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Error Amplifier operation for a TL494

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si2030

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Hi there,

For the life of me I cannot understand the way the error amplifiers work in a TL494 (and for that matter any of the other PWM).

Here is a schematic from the Texas Instruments Application report "Designing Switching Voltage Regulators With the TL494" page 25:

View attachment 61993

For my purposes I have a Vout ranging from 0 to 30 volts which I want to manually adjust and hod at that voltage.. Right now I can manually adjust the pulse width using the dead time pin. However I get the impression that the error amps are used to manage constant current for a particular voltage... I am not sure about this though or how to implement this.

How does the error amp work and in particular How should I use it to work with voltages ranging from 0 to say 30 volts as the REF only goes to 5volts? I know I can use a resistive divider but not being sure about what the error amplifiers actually do and how they work etc I am in the dark as to creating a suitable circuit for using them.


Kind regards

Simon
 
You are using the dead time pin to change the duty cycle.
You can also use the feedback pin to change the duty cycle.

The error amplifier usually compares the reference and some division of the output voltage. If the output voltage is correct then the error amplifier output (feed back pin) does not move. If there is an error then the feedback pin moves up or down to change the duty cycle to move the output voltage to where it should be.
 
If you want to use a pot to set the output voltage, this should work. Due to component and reference tolerances, you might need to trim R3 or R8.
 
So assuming this is used for a regulated power supply (my intention) and you set the voltage to some value say 16 volts..... do you use feedback or dead time to set this voltage... and what happens if you then draw a load off the output wouldn't that alter the voltage as the current draw increases... is this where the error amp is used?
 
So assuming this is used for a regulated power supply (my intention) and you set the voltage to some value say 16 volts..... do you use feedback or dead time to set this voltage... and what happens if you then draw a load off the output wouldn't that alter the voltage as the current draw increases... is this where the error amp is used?
In the circuit I posted, the error amplifier compares the voltage on the wiper of the pot (0 to 2.5V) with 1/12th (R9/(R8+R9)) of the output voltage. If they differ, the difference is amplified and the duty cycle is adjusted, which in turn adjusts the output voltage to maintain balance.

CAVEAT: I have no experience with TL494. I'm basing my comments on my interpretation of the datasheet.
 
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So assuming this is used for a regulated power supply (my intention) and you set the voltage to some value say 16 volts..... do you use feedback or dead time to set this voltage... and what happens if you then draw a load off the output wouldn't that alter the voltage as the current draw increases... is this where the error amp is used?
Yes. Feedback from the output to the error amp keeps the output voltage (essentially) constant even if there are changes in the input voltage (which affects the required duty-cycle) or output load current.
 
Code:
Hi Roff or is it Ron... not sure, and others

Ok I have had a look at hte circuit you have posted and I think I am slowly getting it however I still feel somewhat obtuse. From what I can make out The error amp is either on or off.. it never modifies the actual pulse width of the TL494 it comes on with a pulse width set by the dead time pin and when its on the pulses cause the switching which in turn causes the output voltage to go up and at some point the vout on the bridge divider floats above the reference voltage. the TL signal turns off and the mosfets go silent - no energy is thus flowing into the filter caps at the output and so the output drops... and keeps dropping until the it floats below the reference. this turns on the swtiching and the voltage starts to grow again... and it cycles.

So based on the circuit provided, We adjust REF using R4. TL494 will either turn off the switching pulses until Vout floats down to a value where 1/12 of that value = the newly adjusted REFERENCE voltage. (see table below)....

Does this sound correct??


Further, what the heck does the 510 ohm resistor and the 51k ohm resistor do?

Simon


REF = 5 VOLTS
Vref - VOLTS
Code:
R3	R4     POT+R3	(R4/(R3+34))*REF  (R9/(R8+R9))*30
1000	0	0	0		
1000	50	0.05	0.24		2.86
1000	100	0.09	0.45		5.45
1000	150	0.13	0.65		7.83
1000	200	0.17	0.83		10.00
1000	250	0.20	1.00		12.00
1000	300	0.23	1.15		13.85
1000	350	0.26	1.30		15.56
1000	400	0.29	1.43		17.14
1000	450	0.31	1.55		18.62
1000	500	0.33	1.67		20.00
1000	550	0.35	1.77		21.29
1000	600	0.38	1.88		22.50
1000	650	0.39	1.97		23.64
1000	700	0.41	2.06		24.71
1000	750	0.43	2.14		25.71
1000	800	0.44	2.22		26.67
1000	850	0.46	2.30		27.57
1000	900	0.47	2.37		28.42
1000	950	0.49	2.44		29.23
1000	1000	0.50	2.50		30.00
 
No. The error amplifier is an op amp that is normally used with feedback resistors (and often capacitors for loop compensation) to operate in the linear region. It adjusts the duty-cycle to maintain the required output voltage, just as if you were looking at the output voltage and turning a pot to adjust the duty-cycle until you got the desired voltage.
 
Just as crutschow mentioned the error amplifier work in the linear region just as regular OP amp. We have exactly the same situation in linear voltage regulator.
As for 51K and 510 resistor. They set the open loop gain and also they increase the stability of the error-amplifier circuit by reducing the gain to 101.
And look at real design voltage regulator 0...25V up to 5A using TL494.
 
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Think of a simple OP amp with a reference (+ input) and a value (- input) with no feedback. In this situation, the OP amp acts as an error amplifier. (A-B)*(open loop Gain). When the error is zero. input (A) = ouput(B) it's done.

The only real difference is that the output has to be scaled to in range of the error amp. So if the error amp has a range of 0-5V and the output is 0-50, the output has to be divided by 10 before being used by the error amp.

This is a really fancy ERROR amplifier, the PID controller. https://en.wikipedia.org/wiki/PID_controller Kp * e(t) is the resulting output. Kp is known as the proportional gain and e(t) is known as the error; The measured value - the setpoint. The integral term helps make the setpoint and the measured value agree and the derivative portion deals with overshoots.
 
Hi,

Just a couple of things to note here.

The 51k and the 510 ohm resistor limit the gain of the error amp to around 100 to increase stability. This is unfortunate because the error amplifier is what controls the regulation, and the more gain the error amplifier has the better the load regulation will be. Typical gains in other systems would be 1000 or more.
Why they chose to put that on the data sheet im not sure, and that doesnt mean that every system has to be like that, but it could be because of the unusual error amplifier system they chose to build into this chip.
So if you find that you have bad load regulation, you may think about trying to increase the gain or even add your own external error amplifier, which would just be an op amp, with the on chip error amps disabled through proper external connections.

It would be nice at this point to see the rest of the circuit so that we could predict the overall performance with more accuracy. Output transistors, etc.
 
Current Circuit as it stands...

Hi MrAl, Others,

Sorry for the long post...

I am still struggling with the idea of how the error amp works - in particular how to confidently bias it...

That being said, my impressions now on how it works are as follows:

There are two control signals one from the dead time and another from the 2 error amps. The system only recognises the highest one and uses that to set the PWM. If dead time is set to zero and also the error amps then there is a 0.7volt value as a result (OF dead time signal) and this is whats used - this provides the basis for the minimum 3% dead time.

The error amp is biased for non inverting operation. The difference between the two signals entering the (+) and (-) (with the inverting signal bigger than the non inverting signal) is amplified and forms the basis for the control signal which when applied to the saw tooth wave creates the pulses.

I am guessing the non inverting pin acts as a reference. If the non inverting signal goes below the inverting signal the difference shows up as an increased control signal.


OK my circuit as it stands...

View attachment 62041

the idea here is that I am using the TL494 as a pre-regulator power supply for a linear output stage using a LT1083. In the circuit provided its assumed that mains has already been transformed down to 30 volts. (I did this to speed up the simulation - I have designed and built a transformer that should produce the voltage of ~35 V - Freq 45 kHz) Anyway its run through a filter and that forms the basis of the LT1083 input. The central idea here is that if you adjust the LT1083's voltage this will cause a difference (ultimately) between the input and the output which are set to have a specific gap - input ~2-3 volts higher than the LT1083 output at all times when in a steady state. To achieve this I set the inverting pin of the error amp to be approximately 1/12 of the output voltage and the the non inverting pin of the error amp to be approximately 1/14 the input voltage. That way when they reach an equivalent value there is a discrepancy of ~2-3 volts between the input and output...

Thats the plan!

Referring to the above circuit with the simulation traces. To make things more obvious I used a voltage controlled resistor to set two extremes for the LT1083 adjustment - PINK trace in ohms is this adjusting resistor below the LT1083. The voltage into the LT1083 is indeed higher than the output and it does track the output.. RED trace is the input and TEAL trace is the output. The BLUE and GREEN traces at the bottom are the values for the inverting and non inverting respectively.

In the sim there is a fairly slow response.. takes nearly 0.8 seconds to arrive at a new stable position..

A fair mouthful so far.... However I'd like to ask if this is a legitimate concept for tracking.... (Will it work? Whats wrong with it?) and how to properly bias the error amp - that is working out the values confidently... because there is nothing out there that describes this clearly and in depth.. the best I got was vague sub circuit with no values and no calculations etc.. short of buggering around on a work bench I cannot see a way of building this properly.

Further how does applying a gain to the difference effect the control signal.... does this mean the two values track much closer together because as the difference grows by the gain this causes the change in the control signal to be that much faster?


Sorry for the book but I guess if I describe it properly someone might take me up on this and help describe the gaps in information.

Kind Regards

Simon
 
Your schematic is certainly not how to use a TL494. So to explain how it is not working right is not easy.

I think you want 30 volts bucked down to drive your linear regulator. Because you are using LTSpice look for a LT buck regulator that can handle the current. Once you have the simple buck working then we can show you how to make the error amp work. Look for an example circuit on LTSpice.
 
Hi again,


Oh ok, so we get more of the picture now :)

First thing:
What are you doing with the output there with D1 and D2 and other parts? It looks like you have a bad cross between a half bridge drive and a synchronous drive. It's either one or the other, either you want a half bridge or you want a synchronous buck. From the looks of your circuit and the fact that you dont need a negative output, it appears that you want a synchronous buck.
To get a synchronous buck, remove diode D1 and short D2. That puts the drive right on the inductor left hand side pin and you end up with a buck circuit.

That's just to start you off.

I think you might have a better idea of how the error amp works now. The output ramps up when there is a difference between the output and reference, and that changes the duty cycle of the switcher which in turn raises or lowers the output voltage.
The mathematical model for this is quite easy to understand and is worthwhile to look into. For one thing, it will show how the gain of the error amp affects the output regulation. You can start with a linear regulator with a reference and feedback.
The basic control equation looks like this:
Vout=Vref*G/(1+G*H)
where G is the forward gain and H is the feedback gain. For the simplest case G is the gain of the error amp and H is 1 and Vref is 1. With this setup a perfect regulator would produce Vout=1v for Vref=1, but the gain G always causes an error. With large G this error goes down. G=101 may be enough if you have low plant losses or you dont expect super perfect output regulation. This brings up the question of what performance you expect from this circuit.

Shown in the attachment is a basic feedback circuit with error amp with gain of 100. A linearized buck circuit looks similar to this.
 
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LTC1624 Will buck 30 volts down to what you need. See the app note in post #15. A transistor or P-mosfet us used to detect voltage across the linear regulator, then drives the feed back pin (error amplifier).
 
Hi,

Did you fix the diode connections?
 
An Update.

Hi MrAl, others,

Ok so I spent the entire day reworking this (and learning in the process). I have already physicall built the switching transformer and still wanted to use it so I thought I would keep the TL494 but instead of using it to actively control the voltage and current it now regulates (via the error amp) a voltage around 30 volts. I looked up the correct way to rectify a centre tapped secondary and have used that.

I had another close look at the TL1083 datasheet and it had another pre-regulator circuit which I have used in this. I have also added a second TL1083 to increase the allowable current to 15Amps although I have set it to only work to 10 amps.

I also included a current limit adjustment that can be adjusted up to 10 amps. This is built using the LM324.

Here is the picture of it as it stands so far.

View attachment 62082

The top trace is the value of the load resistor which I set to a function of time. This way I can continuously vary the resistance past the point of 10amps and see if the current limit actually works... The second trace shows the base of the current limit transistor which has turned on twice at the low point for load resistor value.

The third trace shows the tracking of the output voltage by the input voltage... which actually works here.

The forth trace is the output of the TL494 after rectification.

The bottom trace shows the two inputs for the LM324 that turns on the current limit transistor... as can be seen the pink reaches the teal line and activates the transistor.


So that's where I'm at... not sure if this is any better but it seems to work... is there anything that looks clangingly wrong...

Simon
 
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