Questions on Transformers - Page 1
Forum section: electronics / beginners

author: laingalion
author_rank: Contributor
timestamp: January 03, 2015, 05:45:08 am
content: Transformers (and magnetics in general) have always confused me.Here are some questions I have:1.In a situation where a well made transformer is used in the appropriate application, its assumed that roughly all the power is being transferred from the primary to the secondary.In this case, the impedance measured from the primary is simply proportional to the impedance measured on the secondary and the turns ratio. Correct?However, from my understanding, when the secondary is an open circuit the transformer appears to be just an inductor from the primary side. Power will be stored in the form of a magnetic flux in the transformer and be returned with a phase shift back to the primary. Now the impedance measured from the primary is no longer proportional to the secondary impedance and the turns ratio.My mind is having a difficult time grasping the transition between the situation with nearly full power transfer and the situation with infinite impedance on the secondary (no power transfer).A) I assume there must always be some magnetic flux being returned to the primary depending on the secondary impedance (and maybe turns ratio?). Is there an equation for this?B) How do you calculate the impedance seen from the primary as the secondary transitions from zero impedance to infinite impedance?C) When the secondary is at short circuit - zero impedance, does the primary also see zero impedance or does it see the impedance from the inductance of the transformer?2.I made my own small transformer using a ceramic ferrite pot core (Using it around 25 kHz).For the sake of impedance tuning, I created an small air gap (2-3 sheets of paper thick) between the top and bottom halves of the pot core.From what I've read, the majority of the magnetic flux is stored in the gap rather than the core. I understand the gap is a source of leakage flux.A) When the magnetic flux in the gap collapses, does the power get sent back to the primary or the secondary OR is the power divided between both the primary and secondary proportionally to the turns ratio?B) Does the magnetic flux in the air gap collapse at a 90 phase shift like the flux in the rest of the transformer or does it have some funny properties to it?I've fried over a dozen handmade inverters so I'm trying to understand if theres any funny sudden overcurrent business going on that I'm not calculating.Thanks in advance!P.S. For anyone also trying to gain a better understanding of transformer this blog helped me a lot:http://ludens.cl/Electron/Magnet.html

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 03, 2015, 06:46:35 am
content: The magnetizing current is proportional to applied voltage, regardless of load: at the primary, it appears in parallel with the secondary reflected impedance.The equation for parallel impedances is:Ztot = Z1 * Z2 / (Z1 + Z2)So that, when the secondary impedance is very small, it dominates.If Z2 << Z1,Ztot ~= Z1 * Z2 / (Z1 + <a small number>)~= Z1 * Z2 / Z1= Z2Or vice versa (when Z2 >> Z1, Ztot ~= Z1).As for flux, flux is a "total loop" phenomenon. Flux cannot be "stored" in an area. Flux is carried in a loop or path. The flux is constant at all points around the loop, though the fluxdensitycan vary (if the cross sectional area of the loop varies).Magnetization current arises due to energy storage in an air gap. This is the purpose of the air gap: to increase magnetizing current. This is generally undesirable for transformers, but necessary for coupled inductors and single winding inductors (which must store energy).Magnetic flux, give or take a constant (due to number of turns, core area, geometry, etc.), is proportional to the integral of voltage. If you apply some random AC waveform, peak flux will be inversely proportional to frequency. If the waveform is a sine wave, yes, it will vary as the cosine wave, because integral(sin) = -cos (plus a constant -- which we're assuming is zero, because the waveform is AC only -- which isn't always the case, such as in push-pull inverters, where you can get "flux walking").The reflected impedance of a short is limited by the leakage inductance of the transformer. This depends almost entirely on the geometry of the windings, and not on the core. Literally, it is the flux contained in the space between windings and core. When the windings are very closely coupled (bifilar construction, interleaved layers, etc.), leakage is small. When the windings are very distant (opposite legs of a C core?), the leakage is very high. This can be used to advantage in some circuits (resonant converters generally depend on leakage inductance, whether incorporated into the transformer design, or added as an explicit inductor), but is typically a disadvantage (causing excess power loss and peak voltages in flyback converters, for example).Tim

author: laingalion
author_rank: Contributor
timestamp: January 10, 2015, 03:41:02 am
content: This makes a lot more sense now, especially parallel impedance, magnetic flux, and air gap.I really appreciate the responseI want to use this transformer to power a piezoelectric transducer. I was reading somewhere that I should add the airgap in the transformer so that the inductance of the transformer is tuned with the capacitance of the transducer. However, this requires some airgap and therefore losses.If I take away the airgap on the transformer I get ridiculously more inductance on the secondary (this is a step up transformer) and the transducer is no longer tuned.The only thing I care about is the total power out. Do you think its better to have an non-tuned system with less losses on the transformer or a tuned system? Which would get me more power out?Is there a way to get the best of both worlds?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 10, 2015, 02:22:37 pm
content: What are you driving it from?A transformer is just a transformer. As long as it has enough current and voltage capacity, you can put the inductance on the primary or secondary side, if the load requires tuning. It's usually better to have it on the secondary side (with a dedicated inductor, or combining it with the transformer's magnetizing inductance by using a gap), because that reduces reactive current passing through the transformer (or the primary at least).So, yes, tuning at the load reduces losses in the transformer (or reduces the size and cost of the transformer for the same amount of loss).If the only thing you care about is total power out, you should probably use a better circuit, such as a series resonant power oscillator. This way, the supply voltage can be lower, without needing a transformer at all. But you haven't mentioned what voltage or current range any of these components operate at, so it's impossible to say which topology would be best suited.Tim

author: laingalion
author_rank: Contributor
timestamp: January 10, 2015, 10:46:33 pm
content: Ah, sorry I did forget to mention some key points.I'm using a 24 volt lithium battery with roughly a 10-15 amp draw. (The transducer is putting a lot of energy into water)I'm using a homemade (amateur design) full bridge inverter which feeds a ceramic ferrite pot core at frequencies 20 kHz to 30 kHz.When you say a "A transformer is just a transformer." do you mean that it does not contribute to the tuning if it has no air gap?From what I've tried, if I were to remove the air gap altogether the inductance on the secondary will be too high.This is another point which confuses me about transformers. Even in an ideal transformer, doesn't the transformer add a lot of inductance to the circuit?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 10, 2015, 11:25:28 pm
content: No, it doesn't add a lot of inductance, it connects it in parallel. Check out equivalent transformer models (linear non-ideal, with magnetizing and leakage inductances) and complex (AC) impedances in series and parallel.Tim