Current mode transformer input amplifier.
By using a transformer in current mode, its low frequency response can be extended considerably and its distortion can be reduced by an order of magnitude or more.
To operate a transformer in current mode, its input and output signals must be a current so there is no voltage across the windings and no field across the core.
Performance can be increased even further by loading the transformer with a negative resistance in form of a NIC ( Negative Impedance Converter ).
A NICs basic function is that it can convert a component value to its negative equivalent.
It is only used and described for a resistor here, but it also works for reactive components.
The circuit is only useful for high-level applications as the noise is higher than for voltage-mode transformers.
As it is not trivial to calculate the component values, a spread-sheet and some LTSpice simulations can be downloaded here.
Fig.1: Negative resistance.
GEN is a current generator and Rin is the input resistance.
Vin = Iin * Rin
for Iin = 1 A and Rin = 1 Ω, Vin is 1 V.
for Iin = 1 A and Rin = -1 Ω, Vin is -1 V.
A negative resistance simply means that the current/voltage phase is the opposite phase than in a positive resistor.
There are a few components that have a negative resistance ( f.ex. neon lamps and tunnel diodes ), but most are only linear over a very limited part of their operating range.
It is possible to emulate a negative resistance using feedback as shown in fig.1(b).
Let Ra = Rb = Rc = 1 Ω and Iin = 1 A.
The voltage across Rc is then 1V, so Vout is 1 V lower than Vin.
The feedback loop maintains Vin at the same potential as Va so the voltage across Rb is the same as across Rc.
Since Ra and Rb are same value here, the voltage on Vin is -1 V and the voltage on Vout is -2 V.
This gives an imput resistance RinA1 = Vin / Iin = -1 V / 1 A = -1 Ω.
Fig.2: Voltage mode transformer.
Lp and Ls are the transformer winding inductance.
Rp and Rs are the transformer winding resistance.
Ll is the transformer leakage inductance.
n is the transformer winding ratio.
For this transformer Lp = 2.8 H, Rp = Rs = 66 Ω, Ll = 14 mH and n = 1.
The secondary inductance is:
The primary and secondary side resistances are:
The primary side resistance transformed to the secondary side is:
The voltage gain from Vgen to Vout is:
The high-pass filter resistance is:
The high-pass filter frequency is:
The low-pass filter resistance is:
The low-pass filter frequency is:
Fig.3: Current mode transformer input amplifier.
When Ra is 0, the transformer secondary is loaded by the virtual ground on A1's inverting input and its primary is a short circuit.
Rinp is added to convert the voltage Vin to a current.
By increasing the value of Ra, RinA1 becomes negative.
When RinA1=-Rs, VLs is zero and the field in the transformer is zero. This will reduce the transformer distortion to almost zero, but is not possible for stability reasons.
One issue here is that Rs is a piece of copper wire which has a positive temperature coefficient. This may cause the circuit to oscillate at low temperatures unless it is designed with a reasonable margin.
This circuit employs both negative and positive feedback. In order for the circuit to be stable, the negative feedback must dominate:
You can find the calculations for the circuit in the spreadsheet cmtia_Calc.ods. Calculations includes component tolerances and temperature coefficients.
Some texts for NICs in this application suggest adjusting the circuit for minimum distortion. Unless you have precise control over the operating temperature or have a temperature compensating winding in the transformer this is not a good idea. Changing the temperature by a few degrees may cause the circuit to oscillate.
where AvA1 is A1s open-loop gain and B is:
If A1 has sufficient loop-gain, this can be simplified:
The primary and secondary side resistances are:
The voltage gain from Vgen to Vout is:
The lower and upper cut-off frequencies are calculated like the example in fig.1.
|1||600 Ω||600 Ω||-7 dB||19 Hz||15 kHz|
|1||40 Ω||600 Ω||-2.2 dB||5.2 Hz||8.8 kHz|
|2||600 Ω||10 kΩ||0||10 kΩ||10 kΩ||-0.61 dB||3.7 Hz||122 kHz|
|2||600 Ω||10 kΩ||48.7Ω||10 kΩ||10 kΩ||-0.53 dB||0.98 Hz||121 kHz|
With the component values used here, the circuit will have a considerable DC gain that can cause problems due to OP-AMP offset voltage.
Fig.4: Basic NIC input amplifier schematic with OP-AMP offset voltage.
The voltage gain from Vos to Vout is:
With the component values shown above this is a DC-gain of 585 V/V.
A TL071 has a maximum input offset voltage of 10 mV so there can be up to 5.85 V DC on Vout due to the offset voltage, leaving only around 6 V for the signal.
A OPA134 has a maximum input offset voltage of 2 mV so there can be up to 1.2 V DC on Vout due to the offset voltage. This may be acceptable for some applications, but be aware that Vos also has a drift with temperature.
It should be noted that a DC voltage on Vout will cause a DC current in the transformer. Depending on the core material this may be acceptable.
Compact audio transformers with a high-permeability ( like µ-metal ) cores are very intolerant to DC currents and distortion will increase considerably.
AC coupling the circuit will fix the DC problem. If you put a very large value for Rs in the equation for DC-gain above, the DC-gain will drop to 1.
Note that I have ignored the leakage inductance for the following calculations.
Fig.5: NIC input amplifier with AC coupling.
AC coupled NIC:
This is obviously a 2. order high-pass filter so it can have a peak in the frequency response and it can introduce ringing on transient signals.
If the filter is ringing, it may cause periodic clipping for signals close to maximum output.
To avoid ringing, the Q should be 0.5 or lower. A Q of 0.71 ( = √0.5 ) may be acceptable in some applications.
The filter Q can be reduced by increasing the value of RinA1 or the value of Cin.
The self-resonance frequency and Q for this circuit is:
Solving for Cin:
This is a Cin of 38 mF with the component values used here. f0 is 0.5 Hz.
It gets worse:
Calculating the circuit for:
Temperature range: 0 °C to 50 °C
Resistor tolerance: ±1%
Resistor TC: ±50ppm/°C
Capacitor tolerance: ±20%
Transformer inductance tolerance: -0/+20%
Transformer resistance tolerance: ±10%
The minimum value for Cin is 45 mF
f0 for these values is 115 mHz to 161 mHz and Q is 0.075 to 0.5.
Cin can be reduced to 6.2 mF by increasing RinA1 to 0.
This is a really inaccurate filter, but there is not much to do about it as a very large part of the variation is due to the tolerances of Ls, Rs and Cin and the temperature coefficient of Rs.
These capacitance values are impracticable large for a circuit with the component values used here.
In some cases Cin will have a reasonable value and is the simplest solution to the DC problem.
In Fig.5, Cin is shown in series with the OP-AMP input. I some cases it may be preferable to connect Cin in series with the transformer's ground lead with Cin minus and can on ground to reduce Cin's pick-up of nearby signals ( noise ).
I have not tried it in this application, but in some other applications it works well.
Fig.6: NIC input amplifier with DC servo.
NIC input amplifier with DC servo:
The value of Rd should be large enough to reduce noise from A2 feeding into the input of A1, but low enough to allow A2 to cancel the offset voltage from A1.
Vz is also part of the AC feedback loop of the circuit, so only half the output range of A2 is used to cancel the offset and the remaining half is used for AC feedback.
Simulate the circuit and make sure that the AC voltage on Vz is 6 dB or more below the voltage on Vout. If not, increase the value of Re.
VosA1 is the input offset voltage of A1 and VoutA2 is the maximum available output voltage from A2.
This calculation assumes that Rb >> Ra and Rd >> Ra.
Pick the nearest lower resistor value.
With OPA134 ( A1 ) and TL071 ( A2 ) 91 kΩ is a suitable value.
Calculate the voltage gain from Vz to Vout ( sorry, I am too lazy to reduce this to something readable - use the spread-sheet ):
Select Ce ( 1 µF ) and calculate Re:
where Rhp and Cin are the values calculated for fig.5.
For a Q of 0.5 this is an Re of 200 kΩ using the nominal component values and a temperature of 25 °C.
Including component tolerances and a temperature range like for fig.5, Re is 2.2 MΩ.
Do simulate the circuit to make sure you have not made any mistakes.
The simulation should extend to where both the lower and upper roll-off is at least 20 dB down - not just the audio frequency range.
f0 for these circuits can be very low. Measuring oscillation at tens of mHz ( a period in the order of 100 s ) is very time consuming.
Note that the transformer used here has a very low inductance compared to transformers designed for the entire audio range.
The minimum noise that can be achieved from this circuit is limited by the value of Rinp + Rc ( the noise contribution from Ra and Rb can be reduced by lowering their values ).
The noise limit for Rinp = 10 kΩ and Ra = 10 kΩ is -109 dBu in a 20 kHz bandwidth with a 1:1 transformer.
Fig.7: Primary circuit.
Fig.7(a) is the principle schematic of the primary circuit.
The input current is set by the value of Rinp and Rp:
Ipri = Vin / ( Rinp + Rp )
Rinp is normally much larger than Rp, so Rp can be ignored.
For a 1:1 transformer, Is = Ipri.
As the transformer is used in current mode, the transformer output current is the inverse of the winding ratio.
If the primary is 1 turn and the secondary is 2 turns, Is = ( 1 / 2 ) * Ipri.
Fig.7(b) shows a practical circuit.
Rinp is split in 2 to maintain the best possible signal balance on the primary.
Ct1 or Ct2 is used to obtain the best possible CMRR at high frequencies. The value depends on the transformer and the PCB lay-out ( which should be as symmetrical as possible ).
With a given PCB lay-out Ct1 or Ct2 is normally identical within a batch of transformers.
The typical value of Ct1 or Ct2 is 5 pF to 50 pF.
The input of the transformer is connected "out-of-phase" as the amplifier is inverting.
Depending on the design of the transformer, the polarity of the secondary may be important for best CMRR.
For the LM-NP-1001 ( with the windings side-by-side ) it is not really an issue. The batch I have works best with pin 5 on GND, but the difference between pin 5 or pin 8 on GND is a few dB.
For some transformers with layered windings the difference can be more than 20 dB at 20 kHz.
For best CMRR, the end of the secondary nearest the primary goes to GND.
Neutrik NTM1 transformers works best with pin 4 on GND.
Beyer Dynamics TR/BV 310 series works best with pin 7 on GND.
Unless the manufacturer specifically specifies start and end of the winding ( you can not count on the dot being the start of the winding ), the only way to find out is to measure the transformer's CMRR.
Some transformers are wound with interleaved primary/secondary windings and these should be connected as specified by the manufacturer.
If you need a high-pass filter for the input and have a low source impedance, put a capacitor in series with the 2 Rinp/2.
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