ALM2403-Q1: ALM2403-Q1

Hi Baogui,

baogui qin said:

can you tell me what is the calculation formula between OUTPUT1 and resistance R3, resistance R4, INPUT1, Vmin in the figure?

I put together a simplified ALM2403-Q1 driver so that you can simulate it yourself via Tina. 

R4/R3 = 24.9k/10k = 2.49 V/V --> for inverting op amp configuration, the gain = -2.49V/V, the output is inverted 180 degrees of the input. 

Vbias is to establish the common mode operating DC for the ALM2403-Q1 to operate linearly, and maximize the output swing. 

The input signals are typically independent complementary sinusoidal waveform, which it can be generated from SPWM, DAC, sinusoidal generator or other means. 

So this is differential input and differential output resolver driver using two ALM2403-Q1 power amplifiers (2 amplifiers in one package). 

Simplified ALM2403-Q1 Resolver Driver 08302024.TSC

If you have other questions, I will be happy to explain each part of the circuit. 

Best,

Raymond

Hello Raymond, I used Tina to open the routine you gave me, and I had a clear understanding of the input and output. However, I solved the formula of the output voltage in the routine about the input by using the virtual break and virtual short of the operational amplifier, but the result of the formula I solved was different from that of the circuit simulation. The formula I calculated was as follows:

Where Vin is a sine wave with a amplitude of plus or minus 4, but by bringing the mathematical expression of the input signal into the formula, the output amplitude obtained is not the result of circuit simulation. Could you please tell me which step is wrong? Thank you!

Hi Bapgui,

To derive the transfer function, it will be easier to use superposition method. 

1. Short Vg, capacitor is open for DC, then you have a buffer, which Vout_n1= Vbias

2. Short Vbias source, you have inverting op amp, where Vout_n2 = +/- Vg* (R3/R1) - ignore the C4 for simplicity. 

3. The total transfer function is Vout_n = Vout_n1 + Vout_n2

4. The bottom of the op amp waveform is inverted 180 degrees from the top one. 

Superposition-1 09032024.TSC

Superposition-2 09032024.TSC

Please see the simulation above. If you have other questions, please let me know. 

Best,

Raymonnd

Hi Raymond, 

I get it,but I am confused about why my calculations are not correct, as my teachers taught us how to analyze operational amplifiers in class. In the ideal short circuit condition, the positive input terminal V+ of the operational amplifier is equal to the bias voltage Vbias, and in the ideal open circuit condition, the current through R1 is equal to the current through R3, that is, (Vin-Vbias)/R1=(Vbias - V). Is there a specific set of conditions under which this theory applies? Thank you for your guidance!"

Hi Baogui,

You shorted the capacitive coupling at the input front and alter the transfer function, where Vp and Vbia have to be equal for linear operation (your equation is ok, the question is what is your Vin_n). 

If you simulate the circuit that I enclosed previously, you will understand what I did during the superposition derivation. 

If you have other questions, please let us know. 

Best,

Raymond