I am trying to build a bus transmitter complying to 1553 standards. Its input is a 3.3V square wave at 1MHz (CMOS data) and it has to be amplified to 10V or higher at the same frequency. The load will be a center tapped isolation transformer whose primary side inductance is around 5mH.

My current implementation involves an LT1210 current feedback amplifier which is able to amplify square wave but the catch is high current drawn from the sources (500mA) which leads to high power consumption as the datasheet suggests.

Following is my setup:

enter image description here

Right now to the left of the 10 ohm resistor I get 8.5V peak to peak.

What I need to know is if I am on the right track here. Are there any other operational amplifiers available that could amplify large square wave signal and use less current at the same time, or are there any modifications that I could make on my circuit to draw less current and achieve same levels of voltage? (I clearly don't seem to need 500mA.)

P.S. This is a part of a project which is a 1553 bus transceiver analog front end. HI-1579 chip is very close to what I am intending to build.

  • 1
    \$\begingroup\$ I'd look at the topology of a class-D amplifier. Using 2 MOSFETS and pulling them to ground or supply voltage with your 3.3V square wave. \$\endgroup\$
    – Swedgin
    Commented Dec 3, 2019 at 15:36
  • \$\begingroup\$ Did you really mean to say "center tapped isolation transformer"? Looks like what you have is this, from the LT1210 data sheet. \$\endgroup\$
    – SteveSh
    Commented Dec 3, 2019 at 15:46
  • \$\begingroup\$ How can you get 20 volts (your picture) peak when the positive power supply (again your picture) is 10 volts? \$\endgroup\$
    – Andy aka
    Commented Dec 3, 2019 at 15:54
  • \$\begingroup\$ @Andyaka: Fair point. But if you read the rest of the question, I do mention that I get 8.5 Vp-p. 20 is what I need, so I could indeed increase the voltage source as a solution. However, my main issue is high current consumption. \$\endgroup\$
    – Ams
    Commented Dec 3, 2019 at 16:03
  • \$\begingroup\$ @SteveSh: I have used the same implementation. You can see it in the diagram I posted in my original question. Please take the time and read it, the issue is high current consumption of LT1210, Plus yes I did mean center tapped isolation transformer what about it? \$\endgroup\$
    – Ams
    Commented Dec 3, 2019 at 16:04

3 Answers 3


The cheapest way would probably be using a CMOS hex inverter 40106 which can be operated at up to 15V and biasing the input near below the switching threshold using a voltage divider and either AC-coupling the input signal via capacitor or DC-coupling through another resistor.
The signal will get inverted. If this is a probem just feed it into another inverter (there is a total of 6 in the IC package).
Of course it can't provide the current you need. That could be accomplished by a discrete CMOS inverter stage (discrete P-MOSFET/N-MOSFET-pair). In addition if you use two drivers, one at each terminal of the inductor (= H-bridge configuration), you don't need a center tapped inductor to create pulses of both polarities and you don't need a negative supply rail. Depending on which terminal is high and which is low you can provide the appropiate current direction. See circuit diagram below:


simulate this circuit – Schematic created using CircuitLab The inverter should be a 40106 operated at 15V.

There are also more specialized and still cheap level converter/driver ICs like the MC34152, DS0026 or ICL7667 which could also fit for your application. I think they all can drive up to 1.5A.
In any case I recommend to build and test it first on a prototype board before manufacturing a PCB. That will be much more meaningful than any SPICE tests.

  • \$\begingroup\$ Thank you. I couldnt find MC34152 chips SPICE model. Any suggestion for a way to simulate it in LTSpice or similar? \$\endgroup\$
    – Ams
    Commented Dec 3, 2019 at 22:51
  • \$\begingroup\$ I added two more equivalent or at least similar (and pin compatible) ICs from various manufacturers. Maybe you can find a SPICE model for one of them. \$\endgroup\$
    – Curd
    Commented Dec 4, 2019 at 10:25
  • \$\begingroup\$ I have been looking at the chips you mentioned and several more for spice models and/or schematics but I havent been able to find anything. The reason is that not only I have to simulate, but in future I will be drawing the layout in 130 nm in which I would have to know how it works in transistor level. For example Cadence has ICL7667 instance but I couldnt find its Spice model. What im trying to say is, could you please elaborate more on how I could implement this with discrete elements? e.g. I thought of using a tristate large signal mosfet inverter as the driving stage for the transformer \$\endgroup\$
    – Ams
    Commented Dec 4, 2019 at 20:20
  • \$\begingroup\$ @Ams - You do realize that in many cases SPICE models for ICs are not what is actually inside of the IC, right? It sounds like you are planning to import the SPICE model as a schematic and use that directly for your IC design? Even if the device is simple enough that its SPICE model is the same as its real schematic, I would be very worried about copyright/patent violations. The models are not usually available for you to do whatever you like with them, just to simulate to see if you want to buy the associated parts. \$\endgroup\$
    – Justin
    Commented Dec 5, 2019 at 19:46
  • \$\begingroup\$ @ReinstateMonica I surely wouldn't use a protected/patented design in my IC (thanks for the heads up by the way). However the main reason I am looking for spice macromodels of the chips is to be able to simulate them in simple applications like LTSpice. I cant afford to migrate my current design to cadence for the time meantime (where there are models for the parts such as ICL7667 as I mentioned before). I will be purchasing those ICs for my PCB project eventually. \$\endgroup\$
    – Ams
    Commented Dec 6, 2019 at 2:22

I fear you have misunderstood how op amps work. You have overlooked the fact that zero times any gain remains zero. That is, the portion of the input waveform which has a zero value will produce a zero output.

As a result, the op amp (which has a nominal gain of 4) will attempt to produce an output from zero to 13.2 volts. Since Vdd is only 10 volts, the most you could possibly get is a 0 to 10 volt output, rather than a 0 to 13.2. However, this op amp is not what is called a rail-to-rail op amp. That is, its output will not reach Vdd or Vss. Instead, if you look at the data sheet for "VOUT", "Maximum Output Voltage Swing", you will see that with VDD/VSS of +/- 15 volts, the op amp is only guaranteed to put out +/- 10 volts. The output swing is not specified at +/- 10 volts, but something on the order of +/- 5 to 7 volts is probably a reasonable projection. In the event, you're getting 8.5 volts, so that's better than you deserve.

What you need to do is produce an offset, which will produce a 0 output with 1.65 volts in. At the same time, you'll need to increase the gain from its current value of 4 (1 + 845/247) to at least a gain of 6. You can try something like this:


simulate this circuit – Schematic created using CircuitLab

Note that the new op amp must be able to output about +/- 10 mA at much greater than 1 MHz (why?) in order not to distort the waveform, and another LT1210 would seem like an excellent candidate.

Furthermore, the output will now swing from about + 8.5 to - 8.5 unless you increase VDD and VSS, for exactly the same reason it is now only swinging to +8.5. With this op amp and these power supplies, you simply cannot expect (or even hope for) a greater output swing.

  • \$\begingroup\$ Thank you, maybe I misrepresented what my main concerns were in my original post. Ofcourse I cant exceed the supply voltages. Anyway here my main concerns were high current drawn (around 500mA) by LT1210 from +/- 10V supplies even for 8.5Vp and driving the center tapped transformer primary winding (5mH). I will try your circuit \$\endgroup\$
    – Ams
    Commented Dec 3, 2019 at 23:36
  • \$\begingroup\$ Also, be aware that driving a transformer with a large DC component (the average for your 0-8.5 volt output is 4.25 volts) is likely to cause saturation problems, and this may be the cause of the large current draw. \$\endgroup\$ Commented Dec 4, 2019 at 2:10

I would look at using a low-side gate driver IC, which will give you level shifting and an output power stage capable of operating at 10V or more. Also, they are inexpensive.

For example, the Microchip (nee Microsil) MIC44F18/19/20 will drive 100mA with a couple hundred mV drop, and typically draws less than 10mA idling at 1MHz.

enter image description here


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