# Antenna Simulation in LTspice

Discussion in 'Electronic Design' started by rickman, Feb 28, 2013.

1. ### rickmanGuest

I am working on a simulation for a loop antenna in LTspice and I can't
figure out why the signal strength features are what they are. The
model uses a pair of loosely coupled inductors to model the transmitter
and antenna loop with a separate pair of tightly coupled inductors to
model the coupling transformer. A cap on the primary circuit is the
tuning cap and a cap on the secondary is parasitic effects of the
circuit board leading to the inputs on the IC.

There is a resonance near the frequency I would expect, but it is not so
close actually. I can't figure why it is about 5% off. There is a
second resonance fairly high up that I can't figure at all. None of the
component values seem to combine appropriately to produce this peak.
When looking at the tuning capacitor voltage there is an anti-resonance
that is exactly at the frequency corresponding to the secondary
resonance with the transformer and the parasitic capacitance. That
makes sense to me, but it is pretty much the only part that jibes with
what I can figure out.

I have uploaded a zip file with the schematic and a measurement file.

http://arius.com/temp/Antenna_trans_LTspice.zip

2. ### Tim WilliamsGuest

I expect if you reflect the CT secondary stuff (don't forget Lsec) back to
the primary, your answer will appear. Offhand I can't reason out which
sum of L and C makes the resonance, but it's a four pole series-parallel
resonant circuit, analysis should lay it bare.

Tim

3. ### rickmanGuest

the primary that means the primary inductance would be doubled? I
believe the coupling of the two coils means they are one and the same
for the purposes of the circuit analysis, no?

You can't reason which sum of L and C makes the resonance and I can't
either. The calculation is off by about 5% and I can't explain that. I
can explain a null at about 290 kHz. That is the resonance of the
secondary with the secondary capacitance. I can't explain the other
peak at 363 kHz at all. A higher frequency would imply a smaller L
and/or C. How do you combine them to produce that? Consider the two
caps to be in series???

4. ### Tim WilliamsGuest

Sure. If you bring the 10p over to the primary, it looks like 10p * (30m
/ 5u), or whatever the ratio was (I don't have it in front of me now), in
parallel with the primary. (I misspoke earlier, you can safely ignore Ls,
because k = 1. There's no flux which is not common to both windings.)

Inductors effectively in parallel also increase the expected resonant
frequency. If you have this,

.. L1
.. +-----UUU--+------+------+
.. | + | | |
.. ( Vsrc ) === C > R 3 L2
.. | - | > 3
.. | | | |
.. +----------+------+------+
.. _|_ GND

You might expect the resonant frequency is L2 + C, but it's actually (L1
|| L2) = Leq. If L1 is not substantially larger than L2, the resonant
frequency will be pulled higher.

Incidentally, don't forget to include loss components. I didn't see any
explict R on the schematic. I didn't check if you set the LTSpice default
parasitic ESR (cap), or DCR or EPR (coil) on the components. Besides
parasitic losses, your signal is going *somewhere*, and that "where"
consumes power!

The actual transmitter is most certainly not a perfect current source
inductor, nor is the receiver lossless. This simulation has no expression
for radiation in any direction that's not directly between the two
antennas: if all the power transmitted by the current source is reflected
back, even though it's through a 0.1% coupling coefficient, it has to go
somewhere. If it's coming back out the antenna, and it's not being burned
in the "transformer", it's coming back into the transmitter. This is at
odds with reality, where a 100% reflective antenna doesn't magically smoke
a distant transmitter, it simply reflects 99.9% back into space. The
transmitter hardly knows.

In this example, if you set R very large, you'll see ever more voltage on
the output, and ever more current draw from Vsrc. You can mitigate this
by increasing L1 still further, but the point is, if the source and load
(R) aren't matched in some fashion, the power will reflect back to the
transmitter and cause problems (in this case, power reflected back
in-phase causes excessive current draw; in the CCS case, reflected power
in-phase causes minimal voltage generation and little power transmission).

Power is always coming and going somewhere, and if you happen to forget
this fact, it'll reflect back and zap you in the butt sooner or later!

Tim

5. ### leggGuest

<snip>

Pulling out the old reactance paper, there are a couple of expected
interactions using the values present:

Around 50KHz (89.42uH+48uH) with 50.42nF (L3+L1) with C1
Around 290KHz (89.42uH+48uH) with 6.25nF (L3+L1) with C2*N^2
Around 360KHz 48uH with 6.25nF L1 with C2*N^2
nL1/nL2=N=25

The mid-resonance is a dip or rejection.

What's the issue?

RL

6. ### rickmanGuest

Reflecting the capacitance through the transformer changes it by the
square of the turns ratio assuming the coupling coefficient is
sufficiently high. I am simulating K at 1.

This is also true for the inductance, but in the opposite manner. So
going from the 25 turn side to the 1 turn side, the effective
capacitance is multiplied by 625 and the effective inductance (or
resistance) is divided by 625. In fact, in LTspice you indicate the
turns ratio by setting the inductance of the two coils by this ratio.

I see now that the reflected secondary capacitance is in parallel with
the primary, rather than in parallel with the primary capacitor. That
explains a lot... I'll have to hit the books to see how to calculate
this new arrangement. I found a very similar circuit in the Radiotron
Designer's Handbook. In section 4.6(iv)E on page 152 they show a
series-parallel combination that only differs in the placement of the
resistance in the parallel circuit. It need to be placed inline with
the inductor... or is placing it parallel correct since this is the
reflected resistance of the secondary? I'll have to cogitate on that a
bit. I'm thinking it would be properly placed inline with the capacitor
in the reflection since it is essentially inline in the secondary.
Either way I expect it will have little impact on the resonant frequency
and I can just toss all the resistances simplifying the math.

I do see one thing immediately. The null in Vcap I see is explained by
the parallel resonance of the secondary cap with the secondary inductor.
If you reflect that cap back to the primary in parallel with the
primary inductor (resonating at the same frequency) it explains the null
in the capacitor C1 voltage I see. C2' (reflected) and L1 make a
parallel resonance with a high impedance dropping the primary cap
current and voltage to a null. This null is calculated accurately.

What I need to do is change the impedance equation from Radiotron to one
indicating the voltage at Vout relative to the input signal. I think I
can do that by treating the circuit as a voltage divider taking the
ratio of the impedance at the input versus the impedance at the primary
coil. No?

I see, L1 and L2 are in parallel because the impedance of Vsrc is very
low. That is not the circuit I am simulating however. The loop of the
antenna and the loop of the inductor are in series along with the
primary capacitor. I'm not sure what the resistor is intended to
represent, perhaps transformer losses? The resistance of L1 was added
to the simulation model along with the resistance of the secondary coil
which you have not shown... I think. It seems to me you have left out
the tuning capacitor on the primary.

Interesting point. My primary goal with this is to simulate the
resonance of the tuning so I can understand how to best tune the
circuit. In many of the simulations I run the Q ends up being high
enough that a very small drift in the parasitic capacitance on the
secondary detunes the antenna and drops the signal level. It sounds
like there are other losses that will bring the Q much lower.

I would also like to have some idea of the signal strength to expect. My
understanding is that the radiation resistance of loop antennas is
pretty low. So not much energy will be radiated out. No?

You make it sound as if in the simulation, even with a small coupling
coefficient all the energy from antenna inductor will still couple back
into the transmitter inductor regardless of the K value. Do I
misunderstand you? It seems to result in the opposite, minimizing this
back coupling. Or are you saying that the simulation needs to simulate

Actually, my goal was to build the receiver and I realized that my
design would require the largest signal I could get from the antenna. I
never realized I would end up having to learn quite so much about
antenna design.

I've been planning to create a PCB with lots of options so I can test a
number of configurations. Nothing about the simulation makes me doubt
the utility of this idea.

One thing that continues to bug me is that nothing I have seen gives me
a hint on how to factor in the distributed capacitance of the antenna
shield. I am using RG6 with 16 pF/Ft and likely will end up with 100
foot of coax total. At some point I'll just have to make some
measurements and see what the real world does.

7. ### Tim WilliamsGuest

You'll be much better off simply using the conventional radio approach
than trying to simulate everything, especially when circuit equivalents
are nebulous like this.

After all, if you can't quite tell what it *should* look like, how would
you know if you could implement your model once you've found a
satisfactory result?

What kind of antenna are you looking at, loop? The first thing to know
about a loop is, if it's a very small loop (I'm guessing, at this
frequency, it is), its radiation resistance is very low, meaning, you can
treat it as a nearly pure inductance (Q > 10 I think is typical), and its
bandwidth (even with a matched load) will be correspondingly narrow.

The nature of the incoming signal could be modeled as a voltage or current
source; how doesn't really matter, because it isn't really either, it's a
power source that couples in. Again, you don't have voltage without
current and vice versa, it's all about power flow, and the matching that
allows the power to flow.

Since the loop is inductive, your first priority is to resonate it with a
capacitor at the desired frequency. This will require a very precise
value, and even for a single frequency, may require a variable capacitor
to account for manufacturing tolerances. In the AM BCB, a Q of 10 gets
you 50-160kHz bandwidth, so you only get a few channels for any given
tuning position. And if the Q is higher, you get even fewer.

Now that you've got a high Q resonant tank, you can do two things: couple
into the voltage across the capacitor, or the current through the
inductor. You need only a small fraction of either, because the Q is
still going to be large. This can be arranged with a voltage divider
(usually the capacitor is split into a huge hunk and a small variable
part, e.g., 300pF variable + 10nF, output from across the 10nF), a
transformer (a potential transformer across the cap, or a current
transformer in series with the inductor), an inductive pickup (the big
loop carries lots of volts, but you only need a few, so a much smaller
loop can be placed inside the big loop), an impractically large inductor
(like in my example circuit, which models radiation resistance as a
parallel equivalent), etc. Whatever the case, you need to match
transmission line impedance (e.g., 50 ohms) to radiation resistance
(whichever series or parallel equivalent you have).

Once you get the signal into a transmission line, with a reasonable match
(Z ~= Z_line, or alternately, SWR ~= 1), you can do whatever you want with
it. Put it into an amplifier (don't forget to match it, too), etc. Yes,
you're going to have funny behavior at other frequencies, and if you're
concerned about those frequencies, you'll have to choose the coupling
circuit and adjustable (or selectable) components accordingly. But for
the most part, you completely ignore any frequency that you aren't tuning
for, usually enforcing that concept by inserting filters to reject any
stragglers.

Example: suppose you have a loop of 5uH and need to tune it to 500kHz. It
has a reactance of 15.7 ohms. Suppose further it has Q = 20. The ESR
(not counting DCR and skin effect) is X_L / Q, or 0.78 ohms; alternately,
the EPR is X_L * Q, or 314 ohms. The capacitor required is 20.3nF. If we
use a current transformer to match to a 50 ohm line, it needs an impedance
ratio of 1:64, or a turns ratio of 1:8. If we use a voltage transformer,
it's of course 8:1. (A capacitor divider is unsuitable for resonant
impedances less than line impedance, since it can only divide the
impedance down. If the inductance were a lot larger, it could be used.)
To a rough approximation, a smaller inductive loop, of 1/8 diameter of the
larger, I think, would also work.

Tim

8. ### rickmanGuest

290 kHz matches the calculations you just gave. But 290 kHz is the null
(or dip as you call it) from C2 and L2 (or L1 and C2 reflected with N^2).

I thought I wasn't getting the 60 kHz resonance, but I was mistakenly
adding the two capacitances together. So that is closer. Using L3+L1
with C1 I get 60.46 kHz while it is measured at 60 dead on in
simulation. That's nearly a 1% error.

I solved the equations finally. I found some info on the impedance of
series and parallel circuits. With that info I wrote the equation for
the ratio of Vout/Vin and found the roots. Turns out it is not so bad.
The equation is a fourth order, but it has no x^3 or x^1 terms and so
is actually a quadratic of x^2. Solving the quadratic gives the exact
figures for 60 kHz and 393 kHz peaks. Since this is from taking the
square root of x^2, there are also solutions at the negative values... duh!

Reflecting C2 through the transformer to create C2', the two nulls I
found can be calculated by the resonance of L1 and C2' (290 kHz null on
C1) or L1 with C1 and C2' (96500 Hz null on L3).

9. ### rickmanGuest

I don't know what you mean by the "conventional radio approach".

I was simulating a specific circuit for a specific purpose. I got the

Yes, I plan to use a shielded loop. I have found some contradictory
info on the effectiveness of the "shield". One reference seems to have
measurements that show it is primarily E-field coupled in the longer
distance portion of the near-field.

I am aware of the low radiation resistance and have not included that
factor in my simulation. The Q of just the antenna loop is around 100
as calculated from the ratio of reactance to resistance.

A friend in a loop antenna Yahoo group suggested the use of the
transformer coupling with a low k to model the signal reception.

Yes, that is loop antenna 101 I think. It was when I added a coupling
transformer with 100:1 turns ratio that I was told I needed to consider
the parasitics. I have found it is not useful to go much above 25 or
33:1 on the turns ratio. I am receiving a single frequency, 60 kHz.
There is no need for a wide bandwidth. Ultimately, I prefer a Q of >
100 for the higher gain. If it gets too high, the off tuning by
variations (drift) in the parasitic capacitance affects the antenna gain
appreciably.

Transmission line? What transmission line? The antenna is directly
connected to the receiver which has a very high input impedance. Why do
I need to consider radiation resistance? I have not read that anywhere.

I'm not familiar with the concept of voltage transformer vs. current
transformer. How do you mean that?

How did you get the 1:64 impedance ratio and the 1:8 turns ratio? I
don't follow that. Are you saying the line impedance should match the
ESR? Why exactly would it need to match the ESR?

10. ### Tim WilliamsGuest

I trust this resource:
http://vk1od.net/antenna/shieldedloop/
He's got gobs of analytical articles.
High Q isn't the goal, high radiation resistance is -- the bigger the
loop, the better it couples with free space, until it's a wave length
around.

You can go ahead and make a teeny coil out of polished silver litz wire,
and push the Q up into the hundreds, but all you'll see is internal
resistance, hardly anything attributable to actual radiation. Since the
losses dominate over radiation, it makes a crappy antenna. But you know
that from looking at it -- it's a tiny lump, of course it's not going to
see the outside world.

It is true, however, that a small coil, with low losses, will have low
noise. AM radios rely on this, which is how they get away with tiny hunks
of ferrite for picking up radio.

Of course, it doesn't hurt that AM stations are 50kW or so, to push over
atmospheric noise.
Ok, then you can merge the matching transformer, transmission line and
receiver input transformer into one -- an even larger stepup into whatever
impedance it's looking at (what's "very high", kohms? Mohms?) will get you
that much more SNR.
Current transformer measures current (its winding is in series), potential
transformer measures voltage (in parallel).
ESR (and Q) measured on the coil corresponds to radiation resistance
(series equivalent) *plus* internal losses (also series equivalent). You
match by the good old impedance theorem.

~1:64 is 50 ohm / 0.78 ohm, and N2/N1 = sqrt(Z2/Z1), or 8:1 turns ratio.

Tim

11. ### Tauno VoipioGuest

Please note that high Q will destroy the modulation sidebands on
the signal you're listening to.

In aviation, there are non-directional beacons which are transmitting
in a frequency around 300 kHz (1 km wavelength). The antennas cannot
obviously be of efficient length (250 m / 800 ft), so they are short
(20 m / 70 ft) force-tuned to the transmitting frequency. This creates
so high Q that the identification modulation sidebands for the customary
1050 Hz audio do not fit in, and the ID is modulated using 400 Hz audio.

12. ### rickmanGuest

I appreciate the advice from everyone, but much of it is not in the
proper context and way off target. "High" Q is how high? Where are the
modulation sidebands? My point is that I have already considered this.
The modulation sidebands of this signal are on the order of low 10's
of Hz. This signal is modulated at a 1 bit per second rate. I will be
demodulating a 30 Hz sample rate. So a bandwidth of 100 Hz is plenty
which corresponds to a Q of around 500.

I said I was looking for a Q over 100, maybe I should have said a Q of a
bit over 100. By the time it gets to 300 it is to peaky to hold a tune
setting. That is the problem I am concerned with.

Ok, but that is nothing like my application, receiving WWVB.

13. ### rickmanGuest

I'm not clear on why you keep referring to radiation resistance for a
receiving antenna. Does this result in a larger received signal? I am
concerned with maximizing the voltage at the input to the receiver.

I have no idea why you are talking about Litz wire and tiny coils. I
never said I was looking to maximize the Q. I said I wanted a Q of over
100. I should have said, slightly over 100. A higher Q clearly does
increase the voltage on the input in my simulations. Is there something
wrong with my simulations?

Yes, a higher stepup ratio gets larger signal up to a point. That point
is determined by the parasitic capacitance of the receiver input. That
capacitance is reflected back through the transformer and affects the
antenna tuning. In my simulations it creates a filter with two resonances.

Series and parallel with what? I'm not following this. I have trouble
with series and parallel resonance, but I'm starting to get the concept.
Sometimes it is hard to tell how a circuit is being stimulated.

Internal losses of what? How do you determine the internal losses?

Ok, so you were matching the hypothetical ESR to the hypothetical line
impedance.

14. ### Tauno VoipioGuest

I'd still be wary of high Q. The antenna is, by definition, in close
interaction with its surroundings, and a high-Q thing is quickly
detuned.

At those low frequencies, the atmospheric and other outside noise is
far larger than the internal noise of an amplifier, so in my opinion,
the way to go is a loop tuned to 60 kHz with as low Q as easily comes
without extra attenuation and a good pre-amplifier. The preamp can
then contain a tuned interstage tank for interference suppression.

15. ### rickmanGuest

I understand. But this is intended to be *very* low power and I haven't
found an amp I can use that is in the low double digits uW power
consumption range. I plan to use no amp and go straight to digital.

16. ### Tim WilliamsGuest

A Q of a million will get you gobs of "gain", but if it doesn't couple
into free space, it's only the thermal noise of the loss generating that
signal.

An antenna with high (expressed as ESR) radiation resistance might have a
modest Q, but gives far better SNR because it couples to free space.

Raw volts don't matter, you can always throw more amplifiers at it (as
long as they don't corrupt the SNR also!).
Oooh, capacitance! I like capacitance. Capacitance is easy to
cancel...inductors are good at that.

What's a nearby inductor working against that capacitance? The current
transformer in your simulation, if its inductance can be controlled, would
be an excellent candidate. The circuit effectively becomes a double tuned
interstage transformer, like,

http://www.jrmagnetics.com/rf/doubtune/doubccl_c.php
This is two resonators coupled with a cap, but any coupling method will
do. Capacitive, magnetic (putting the coils end-to-end) or
electromagnetic (coils side-by-side) coupling does equally well; normal
arrangements have them all in phase, so in practice, unshielded coils will
need smaller coupling capacitance than designed, etc.

If you line up that 10p resonance with the operating frequency, you should
get gobs more gain. In fact, because the reactances cancel, the driven
impedance will be much higher than you were expecting, and so will the
gain. The CT might go from, say, 1:8 up to, who knows, 1:20? 1:100?

The bandwidth of that coupling (not necessarily of the antenna itself, so
they should be similar bandwidths) is determined by the coupling
coefficient (in the coupled-inductors case, simply k) and Q of the
components.

might be able to estimate the equivalent loss and optimize gain.

Tim

17. ### rickmanGuest

SNR would be good, but I am concerned with maximizing the signal actually.

I think you aren't reading what I am writing. I said I wanted a Q over
100, not 1 million. I don't get why you keep talking in hyperbole.
What you are describing is not even a tradeoff between signal strength
and SNR. If there is no coupling, there is no signal.

antenna. I have a formula for the effective height of a loop antenna
which is what determines the received signal strength at the antenna. It
does not calculate the radiation resistance, it uses the coil parameters
and the wire resistance. Is that a wrong formula?

Maybe you didn't read my other posts. I am not using an amplifier. I
am running the antenna and coupler output directly into a digital input.

The receiver input is high impedance, approximately 10 MOhms with a low
capacitance between the differential inputs of not more than 10 pF.

Your description of what is happening is very terse and full of
shortened terms that I don't understand. What do you mean "line up that
10p resonance with the operating frequency"? I assume you are referring
to the 10 pF input capacitance. How does this get "lined up" with
anything?

When you talk about reactances canceling, that sounds a lot like a tuned
circuit at resonance. That is what I *am* doing and where this thread
started. One problem with that is the lack of precision or stability of
the parasitic capacitance. Any idea how to deal with that?

Have you looked at the simulation data I had posted? I think you are
describing exactly the circuit we are simulating which I believe is an
accurate representation of the circuit I plan to build. Is that not
correct?

18. ### Jim MuellerGuest

<snip>

An electric circuit consists of a source of power, a load, and something
(like wires) connecting them. Transformers can be used if the source is
providing alternating current. A voltage transformer is connected in
parallel with the load so that the source, the transformer, and the load
all see the same voltage. It can also be used to match a load to a
source. A common example of a voltage transformer is the power
transformer in a piece of equipment that changes the AC line voltage to
whatever other voltages are required by the equipment.

A current transformer, on the other hand, is connected in series with the
load so that the source, load, and transformer all have the same current
flowing through them. The most common use of a current transformer is to
measure the current flowing into a load. A clamp-on ammeter is a common
example.

Historical examples of voltage and current transformers are the "picture
tube brighteners" that were commonly used in TV sets to prolong the
useful life of the CRT. There were two types, parallel and series. The
parallel types were used in transformer operated TVs and consisted of a
step-up transformer to raise the heater voltage of the CRT above normal
to increase emission. The series type was used in sets with the tube
heaters in series and consisted of a step-down transformer that raised
the heater current above normal. Of course, raising either the voltage
or the current also raised the other. These were, respectively, voltage
and current transformers.

A loop antenna is a distributed source with the voltage being generated
along the length of the wire and also having a magnetic field so that it
can be used as part of a transformer. This blurs the distinction between
a current and voltage transformer.

19. ### rickmanGuest

Yes, I have done my homework on the WWVB signal. I am at the fringe of
the 100 uV/m contour. I would very much like to see the signal on an
oscilloscope when I test this. They have a receiver not far from here
in Gaithersburg, MD and the signal is often strong during the day. So
much so that I don't follow why they say there is this day/night signal
strength fluctuation. It looks much more random to me.

The WWVB signal is not truly on-off keying. I believe they use a 10 dB
modulation factor for the AM signal. This is close to on-off I agree.
But they also phase modulate the signal and I will be demodulating both
to see which one works best in my design.

The ADC in my design is truly one bit. It is an LVDS input on an FPGA.
I looked at delta-sigma (or is it sigma-delta? conversion and got
code from the chip vendor for a simplistic implementation. I don't
think I have the power budget for that and am using a much simpler 1 bit
ADC at 4x the carrier rate. The bit stream is multiplied by quadrature
carriers at 60 kHz and each stream summed for 1/30 of a second to
implement what can be considered a DFT bin, a decimated FIR filter or a
decimated down conversion; take your pick, they are all mathematically
the same in this case because the sampling is synchronous to the carrier
(or very close to synchronous).

What comes out the other end of this processing gains nearly 40 dB in
SNR. My simulations show a recoverable signal when it is more than 20
dB below the noise.

Of course, I have not tested this yet on a real signal. I want to run
some tests on the antenna and coupling transformer to verify the
simulation. Then I will start working with the FPGA to see if I can
make the LVDS input do what I want. I have ideas on how to bend digital
circuits to do my bidding. This LVDS input is why I want as large a
signal as possible from the antenna. With the high impedance input on
the chip I should be able to boost the signal pretty well with just
passive devices and signal processing.

The loop antenna is rather large. I would like to end up with something
smaller. Once I get this working with a shielded loop antenna I will
check out the ferrite core antennas. My understanding is that they
don't produce as much signal.

I'm not sure how you came up with 2 Hz for the bandwidth. In this case
the bandwidth is not just twice the bit rate. I believe the stated
"system" bandwidth is around 5 Hz (from a 1995 paper prior to addition
of the phase modulation). Regardless, I am sampling at 30 Hz and if I
expect to see significant changes in phase or amplitude within one
sample time, I need an appropriate bandwidth.

Even so, that is not the limiting factor. The limiting factor is the
difficulty in holding tune with drift in passive component values. The
Q can be raised by increasing the turns ratio on the transformer, but it
becomes so sensitive to the parasitic capacitance that the sensitivity
drops 10 dB with a 1 pF change.

Thanks. I will take a look at that.

I will be needing a time code simulator. I designed a commercial
product that works with the IRIG-B time code which is similar. The
functionality is not hard, it is just a matter of generating the data,
encoding it into the modulation pattern, then impressing the carrier
with the modulation. Working in an FPGA this sort of stuff is easy.

The trouble is if you make the same mistake in both the generator and
receiver they work just fine in simulation, but not with other
equipment. lol

I'll take a look at this link.

I might look into that. Certainly it can't hurt to get more input.

20. ### rickmanGuest

Is this a current transformer or a voltage transformer?
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