Need Help Understanding TL494 Voltage Error Amplifier and Adjustable Voltage Con

[mawyatt]

Yes #41.

See post #28, its shown correctly there. In this schematic you can see the current sense comparator (#2) and the error amplifier both have series diodes, this is a diode OR which allows either to set the PWM comparator level.

I suspect your are missing my point. I see all that you are saying, including the 'OR' of the two error amp outputs.

What I'm suggesting is that connecting the + input of the PWM comparator to ground through this current source means it will always have 0V voltage (at least so conceptually), making everything else connected to it useless (which I believe is impossible to be true).

The current source will simply draw the current from either of the diode OR connections. It will not force the voltage to ground, and simply assume whatever voltage is supplying the current. If its below the current limit then the feedback voltage side will supply the 0.7ma current and the voltage will be Vo1, when the current limit is reached then the current sense comparator overrides and the voltage (diode OR) and now this is the comparator high output minus a diode drop.

Best

[max.wwwang]

But since there isn't a voltage drop across a(ny) current source[Edit: unfortunately, this understanding is wrong], and its left hand side is at 0V (grounded, which is connect to the + input), wouldn't the voltage on the + input also be 0V (and always so)?

Have I made my point clear enough? Or am I still missing something?

Please do let me know if my way of thinking is either wrong, or just incompatible with the prevailing practice of using block diagram only to illustrate ideas roughly without that level of conceptual precision?

[Edit: indeed I was wrong. Thanks to @mawyatt! ]

[Andy Chee]

My conceptual interpretation is that of diode-OR logic;

https://en.wikipedia.org/wiki/Diode_logic#Active-high_OR_logic_gate

https://upload.wikimedia.org/wikipedia/commons/6/6e/Animated_wired_OR_diode_logic.gif

Obviously in the TL494 application we are dealing with continuously changing voltage level, rather than discrete binary voltage levels seen in logic.

Nevertheless, the concept is the same.

Consider what happens when both opamps are high and outputting/sourcing, say, 50mA each.  That 100mA is going to overwhelm the 0.7mA, thus causing the voltage at that node to go high.

(50mA is an exaggeration, but hopefully you get the idea)

[MariuszD]

I don't understand why it's a problem in us looking at the same circuit from different perspectives. I don't get where my "mistake" is.
As I said – and this is my personal view or preference (which by no means applies to anybody else) – I don't care what it is called as long as I have understood how it behaves. It's a bit like the question of whether someone should be called CEO, president, or boss – choose whatever you fancy. 

Only when we all use a common language is understanding possible. Let's assume that one day you will be constructing a similar circuit and you ask about it on a forum.

So its voltage gain is (as discussed above, dependent on Vi1):

If you write that the gain depends on vi, no one will understand you. People will have a voltage-controlled amplifier in mind and will be wondering how someone attempted to use a VCA in a DCDC converter. People will be asking questions, and it will probably take a dozen posts and a few days to reach an agreement.

point is when the input voltage is about 2.358V, when the gain is 0.

If you write that the gain dropped to zero, similarly, no one will understand and they will be looking for an error in the circuit, an error that doesn't exist. If you used the word saturation, everyone would understand without discussion.

Your example of how we refer to a CEO is not adequate because you are using names that everyone knows. A better example would be if if you used the word tsfsd and expected people to understand that it refers to the CEO.

As I said – and this is my personal view or preference (which by no means applies to anybody else) – I don't care what it is called as long as I have understood how it behaves.

It's not a problem until you try to talk to someone about it.

[mawyatt]

But since there isn't a voltage drop across a(ny) current source, and its left hand side is at 0V (grounded, which is connect to the + input), wouldn't the voltage on the + input also be 0V (and always so)?

Have I made my point clear enough? Or am I still missing something?

Please do let me know if my way of thinking is either wrong, or just incompatible with the prevailing practice of using block diagram only to illustrate ideas roughly without that level of conceptual precision.

You need to revisit how ideal current sources behave, then practical current sources. An ideal current source doesn't care about the voltage across it, the current will be "forced" thru anything regardless and produce whatever voltage results. A practical current source has limits which includes voltage levels and polarity.

Best

[max.wwwang]

You need to revisit how ideal current sources behave, then practical current sources. An ideal current source doesn't care about the voltage across it, the current will be "forced" thru anything regardless and produce whatever voltage results. A practical current source has limits which includes voltage levels and polarity.

Thanks for your advice. It's very helpful. This showed to me the challenge in truly grasping the concept of a current source.

A quick and simple thought experiment revealed my mistake. Supply a resistor R with an ideal current source. With this given current source, the voltage difference between its two ends depends entirely on the value of R and can be anything between 0 to infinity.

The say it a challenge for one reason at least. "Short' the two ends of an ideal current source does not result in a short circuit. There will only be an finite (as specified) current flowing through the apparent 'short' circuit.

I knew very well part of the concept that, no matter what, an ideal current source will make sure the specified amount of current flows through the line it is in, and certainly there will be limitations with one in practice. I only made a wrong assumption that the 'voltage drop' across it will always be zero.

Now go back to what I said was an error in the block diagram. The earthing symbol is NOT an error. But I still think it's ok, or even better, without it. For one reason, we can replace it with, say, the reference voltage 5V without any issue or any difference. Please correct me if, again, I'm wrong. (Believe me, saying this is not attempting to make myself look any good, or better. The circuit is all that I'm interested in.)

[mawyatt]

Sure the ideal DC current source can be connected to anything, zero, 5, -5, 1000, -1000V and so on. Doesn't matter, linear or non-linear load, the current source will supply the DC current whatever the load conditions and load voltages are...it's ideal

Practical current sources have a voltage range and polarity that needs attention. For example, you can't force a DC current into a +5V Voltage Source (Vref) unless the actual current source is biased above 5V somehow. You can sink DC current from Vref but you can't source current into Vref unless you use a higher than Vref bias in the current source.

Best

[ommsiva]

Dear All,

1)My understanding is that the op-amp employed in the lower part of the schematic is operating in an inverting configuration. But why is it configured this way? The only reason I can think of is that it ensures the voltage at inverting terminal  always remains lower than the non-inverting input. Please correct me if I am wrong.

2)The op-amp and diode form a super-diode configuration. When the non-inverting input is higher, the diode turns ON; otherwise, the pulse width has to be increased. However, I still cannot literally understand why there should be a 0.7 mA current source.

3)My understanding is that R1 = 1 kΩ should be made variable to adjust the current min-10A, and R8 should also be made variable to adjust the output voltage . Is this correct?

[MariuszD]

1)My understanding is that the op-amp employed in the lower part of the schematic is operating in an inverting configuration. But why is it configured this way? The only reason I can think of is that it ensures the voltage at inverting terminal  always remains lower than the non-inverting input. Please correct me if I am wrong.

I will analyze the diagram to see how the voltage changes at each point. It is important to check whether the feedback in the loop is positive or negative.

First, let's assume that the output voltage drops due to an increase in load (1.), which causes a drop at the amplifier's input (2.) and a drop at the amplifier's output (3.). The decrease at the amplifier's output causes an increase in the duty cycle (4.), which in turn causes an increase in the output voltage, counteracting (1.), meaning the feedback is negative. Since (2.) (3.) change in the same direction, the amplifier does not invert.

2)The op-amp and diode form a super-diode configuration. When the non-inverting input is higher, the diode turns ON; otherwise, the pulse width has to be increased. However, I still cannot literally understand why there should be a 0.7 mA current source.

Assume that the output voltage at the cathode of the diode is 0.5V and analyze in which direction and where the current flowing thru Rf should go.

[nonius_]

2)The op-amp and diode form a super-diode configuration.

I'm not sure what a 'super-diode' is. But as the datasheet of the TL494 explains, the outputs of both error-amplifiers are OR-ed together. If the voltage of EITHER the 'voltage-regulator' OR the 'current-limiting' opamp becomes too high (or both at the same time), the output signal (after the diodes) becomes high and lowers the dutycycle of the PWM-circuit.

As to 1): you want a circuit that, when the input signal goes above a certain threshold, the output goes high. Thus you want a non-inverting amplifier. And that's how both error-amplifiers are configured in the schematic in your startpost: as non-inverting amplifiers (with negative feedback).

3) if you want adjustable voltage you could add a potmeter at R8/R9. For adjustable current, I'd make the value of R1 adjustable from 0-1kohm.

In one of my modified PC AT(X) PSU's I've implemented it differently though; see attached schematic. But that was for the modification of an existing AT(X) PSU; in your situation, I'd try adjusting the reference-level of the 'current'-error amplifier.

Keep in mind that by adjusting voltage and/or current, the overall loop-gain alters which could lead to instability (ringing, oscillation). I've managed to get all my homebuilt TL494 power supplies stable-ish, but I strongly doubt their control-loops are optimal. My excuse is that by education I'm an economist and mechanical engineer, not an electrical engineer.

If you want to learn more about control loops, Manfred Mornhinweg/XQ2FOD has written a relatively simple explanation about it: https://ludens.cl/Electron/loops/loops.html

[ommsiva]

"Assume that the output voltage at the cathode of the diode is 0.5V and analyze in which direction and where the current flowing thru Rf should go."

Solution: Anode will be positive by 0.7V, means output of opamp should 1.2V. the opamp is acting as a current source. current flows to Rf through the diode.

[MariuszD]

"Assume that the output voltage at the cathode of the diode is 0.5V and analyze in which direction and where the current flowing thru Rf should go."

Solution: Anode will be positive by 0.7V, means output of opamp should 1.2V. the opamp is acting as a current source. current flows to Rf through the diode.

The current flows from point V2 to Vo1, it cannot flow to the amplifier's output because the diode prevents it, so where will it flow?

[MariuszD]

I continued testing converter #17 and the response to a sudden load change is fine. (tested at 15V 1A) However, the current regulator is oscillating. Set to 5A, it gives 10A peaks (oscillogram from the voltage on a 0.1Ω resistor).

[max.wwwang]

[Edit: see revised analysis in #74.]

I should correct myself again on what I said earlier that the internal 0.7mA current source will 'prioritize' the output of the opamp (through the diode). (My apologies – it's a learning journey to myself as well.)

Looking at the equivalent circuit of the ideal opamp, its output is modelled as a current source (rather than a voltage source). This means that there is no limitless current to be source from its output that can be prioritized (due to low impedance), and this opamp output and the 0.7mA outside the will need to reconcile with each other, which – when the reconciliation fails – has to be accommodated by the feedback loop, as far as it is capable of.
[Picture courtesy of Microelectronic Circuits]

One question arises.

Suppose the voltage on pin 1 goes very low, for example 0V (which is possible if the upper end of R8 breaks open), the opamp will try hard to match this voltage at pin 2. But the best it can get will be 0V at pin 3, which only gives pin 2 voltage of 2.36V. (This gives a current of 46uA through RF.) Due to the difference between 0V on the non-inverting input and 2.36V on the inverting input, the opamp will desperately try to sink current, but only in vain due to the diode. This means it will be saturated with its output at 0V, with no current through the diode.

Now there is a shortage of 0.7-0.046=0.654mA in the required 0.7mA. I presume the result will have to be, due to compliance constraints, that the 0.7mA current source becomes a 46uA current source, the best it can do to 'comply' (because obviously, it's not ideal). Do you think so? (This is the question.)

This also demonstrates that, due to the diode, the opamp may operate as a voltage comparator (i.e. saturated) with certain input voltages at pin 1, when the golden rule of virtual short no longer applies.

(All pin identifications are those of the TL494 IC, which are different from those in the ideal opamp model above.)

Edit:
Based on this understanding, there are the following distinct operating modes corresponding to the input voltages (or voltage ranges) at pin 1:

• Pin 1 voltage <2.36V / opamp saturated (working in the voltage comparator mode) / voltage at pin 3 = 0V / 0.7mA current source is forced to comply. (In this mode the voltage at pin 2 will refuse to go under 2.36V)
• As soon as Pin 1 voltage becomes >=2.36V / opamp working as opamp with negative feedback (virtual short applies) / voltage at pin 3 = 0V / there will be a fight between the opamp and the 0.7mA current source, where the 'loser' (most likely the opamp) will be forced to comply.
• Pin 1 voltage =2.5V / as above / as above (this is the desired point)
• As soon as Pin 1 voltage goes >2.78V / as above / voltage at pin 3 reaches 5V / the required 0.7mA will most likely be satisfied by the opamp, which also supplies the current through RF. (This is when the PWM outputs are turned off – or earlier, depending on the exact voltage threshold)

(For all pin 1 voltages above 2.36, the opamp will always in the opamp mode, and the virtual short golden rule applies.)

1)My understanding is that the op-amp employed in the lower part of the schematic is operating in an inverting configuration. But why is it configured this way? The only reason I can think of is that it ensures the voltage at inverting terminal  always remains lower than the non-inverting input. Please correct me if I am wrong.

As demonstrated above, the error amp 1 subcircuit (all included as a black box with pin 1 voltage as input and pin 3 voltage as output) works as either a nil gain converter (output = 0V) or non-inverting converter (positive output voltage, with positive input voltage), depending on the input voltage level. I still don't understand why its categorization is so important to you, but fine.

(Note that this is a revised interpretation from what I gave earlier, based on a refined understanding and an analysis at a more detailed and more accurate level.)

why is it configured this way? This is a hard one, depending on what exactly you mean, and also depending on the mode of thinking we are in. We are analyzing an existing circuit, with all components specified. So we are in a sort of 'reverse-engineering' mode of thinking. Working from all the given information, based on the established theories, we get the desired outcome (or verify if the outcome either by analysis, simulation, or prototyping, is what we expected). That's why. But if we are designing this circuit from scratch, that's an entirely different story.

2)... I still cannot literally understand why there should be a 0.7 mA current source.

To answer this in some way -- first remember why 0.7mA was chosen (as a design choice) must have been due to many factors, some of which are possibly not relevant to, or seen in, this example. Then, as demonstrated above, 0.7mA should be taken as its nominal capacity, and it can often be restrained to comply so will be supplying a less current (it's not an ideal current source).

3)My understanding is that R1 = 1 kΩ should be made variable to adjust the current min-10A, and R8 should also be made variable to adjust the output voltage . Is this correct?

Yes

My advice in short:
1) learn and understand the golden rules of opamp;
2) read the datasheet, again and again, over and over.

[MariuszD]

Looking at the equivalent circuit of the ideal opamp, its output is modelled as a current source (rather than a voltage source).

That's not true. This is a voltage source. But in SMPS, you will also encounter amplifiers with a current source (operational transconductance amplifier).

This means that there is no limitless current to be source from its output

Real voltage sources always have limited current output. But it does not affect the basic circuit analysis. Even if it had unlimited current output, it wouldn't change the operation of the amplifier.

Now there is a shortage of 0.7-0.046=0.654mA in the required 0.7mA. I presume the result will have to be, due to compliance constraints, that the 0.7mA current source becomes a 46uA current source, the best it can do to 'comply' (because obviously, it's not ideal). Do you think so? (This is the question.)

True. A real current source is a transistor whose current decreases when the voltage across it drops below a certain level.

This also demonstrates that, due to the diode, the opamp may operate as a voltage comparator (i.e. saturated) with certain input voltages at pin 1, when the golden rule of virtual short no longer applies.

We talk about a comparator when the output can only take two states, like in the MC34063. A saturated operational amplifier is still an operational amplifier.

As soon as Pin 1 voltage becomes >=2.36V / opamp working as opamp with negative feedback (virtual short applies) / voltage at pin 3 = 0V / there will be a fight between the opamp and the 0.7mA current source, where the 'loser' (most likely the opamp) will be forced to comply.

When the voltage exceeds 2.36V, the diode conducts and the amplifier delivers 0.7mA. Since the amplifier has a low output resistance, the 0.7mA load does not affect the output voltage.


   

As soon as Pin 1 voltage goes >2.78V / as above / voltage at pin 3 reaches 5V / the required 0.7mA will most likely be satisfied by the opamp, which also supplies the current through RF. (This is when the PWM outputs are turned off – or earlier, depending on the exact voltage threshold)

The useful range of the PWM modulator is 0.5V-3.5V, which corresponds to an input voltage of 3.39-2.55V.

As demonstrated above, the error amp 1 subcircuit (all included as a black box with pin 1 voltage as input and pin 3 voltage as output) works as either a nil gain converter (output = 0V)...

Saturation occurs during transient states. It's like the needle of a gage moving beyond the scale; it's not a different mode of operation, we are simply outside the range of normal operation.

...or non-inverting converter (positive output voltage, with positive input voltage), depending on the input voltage level. I still don't understand why its categorization is so important to you, but fine.

If you use the TL494 as a converter controller, it matters. If you only used the amplifier, it wouldn't matter.

[max.wwwang]

Looking at the equivalent circuit of the ideal opamp, its output is modelled as a current source (rather than a voltage source).

That's not true. This is a voltage source. But in SMPS, you will also encounter amplifiers with a current source (operational transconductance amplifier).

This was not my invention. The picture comes from Microelectronic Circuits (source added in the post above). This is of course not saying that is the only way of modelling the ideal opamp.

[Andy Chee]

This was not my invention. The picture comes from Microelectronic Circuits (source added in the post above). This is of course not saying that is the only way of modelling the ideal opamp.

I think that was a clarification of your description of the picture.

The diamond symbol does NOT mean current source.  The + and - symbols within the diamond indicate it to be a voltage source.

[max.wwwang]

The diamond symbol does NOT mean current source.  The + and - symbols within the diamond indicate it to be a voltage source.

You are right here. Having checked out the book again, the output is indeed modelled as a voltage source. I was misled by the different symbol used for A(v2-v1) from those for v1 and v2 in the same picture. Thanks for your correction. 

And things are more lined up now. An ideal current source has 0 output impedance, which is consistent with 0 output impedance of ideal opamp. This means the output of ideal opamp cannot be modeled as an ideal current source (which has infinitely high output impedance).

I will update my analysis in #63.

[MariuszD]

This is of course not saying that is the only way of modelling the ideal opamp.

You should learn the basics of electronics better. If you understood what a voltage source is and what its properties are, you would immediately catch your mistake.

To simulate the output of an amplifier that has a low output resistance, a voltage source with a low output resistance is ideally suited. Using a current source to simulate voltage source would be cumbersome. So you won't encounter the output of a classical operational amplifier modeled by a current source.

Some controllers like the SG3525 have a transconductance amplifier as the error amplifier. There, the output is current-based and the output resistance is high. If you can't distinguish between a current source and a voltage source, analyzing this circuit will turn your understanding of how an amplifier works upside down.

[Andy Chee]

Some controllers like the SG3525 have a transconductance amplifier as the error amplifier.

And on a tangent, the TL431 is often considered a transconductance amplifier.

What do you get when the control loop combines a TL431 with a TL494?    (an extremely common configuration found in ATX power supplies in years gone by)

[mtwieg]

1)My understanding is that the op-amp employed in the lower part of the schematic is operating in an inverting configuration. But why is it configured this way? The only reason I can think of is that it ensures the voltage at inverting terminal  always remains lower than the non-inverting input. Please correct me if I am wrong.

To be clear, which schematic are you referring to?

2)The op-amp and diode form a super-diode configuration.

Not really "super-diodes" (i.e. precision rectifier circuits). The diodes are there in order to allow either opamp (which ever one has the higher output) to take control of the voltage at pin 3. Without the diodes, the opamp outputs would be short circuited together, which would be problematic for multiple reasons.

When the non-inverting input is higher, the diode turns ON; otherwise, the pulse width has to be increased.

Correct, non-inverting input increasing causes the opamp output to increase, which may cause pin 3 voltage to increase, causing the PWM duty cycle to decrease.

However, I still cannot literally understand why there should be a 0.7 mA current source.

Because of the diodes on each opamp's output, it's only possible for the opamps to source current into pin 3. The current source is there to help pull pin 3 low when neither diode is forward biased. Without that current source, the error amplifier circuits would not behave as expected.

[ommsiva]

Dear Sir,

I am refering to attached circuit.

[max.wwwang]

I see what you mean with super diode now. It’s interesting. Also precision rectifier.

No, as suggested, the opamps and the diodes here are not intended to work as super diodes.

[I like the term 'super diode', partly because I prefer the term "ideal diode" to be reserved for a conceptual diode that is similar to super diode but with a non-zero voltage drop, i.e. a one-way switch with a voltage drop. This makes sense even with respect that a super diode is in a way superior than an ideal diode (though some might find it uncomfortable to have something superior than ideal).  ]

[mtwieg]

Dear Sir,

I am refering to attached circuit.

In this case that opamp circuit is acting as a difference amplifier. Its output will be roughly (Vout-Vref)*10. Though this will be only be the case if the other opamp is not in control of pin 3.

[max.wwwang]

[Picture courtesy of Microelectronic Circuits]

This is a revision of my analysis in #63 of the voltage regulator feedback subcircuit. It is based on the following assumptions or premises:
1) ideal opamp with a voltage source as its output;
2) 'ideal diode' (my term – meaning an ideal one-way switch but with a non-zero voltage drop when on, which is assumed as 0.7V here [Edit: actually this does not matter; furthermore, this assumption is not necessary.]);
3) single voltage supply for the opamp, meaning its output will not go under 0V (as constrains for the 0.7mA current source when it has to 'comply');
4) the current feedback opamp is not firing up.

All critiques are welcome.

It's based on this equivalent circuit:


An ideal opamp will be capable of sourcing or sinking any current required by a current source connected to its output. Due to the diode however, only sourcing is possible. And due to the 0.7V voltage drop, sourcing is possible only if/when the voltage source at the opamp's output is 0.7V or above.

This means, still, my previous statement "the 0.7mA current source will prioritize the opamp outputs" is not true. As long as the opamp output voltage remains <0.7V, this current will come solely from the path of RF.

Consider some typical scenarios:

1. When there is a 0.7mA current flowing from +2.5V through RF, just meeting the needs of the current source.
Theoretically this will give a voltage of 0.358V at pin 2 and -35.342V at pin 3 (or output voltage). Due to constraints (output voltage is impossible to go under 0V), the result will have to be the following scenario –

2. A current of 46uA is flowing through the same path, giving 2.358V at pin 2 and output voltage of 0V.
The current source is forced to comply and becomes a 46uA current source. This happens whenever pin 1 voltage is under 2.358V (when the diode will remain off).

Still, there will be the following modes of operating based on pin 1 voltage (or input voltage):

• Input voltage <=2.358V / the diode is off / the 0.7mA current source is forced to comply and downgrades to 46uA current source / the opamp is saturated (effectively working in a voltage comparator mode) / output voltage = 0V .
• As soon as input voltage becomes >2.358V / the diode turns on / the demand of the 'starving' 0.7mA current source is immediately met (with the shortage met by the opamp) / the opamp working as opamp with negative feedback (the virtual short rule applies) / output voltage starts to rise from 0V.
• Before input voltage reaching 2.5V / the diode remains on / the 0.7mA current source is pumping 0.7mA through (supplied by both the opamp and through RF) / the opamp working as opamp with negative feedback (the virtual short rule applies) / output voltage keeps rising (now between 0V and 2.5V).
• When input voltage reaches 2.5V / the diode remains on / the 0.7mA current source is pumping 0.7mA through (but now supplied soley by the opamp, no current through RF) / the opamp working as opamp with negative feedback (the virtual short rule applies) / output voltage reaches 2.5V. (This is the desired point.)
• As soon as input voltage goes >=2.642V / the diode remains on / the 0.7mA current source is pumping 0.7mA through (but the opamp is now supplying the current for both of the 0.7mA current source and through RF) / the opamp working as opamp with negative feedback (the virtual short rule applies) / output voltage reaches 5V. (This is when the PWM outputs are turned off – or earlier, depending on the exact voltage threshold.)

Here is what I get as the relationship between V1 (input) and V3 (output):

This shows that the voltage regulator subcircuit is either a nil gain converter or a non-inverting converter, depending on the input voltage.