Need Help Understanding TL494 Voltage Error Amplifier and Adjustable Voltage Con

[Andy Chee]

Besides, is anyone able to tell what exactly that part of the circuit is for (the part in the green box, with the input into pin 1 included)?

All voltage regulators (linear and SMPS) compare a sample of the output voltage to a fixed voltage reference.

When the output voltage goes higher than the fixed voltage reference, the circuit cuts back the output.
When the output voltage falls below the fixed voltage reference, the circuit increases the output.

Regulation!

It is conceptually no different to a thermostat temperature regulator.

[max.wwwang]

Besides, is anyone able to tell what exactly that part of the circuit is for (the part in the green box, with the input into pin 1 included)?

All voltage regulators (linear and SMPS) compare a sample of the output voltage to a fixed voltage reference.

When the output voltage goes higher than the fixed voltage reference, the circuit cuts back the output.
When the output voltage falls below the fixed voltage reference, the circuit increases the output.

Regulation!

It is conceptually no different to a thermostat temperature regulator.

Thanks. I’m well aware of that. What I’m asking is, how is regulation done exactly and specifically in this case?

[Andy Chee]

Thanks. I’m well aware of that. What I’m asking is, how is regulation done exactly and specifically in this case?

In this specific case, you actually need the datasheet to see what else is internally connected to the junction of the output of the two opamps.

[max.wwwang]

To facilitate analysis, attached is the same circuit based on the more detailed block diagram (with the Error Amp 1 sub-circuit copied in the box).

[MariuszD]

Looking closer at the (lower) op-amp and its voltage gain, I ended up with something in the form of:
$$V_o=a V_i + b$$
So the gain is dependent on the input voltage, not a neat constant. 

It is not possible for the gain to depend on the voltage; moreover, the formula you wrote does not include such a dependency.

[MariuszD]

Looking closer at the (lower) op-amp and its voltage gain, I ended up with something in the form of:
$$V_o=a V_i + b$$
So the gain is dependent on the input voltage, not a neat constant. 

It is not possible for the gain to depend on the voltage; moreover, the formula you wrote does not include such a dependency.

@max.wwwang @ommsiva

You skipped learning about simple regulators and now you have difficulty understanding it with a complex example. In the SMPS design materials, you won't find this knowledge because it's something everyone learns earlier.

Let's take a simple example of a linear regulator.


I deliberately set a low amplifier gain and a high resistance R5 so that you can see how it works.

When the load current I1=1A appears, the voltage at the reg_out point drops. the voltage at point in_n also drops and the difference at the amplifier's input increases. The amplifier amplifies this difference by 100x, and at the amp_out point, a voltage rise appears that compensates for the drop caused by the load.

Real amplifiers have much greater gain, but the principle is the same.

Second example without an amplifier, the voltage drops almost to zero and the current does not reach 1A.

[ommsiva]

Dear Sir,

I also Meant the Same. Thank you for your illustration.

[MariuszD]

Take my simulation from LTSpice, try to run it and change the parameters.


What does the 51k resistor do in the first schematic? It reduces the gain of the error amplifier. In my example, I added R6 which serves the same function. What does it change? The gain is lower, the regulator works worse, and the voltage drop under load is greater. The likely reason for adding this resistor is to avoid oscillations, but this is not a sensible/recommended approach; it is advised to ensure that the gain for DC is maximized. And to achieve stability, the amplitude-phase characteristic is shaped only in the high-frequency range.

[ommsiva]

Dear Sir,

Thank you.

My understanding is that by adjusting the feedback resistor to 10 kΩ, I can obtain a 5 V output in your simulation files.

Similarly, referring to my first question regarding output-voltage adjustment, my inference from this simulation is that the output voltage can be adjusted by changing R8. Is my understanding correct?

I have also attached the simulation results for your reference.

[armandine2]

..build a tester, might help in your understanding of the chip - though I failed to follow Richard's instructions, you may have more joy

[MariuszD]

My understanding is that by adjusting the feedback resistor to 10 kΩ, I can obtain a 5 V output in your simulation files.

At this moment, you don't have a functioning circuit, just a saturated transistor.

Notice that in my schematic, the power supply V2 has only 5V, which is too low for such a regulator to output 5V. Change V2 to a higher value, for example, 30V.

At the input in_p, there is 0.5V; to obtain 5V at the output, the divider R1,R2 must divide by 10, for example, 1kΩ and 9kΩ.

Since the amplifier has low gain, it won't give exactly 5V. If you go into the amplifier properties and change Avol to 100k (100000), which is closer to real amplifiers, it will be 5V.

Similarly, referring to my first question regarding output-voltage adjustment, my inference from this simulation is that the output voltage can be adjusted by changing R8. Is my understanding correct?

That's right, you can adjust the voltage by changing R8, but with this method, you won't get less than 2.5V because that's what you'll get if you set R8=0.

Derive the formula for voltage as a function of R8.

By replacing R3 and R4 (in my simulation) with a potentiometer, the voltage can be adjusted from zero.

The 5V regulator with a high-gain amplifier will work like this:


The input voltage of the amplifier is below 1mV, so in simplified terms, it is said that the amplifier strives to achieve 0V at the input.