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I am designing a power supply modulator using an op-amp. I have a DC voltage of 1.6 V when the clock is LOW the output should be at 1.6 V and when CLK is high the output should be at 3.6 V. I am using the below circuit to implement this functionality. Please find the attached circuit.

When the input frequency is low (say 100 kHz) output is coming as expected when it goes to 1 MHz the output is weird. My op-amp (TLV9062, datasheet) has a BW of 10 MHz and BJT(MMBT2369ALT1) has an ft of 700 MHz.Please find the attached circuit diagram and output obtained at 100 kHz and 1 MHz.

May I know where I Went Wrong?

Circuit Diagram

OUTPUT at 100Khz

OUTPUT at 1Mhz

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    \$\begingroup\$ BJT has an ft of 700Mhz That does not mean that the BW of that NPN, when used in as a common emitter, will be 700 MHz. In the 100 kHz plot, zoom in such that the timescale is the same as the 1MHz plot, look at the rising/falling slopes of the curve. Are they different or the same? My point: while in the 100 kH situation the output has enough time to settle to the correct output voltage, it does not have that time in the 1 MHz situation. \$\endgroup\$ May 25 at 10:08
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    \$\begingroup\$ Also do a calculation, calculate how much current is needed to charge/discharge C1 (10 pF) and compare that to how much current the opamp's output can source/sink. Also look up the "slewing rate" property of the opamp. \$\endgroup\$ May 25 at 10:10
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    \$\begingroup\$ Try to understand the performance of each stage individually. Look at the waveform at the collector of T1. Try removing T1 and driving R1 directly with a 0 to 1.6 V square wave. \$\endgroup\$
    – Dave Tweed
    May 25 at 10:17
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I have no idea what a "supply modulator" is but if the load has decoupling capacitors on it, then the voltage slew rate will be pretty low.

Besides that, the opamp has a slew rate of 6.5V/µs, so it will need 300ns to swing 2V (from 1.6V to 3.6V). It has to swing twice per period, so with 1MHz square wave the best you can hope is 300ns ramp up, 200ns constant, 300ns ramp down, 200ns constant. If that's not okay, you need an opamp with higher slew rate.

Also using the BJT as a saturated switch will be slow. You should wire the opamp as an adder instead, and add a constant DC to your input square wave.

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  • \$\begingroup\$ We are calling it a "supply modulator" in our design. if you don't mind could you please explain the math you wrote above."6.5V/µs, so it will need 300ns to swing 2V (from 1.6V to 3.6V). It has to swing twice per period, so with 1MHz square wave the best you can hope is 300ns ramp up, 200ns constant, 300ns ramp down, 200ns constant " \$\endgroup\$
    – HARI T O
    May 25 at 11:59
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    \$\begingroup\$ The math is 2 Volts divided by 6.5V/µs which gives the time 300ns... \$\endgroup\$
    – bobflux
    May 25 at 16:56

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