would introduce even more losses. Note, there's no term for load current in the formula.
Transformer short circuit current
Only the leakage inductance limits the current during a short. It seems that the current through the primary is limited by winding resistance and leakage resistance when the secondary is shorted.
Secondary voltage drop
The field in transformer core goes goes actuaally DOWN a bit when the transformer is loaded. This is due to the effective primary voltage being reduced by (primary current * resistance of primary winding):
Vs = IpRp + BA[omega]Np
- B is the r.m.s (not peak) induction
- A is the area of cross-section of the core
- [omega] is 2[pi]f, of course
- Np is the number of turns.
Other models for transformers
What about the isolation ?
Real transformer provides isolation between the input and output. The model above does not show the isolatio but is sufficent for most of the analysis. Where the isolation is needed in model you can pretend that there is an ideal transformer between the "t" and the load like in the picure below:
1:N ideal transformer
Primary | 0|| Secondary
Side Lm 0||0 Side
One model for ideal transformer wiht isolation
This model displays transformers intuitively, the way we most often think of them:
-> Ip -----R1---L1---+---, ,---L2----R2------ Is ->
Primary | O|| / Secondary
Side, Vp Lm O||O Side, Vs
| O|| \
Lm is the required magnetizing inductance. The perfect transformer
converts Vp to Vs by the ratio of the turns independant of frequency,
and draws a primary current Ip, related to the secondary current Is
by the inverse ratio of the turns. Should you desire to group the
series resistances Rx and the leakage inductances Lx, all on one side,
just move the values from the other side, translated by the square of
the turns ratio. This model also directly matches the simple bench
measurements we can take to characterize a transformer, measuring both
lead resistances, and primary inductance with the secondary both open
(magnetizing inductance) and shorted (leakage inductance L1 + L2*N^2).
Based on 
As to how we sort it out, in the one case that matters a lot to me, we specify the general winding details, a range for R1 and R2, maximum values for L1+L2||Lm (measured from primary with secondary shorted) and L2+L1||Lm (measured from the secondary with the primary shorted), minimum values for L1+Lm (measured from primary with secondary open) and L2+Lm. The vendor gets to choose the number of turns (same for both secondary and primary), the wire, and gets to play with the laminations (a mix of silicon steel and high nickel steel). Then at incoming inspection, we measure all those things. At this point, we have four measurements determining 3 things (L1,L2,Lm) so even though the turns ratio is 1:1, I pretend the turns ratio is 1:n which gives me 4 variables and four equations and I solve the whole mess.
If you heavily load a transformer with a resistive load so that the current draw is large compared to the no load current. You will find the currents and voltages are in phase. They have to move into phase because at 90 phase shift between current and voltage no net power transfer (averaged over one cycle occurs). As you well know, the power companies spend a lot of effort keeping current and voltage in phase ( hence power factor ).
It is true that the slope of a sinewave for both current and voltage is maximum at the zero crossings. I can see how that combined with V=LdI/dt makes it seem that the current and voltage should be 90 out of phase. BUT. That only happens for a unloaded transformer that looks like an inductor. For a resitively loaded transformer you will decrease the phase angle decrease with increasing load. This is easy for you to try, do it!
The reason for this is that, we can really only (simply) apply Amperes' law around a contour enclosing half of each winding. In that situation if you consider Vprimary and d( N*Iprimary - I secondary)/dt you will come up with the situation you have discribed where the difference of these currents and voltages are 90 degrees out of phase. BUT, (N*Iprimary-Isecondary) is much much less than Iprimary ( order of 1%) for a heavily loaded transformer. In that situation the dominate (largest) components currents can be in phase and typically are.
For example let us take a unloaded 1:1 transformer that draws 10ma when on loaded. Let us call this current Iinitial The current and voltage are 90 out of phase. But if we add 1 amp to Iprimary at the same time adding 1amp to Isecondary in phase with the voltage and each other ( or 180 degrees depending on the transformer polarity convention ) then d(Iprimary-Isecondary/1)/dt doesn't change it is still just the unloaded current Iinitial. However, if we look at the total transformer primary current, Iprimary+Iinitial = 1cos(wt)+.01sin(wt) then it is almost perfectly in phase with the voltage Vcos(wt)
Note the currents don't have to be in phase, if we load the transformer output with a large capacitor or a small inductor much larger currents will flow but the phase of the primary current will change accordingly.
If the input voltage and currents are 90 deg out of phase no power goes in. If the input and output voltages are 90 degrees out of phase then everything everyone learned about transformers is totally wrong.
The following assumes an ideal transformer:
If I apply a voltage to the primary I get a voltage on the secondary. The voltage ratio is function of the turns ratio. Now if I connect a purely resistive load to the secondary I get a secondary current given by the secondary voltage and the value of the resistance. The current in the primary is a function of the secondary current and the turns ratio. The output power may be calculated from Vs^2*Rload. The input power may be calculated from Vp*Ip. For an ideal transformer these two numbers are equal. If there is a phase difference between them then this cannot be true. Pin = Pout! Not Pout=Pin*cos(theta).
Text book definitions for ideal transformers:
Vs=Vp*(Ns/Np) Ip=Is*(Ns/Np) Pin=PoutIt seems quite clear to me that there can be no phase differences (at least for the ideal transformer).
Based on 
Measuring B-H curve
You can easily display the B-H curve of a transformer on a scope that can do X-Y display with just a couple of components. A heater transformer (For those that remember valves - or tubes as the locals say) used in reverse works well. Feed it with 6.3v AC from another similar transformer.
R2 senses the current in the primary (The magnetizing force) - it should be selected to give a couple of volts for the X-axis of the display - a few ohms.
R1 and C1 act as a crude integrator, since the voltage across the secondary of the transformer is proportional to the rate of change of the magnetic field rather than the field itself. Select R1 to give negligible loading on the transformer (it could be 100s of K) and C1 so that the voltage across it is less than 5% of voltage on the secondary of the transformer.
R1 -------- ----/\/\/\--|-- Scope Y input )||( | )||( 240/120 = C1 6.3v )||( | )||(____________|___ Scope Ground | |_________ Scope X input | \ / \ R2 / | ------------------ Scope GroundYou cna fo example use 100 kohms resistor and 3 uF capacitor for this circuit.
Other transformer measurement ideas
Here is some basic measurements to get to know most of transfromer parameters:
- 1. Pri/Sec winding resistances can be measured wih multimeter directly.
- 2. Measure open circuit secondary volts with some known primary voltage to get the turns-ratio.
- 3. Short the secondary with an ammeter and plot V-primary against I-secondary. (A scope across the ammeter might be handy to check sec waveshape, just in case.)
- 4. As a crosscheck do V-pri/I-pri with a shorted sec.
Leakeage inductance measurement
Here are leakage inductance measurements done two different ways, which show pretty good agreement.
Secondary equivalent circuit:
O------(Rs+Rp/N^2)-----(Ls+Lp/N^2)-------+ | | | E(t) Rl | | O----------------------------------------+Open circuit voltages on my transformer show Ns/Np = 8.5. Other measured values:
Rp = 144.5 ohm Rs = 2.13 ohm E(t) = 14.17 V Rl = 25.20 ohm Voltage across Rl = 12.20 VSecondary impedance Zs = 14.17 V / (12.20 V / 25.2 ohm) = 29.3 ohm
(Ls+Lp/N^2) = Sqrt( Zs^2 - ( (Rs+Rp/N^2) + Rl )^2 ) = 0 , effectively.
The quantity (Rs+Rp/N^2) + Rl = 29.33 ohm which is not significantly different from Zs .
It appears that within the limits of experimental error, I cannot resolve any total leakage inductance in this experiment. Thus the leakage inductance must be very negligible.
In the second approach, I shorted the primary winding, and applied voltage to the primary.
Secondary equivalent circuit with the primary shorted:
O------(Rs+Rp/N^2)----+ | | E(t) | | | O----(Ls+Lp/N^2)------+This time,
E(t) = 1.463 V Is = 0.342 A Zs = 4.28 ohm(Ls+Lp/N^2) = Sqrt( Zs^2 - (Rs+Rp/N^2)^2 ) = 3 mH
Here the quantity Rs+Rp/N^2 = 4.13 which is different from Zs.
In this case Rp/N^2 = 2, which is consistent with Rs = 2.13 for a well-designed transformer. The primary should have just a little more winding area than the secondary.
The measured leakage reactance (3 mH) is a bit on the high side, but not unreasonable for a laminated transformer. It's too high for a well-designed toroid. Anyway in measurements like theat the accuracy of the measurements needs to taken account.
The current waveforms should be reasonably nearly sinusoidal in both tests, unlike the no-load primary current.
Transformer design and selection for applications
Transformer core type selection
TOROIDS vs. E-CORES ADVANTAGESToroids:
- More compact than E-core design
- Materials cost is lower due to single component
- Tighter magnetic coupling - lower stray flux leakage
- Easier to automate winding process
- Can be mounted by pins on the bobbins
- Easier to isolate electrically multiple windings
- Core can be easily gapped to extend energy storage capability
Power transformer designing principles
Based on 
It has been my suspocion that to save iron and weight, most power transformers are designed to operate right on the ragged edge of saturation, hence all hell can break loose (at least transformer hears more) when you take a product designed for 60Hz service and power it with 50Hz.
Designing a power transformer is quite careful thng if an optimized design is needed. To get a general view of the design of a power transformer I geve here you some approzimate design equations for 50 Hz power transformer using laminated iron transformer E-core:
primary turns = 45 * primary voltage / core area secondady turns = 48 * secondary voltage / core area core area = 1.1 * sqrt ( P )Where:
- core area = cross square ares of the core going through the coil in square centimeters
- primary voltage = AC voltage fed to prmary in volts
- secondary voltage = AC voltage wanted on secondary in volts
- P = transformer power
The wire in primary and secondary must be sized according the allowed voltage drops and heating inside the transformer. As a rule of thumb do not try to push more then 2.5 amperes of current per square millimeter of the wire in coils inside transformer.
The size of the transformer core must be determined based on the transformer total power. The area of the core (as used in equation above) should at least have the value accoring the following equation (can be larger):
core area = sqrt ( transformer power in watts )
Here is a table of wire sized for different currents suitable for power transformers:
Current Wire diameter (mA) (mm) 10 0,05 25 0,13 50 0,17 100 0,25 300 0,37 500 0,48 1000 0,7 3000 1,2 5000 1,54 10000 2,24If you make a transformer using those equations, you mith throughly test it before connecting it to the mains power. Generally nowadays it is a good idea to buy a mains transformer ready made and so make sure that you get a product which is safe to use (fullfills all the safety regulations).
Low frequency transformers
Based on 
For low power low frequency transformers you can generally determine that the turns ratio determines the voltage transfer ratio. For given impedance circuti you need to determine the minimum impedance for a certain transformer coil using the following formula:
L = Z / (2 * pi * f)Where:
- L = primary coil inductance (secondary open circuit)
- Z = circuit impedance
- pi = 3.14159
- f = lowest frequency the transformer must work
The actual number of turns needed to get the necessary inductance depends on the tranformer core model and magnetic material used it. Consult the datasheet of the coil material you are using for more details or it. Other option is to first wire one test coul and measure it. Using the measurement results you can determine how many turns are needed for a specific inductance. General approximare inductance formula (for coils with cores) is useful for this:
L = N * N * aWhere:
- L = inductance
- N = number of turns
- a = a constant value (determine value from coil core data or measure it with test coil)
If you are using iron core and need to transfer some power you can determine the needed core size using formula:
Afe = sqrt ( P / (Bmax * S * f) )Where:
- Afe = core area (cm^2)
- P = maximum transmitted power
- Bmax = maximum magnetic flux in core (Vs/m^2) (usually 4000 G = 0.4 Vs/m^2)
- S = Current density (A/mm^2) (usually 0.5 A/mm^2)
- f = lowest frequency transformer needs to operate (Hz)
Transformers without air gap
And when you know core area you can calculate the number of turns for transformer primary for transformer without air gap in core using the following formula:
N1 = sqrt ( (10^8 * L1 * l) / (u * Afe) )Where:
- N1 = number of turns in primary coil
- Afe = core area (cm^2)
- L1 = primary coil inductance (H)
- l = average length of magnet flow force lines (cm) (lenght of line around coil going through inside the core)
- u = relative permiability of magnetic material (around 500 for typical transformer iron)
You can determine the number of turns on secondary coil using the following formula (expects transformer effiency of 90%):
N2 = 1.1 * U2 / U1 = 1.1 * sqrt (Z2 / Z1) =Where:
- N1 = number of turns in primary coil
- N2 = number of turns in secondary coil
- U1 = primary voltage
- U2 = secondary voltage
- Z1 = primary impedance
- Z2 = secondary impedance
Transformers with air gap
If there is any DC current flowing on transformer primary, the primary inductance is reduced. To compensate the effect of this (in circuits where this is a problem) the core should have a small air gap in the core. In practice the air gap should be selected to be around 1/1000 of the length of the magnetic lines in the core. In this case the following equation can be used to determine the number of turns needed for primary coil:
N1 = sqrt ( (L1 * li) / (Afe * 10^8) )Where:
- N1 = number of turns in primary coil
- Afe = core area (cm^2)
- L1 = primary coil inductance (H)
- li = size of the air gap (mm)
Based on 
Impedance matching transformer selection
Matching is required to ensure maximum power transfer from source to the load. A matched condition exists when:
N = N2 / N1 = sqrt (Zl / Zs)Where:
- N = turns ratio between primary and secondary
- N1 = number of turns in primary
- N2 = number of turns in secondary
- Zs = signal source impedance
- Zl = transformer load impedance
Lp = 5 * Zs / (2 * pi * fmin)Where:
- Lp = prmary inductance
- Zs = source impedance
- fmin = minimum frequency needed to be transferred through transformer
- pi = 3.14159
Selection procedures for pulse matching transformers
It is necessary to check for pulse distortion when selecting the transformer. There is a maximum area of pulse which a given transformer can transmit. This is known as the Et constant. The following formulas will describe how this may be estimated from the known pulse shape
Et = Vp * tpw Lp = R * tpw / Ln (I - D) D = delta / Vp = 1 - exp (-R * tt / Lp) 0 < tt < tpwWhere:
- tpw = the worst-case (maximum) pulse width to be transmitted
- Vp = pulse voltage (voltage from top to bottom)
- delta = how much pulse top is allowed to drop
- tt = time the pulse top is active (tpw - start and end slopes)
- D = droop (usually 10& can be tolerated)
- R = parallel combination of the source and reflected load impedance (for a matched case this is half of the source impedance)
The other distortion which should be checked is droop. The droop in in relation to the pulse time, primary inductance and system impedances. Unless otherwise specified a droop of 10% can usually be tolerated. Here again excessive inductance brings parasitics and their attendant problems.
From the preceeing description we can propose a strategy which should enable us to select correct components in the majority of applications.
- 1. Indentify the sytem impedances Zs and Zl
- 2. Identify minimum operating frequency (fmin)
- 3. Identify maximum pulse width (tpw) and voltage (Vp)
- 4. Calculate turns ratio from: N = sqrt (Zl / Zs)
- 5. Calculate minimum primary inductance from: Lp(min) = 2.5 * Zs / (2 * pi * fmin)
- 6. Calculate minimum Et constant from: Et(min) = Vp * tpw
- 7. Check that droop is acceptable (propably < 10%): D = 1 - exp ( -Zs * tpw / (2*Lp) )
- 8. If droop is unacceptable re-calculate Lp from: Lp = - Zs * tpw / (2 * Lp)
- 9. Select the device that meets the above specifications with the lowest values of leakage inductance and interwinding capacitance.
The approximations made in the formulas that the strategy has its limitations but the errors are usually negligble.
Transformers for thyristor drives
Transformers are used in thyristor drives to provide isolation of control circuitry and voltage/current transformation. In order for the thyristor to switch on the gate must be held high until the current in the thyristor exceeds the holding current of the device. This time depends on the device itself and the load characteristics. A resistive load will have a fast current rise time and hence require a narrower pulse than would and inductive load. Unfortunately the majority of the applications are for motor drives and it is often difficult to define a figure for maximum pulse duration.
It is also important to ensure that the thyristor does not turn on too slowly. This leads to local "hot spots" in the device and premature device failure. This requirement means that the transformer should have as low a leakage inductance as possible.
For applications where pulse-width-modulation (PWM) techniques are to be employed it should be remembered that it is very difficult, if not impossible, to work pulse transformers and more thatn 60% mark:space ratio. The reason for this is that the transformer needs requires time to reset between pulses.
Details on using transformers in electronics designs
Based on 
Low signal distortion
Yes, 'bottom-bend distortion' is something to watch out for in using silicon-iron cored transformers for audio applications outside their specifications. The usual case is using too big a transformer, so that the induction at low signal levels is minute. It can also happen with nickel-iron cores, but only at extremely low inductions indeed.
When the hysteresis curve is first introduced to students, the S-shaped 'initial magnetizing curve' is usually drawn first, then the BH loop. After that, the S-shape of the initial curve is forgotten, but that bottom bend is still there, waiting to bite you!
As for the curve being linear at low intensity, we all know that the B-H curve flattens out at the top, but I think you will find there is a flattening around the origion also.
For example it can happen than when you decreased the primary signal by 80dB, the secondary signal can be decreased for exmaple decreased by 81dB. T Indeed the BH curve does have a flattening near zero. This problem can be reduced by using right size air gap in the transformer core, which makes is possible to get for example linearity over a 80 dB.
Information on transformers used in switched mode power supplies
The output voltage on a high frequency transformer has the same waveform (not voltage necessarily) as the input waveform (leakage etc ignored). In fact the the secondary current may be 'sensed' or measured from the primary, as it commonly is for current mode control systems or even voltage mode controller schemes with over load protection. The secondary voltage and current are completely in phase with the primary voltage and current.
Below is what I hope is clear as typical voltag and current wavefroms for a two phase SMPS forward converter:
|-------| |------ | | | Pvolts ---| |---| |---| | | |--------| /| /| / | / | / | / | / | / | / | / | / / | / | / / | / | / Pcurrent / | / | / / |---/ |---/You will no doubt recognise the inductor current waveform in the primary current waveform above. This is all down to the fact that the input and output voltage and current waveforms are totally in phase (ignoring leakage L C etc etc).
What is the difference between laminated and toroidal transformer ?
There is no dramatic difference between a toroidal transformer and a conventional transformer. Both work in same way. Basically the difference is just the mechanical form of the transformer.
The main difference is that the traditional transformer and toroidal transformer are wound to a different for of transformer core. The traditonal tranformer use typically so called "E"-cores which are made of stacks of iron. Toroidla transformer used a toroidal trnasformer core (shape of "O"). a torrid core provides a closed magnetic circuit and does'nt loose any flux into free space as it would if the same core was in the shape of a rod. lost flux is lost energy, therefore, a torrid will provide higher inductance, tighter coupling , higher efficiency, and higher Q, and on and on. The whole concept is to physically concentrate the flux where it is needed. Also, because the flux is concentrated in the core, components that would normally be affected by being in the proximity of an inductor/transformer, can be mounted closer to a torrid, and a torrid will generally be smaller than an inductor or transformer using more conventional core shapes.
Toroids are usually made from thinner strip of a higher grade of silicon iron and they have a truly continuous magnetic circuit. Those are the basic characteristics that give rise to lower losses and near-zero external magnetic field, which are the usual reasons for choosing an, often more costly, toroid rather than a laminated-core transformer.
In principle a perfect toroidal winding has no external magnetic field, and in practice toroidal transformers do have lower external fields, but transformer designers tend to design toroids to run closer to saturation, which increases the external field, largely eliminating the advantage.
Toroids are popular in hi-fi amplifiers because they allows claims about low external field, and - much more important - because the weight of the wound toroidal transformer is lower than than equivalent conventional transformer.
The "squashed" profile of the toroidal transformer also gives it more surface area per unit VA than a conventional transformer, so it dissipate more heat per unit temperature rise, which the designers exploit by running them at higher current density.
Power transformer details
When a transformer core saturates, it loses its inductive characteristics; primary winding current can then reach extremely high values for several ac cycles. Because transformers remain polarized when turned off, saturation occurrence is a function of the polarity and phase angle of the ac cycle when you switch the circuit on and off.
Transformer-core saturation can cause inexplicable fuse blowing, system crashes, or premature switch and relay failure Besides on transformer saturation inrush current from power-supply can be also caused by filter capacitors initial charge pulse.
By using a resistor, an inrush device, or an inductive input filter in the secondary winding, you can reduce this inrush surge. Another solution is to soft-start the transformer by using a resistor in the primary to limit inrush and saturation currents to an acceptable level.
-  Hannu Miettinen, K�yt�nn�n Elektroniikkaa, Infopress, 1976
-  Various Usenet news articles
-  Various Web documents
-  Newport Components Application Notes book
-  Smart switch cuts transformer turn-on current, EDN April 23, 1998