Linear Power supply - Large current on the trafo side before the caps (schematic inside)

crutschow

Joined Mar 14, 2008
34,280
...................
Right Ok, In the simulation though, when I'm drawing 300mA, the RMS current on the transformers side is about 1.4A RMS. Which is more than 0.6A RMS.
Don't know how you got those values. :confused:
Below is the simulation of your circuit with R6 set to 90Ω to give about a 300mA load.
The measured transformer current is 614mArms.

The output ripple voltage is because the voltage drop across the regulator is too small and it's falling out of regulation.

upload_2016-8-30_17-59-11.png
 

Thread Starter

FurryLemon

Joined Aug 14, 2016
20
Don't know how you got those values. :confused:
Below is the simulation of your circuit with R6 set to 90Ω to give about a 300mA load.
The measured transformer current is 614mArms.

The output ripple voltage is because the voltage drop across the regulator is too small and it's falling out of regulation.

View attachment 111265
Oops. I was calculating the RMS value but forgetting that there is one peak for every cycle, not two.

However, after the diode bridge and before the capacitor the RMS value is larger. Surely then I would need diodes that can take the larger current?
Edit: Actually im wrong, RMS is still the same.
 
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Measure the resistance of the secondary winding with an ohmmeter. Add a resistor of that value in series with the voltage source in your simulation. Crutschow suggests that it will reduce the transformer RMS current a little, but other effects such as leakage inductance will also reduce it.

Does you have a meter having the capability to measure RMS current? Even though you don't have a scope, you could just measure the transformer secondary current; the RMS current is what counts.

Have a look at this thread: http://forum.allaboutcircuits.com/threads/derating-transformers-for-rectifier-use.68839/
 

crutschow

Joined Mar 14, 2008
34,280
..................
However, after the diode bridge and before the capacitor the RMS value is larger. Surely then I would need diodes that can take the larger current?
Edit: Actually im wrong, RMS is still the same.
You need diodes that can handle half the average DC current and the peak current.

RMS current values don't apply to diode ratings since diodes are very nonlinear and not ohmic (the voltage drop increases logarithmically with current not linearly.)
For example, if you double the current through a diode, the power dissipated is increased by only about 2.1 times, not 4 times as an ohmic resistor would experience.
 
You need diodes that can handle half the average DC current and the peak current.

RMS current values don't apply to diode ratings since diodes are very nonlinear and not ohmic (the voltage drop increases logarithmically with current not linearly.)
For example, if you double the current through a diode, the power dissipated is increased by only about 2.1 times, not 4 times as an ohmic resistor would experience.
You raise an interesting point that is often not well understood.

If we look up a graph of instantaneous forward voltage vs. forward current we can see that logarithmic variation. For example, here is such a graph from the General Semiconductor data sheet for a 1N400x rectifier:

1N400X.png

This is plotted on semilog "paper". One of the characteristics of an exponential function is that it plots as a straight line on a semilog graph. We can see that behavior at the lower left corner.

But the overall curve is not straight; why not?

Here's the same graph, but with a straight red line continuing the curve from the lower left. This red line is what we would get if the diode V-I characteristic was logarithmic for all currents:

1N400XX.png

We see that at about .2 amps, the voltage across the diode is greater than it would be if the logarithmic characteristic continued at higher and higher currents. The behavior is like a resistor at high currents where the black curve goes further right of the red (log) curve. This is due to the actual ohmic resistance of the semiconductor material which begins to dominate at high currents.

The spec for rectifiers is given as a maximum average current and as you point out, when the V-I characteristic is logarithmic the power dissipation as current increases is different than when the V-I characteristic is linear (like a resistor).

Typical capacitor input power supplies, such as the one shown in this thread, have peaky current pulses drawn from the transformer. If the rectifier is operated so that the average current is near the maximum rating, the peaky current pulses will undoubtedly be several times the max average current rating. Those peaky pulses will extend into the region where the V-I characteristic is more like a resistor than a logarithmic device.

Thus the power dissipated in the rectifier isn't proportional to the average current only, or to the RMS current only. It's a combination of both. Neither a true RMS meter, or an average (so-called) responding meter can tell accurately the effective diode dissipation. What I do is use a digital oscilloscope that can do waveform math, multiply the instantaneous voltage and current, and average that result over several cycles. This gives the true power dissipation in the diode.
 

ericgibbs

Joined Jan 29, 2010
18,766
hi FL.
You have posted that you have a 22V RMS transformer, I am assuming that you have 2 transformers.???

I guess you know that you cannot connect the two rectifier bridges and regulators to a common transformer secondary.

E

EDIT:
You should find this 'asc' file helpful for transformer sims.
 

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Thread Starter

FurryLemon

Joined Aug 14, 2016
20
hi FL.
You have posted that you have a 22V RMS transformer, I am assuming that you have 2 transformers.???

I guess you know that you cannot connect the two rectifier bridges and regulators to a common transformer secondary.

E

EDIT:
You should find this 'asc' file helpful for transformer sims.
Yes, you are right, I am using one transformer with two separate output coils. Each supply +ve/-ve, is isolated.
 

Thread Starter

FurryLemon

Joined Aug 14, 2016
20
I assume the peak 0.8A transformer limit you are referring to is the peak sine current of the 0.6Arms rating.
There is a transformer limit but it's not the peak current per se, it's the RMS current which results from those high peak currents.

The high peak currents, due to the filter capacitor, cause the RMS transformer current to be about twice the DC output current, thus you should derate the transformer RMS current rating about 50% as compared to the DC output current.
This means you should draw no more than about 0.3ADC from your 0.6Arms transformer.
(You can see this RMS current in the simulation by doing a CTRL left-click on the transformer current waveform title in the plot window.)
I've been thinking about this now. "The high peak currents, due to the filter capacitor, cause the RMS transformer current to be about twice the DC output current, thus you should derate the transformer RMS current rating about 50% as compared to the DC output current". Is this a general rule of thumb? Or can it be described as a relationship (mathematical)? dannyf explained what it was that was happening in terms of the capacitor and current.

Because for example, I would I of known this to be happening without the simulation?
 
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Thread Starter

FurryLemon

Joined Aug 14, 2016
20
why would it be?
Maybe I shouldn't of put it that way.

I'm still trying to understand what you said earlier.

So if the loading is low, the capacitor is not as depleted and it's voltage remains high. That means that the transformer has a shorter period of time to charge up the capacitor, thus the peak current is high.

When the loading is high, the capacitor gets more depleted and the voltage on the capacitor is low. This the transformer has more time to charge up the capacitor so the peak current is lower, everything being equal.
So when the loading is low, this means more current is being drawn. You say the capacitor is not as depleted and the voltage remains high. That means the transformer has a shorter period of time to charge up the capacitor ?

Oh hang on. I may of got it.

So, the transformer is always drawing a current based on the load. That current will be effectively added on to the current supplied by the capacitor. Thats why you get a large peak current on a low load. Is that right?

So if the load was large, less current would be drawn. Less current would be added on to the transformers current.

EDIT: I think 'added' is the wrong term to describe this. I do think I understand this now, I'm trying to think of the right word instead of 'added'. Essentially the transformer has to do the job of supplying current for the load aswell as the charging of the capacitor?

I hope what I'm saying is right.
 
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The Electrician

Joined Oct 9, 2007
2,971
Maybe I shouldn't of put it that way.

I'm still trying to understand what you said earlier.



So when the loading is low, this means more current is being drawn. You say the capacitor is not as depleted and the voltage remains high. That means the transformer has a shorter period of time to charge up the capacitor ?

Oh hang on. I may of got it.

So, the transformer is always drawing a current based on the load. That current will be effectively added on to the current supplied by the capacitor. Thats why you get a large peak current on a low load. Is that right?

So if the load was large, less current would be drawn. Less current would be added on to the transformers current.

EDIT: I think 'added' is the wrong term to describe this. I do think I understand this now, I'm trying to think of the right word instead of 'added'. Essentially the transformer has to do the job of supplying current for the load aswell as the charging of the capacitor?

I hope what I'm saying is right.
You are misunderstanding the word "loading" as used by engineers in the business. You are taking the phrase "loading is high" to mean a high resistance load that draws a small current from the supply, and the phrase "loading is low", or "low load" to mean a low resistance load. Your understanding is just backwards compared to common usage among power supply engineers. When we (engineers) say "loading is low", we mean a load that doesn't take much current from the supply.

Did you read the link I gave you: http://forum.allaboutcircuits.com/threads/derating-transformers-for-rectifier-use.68839/
 

crutschow

Joined Mar 14, 2008
34,280
I've been thinking about this now. "The high peak currents, due to the filter capacitor, cause the RMS transformer current to be about twice the DC output current, thus you should derate the transformer RMS current rating about 50% as compared to the DC output current". Is this a general rule of thumb? Or can it be described as a relationship (mathematical)? dannyf explained what it was that was happening in terms of the capacitor and current.

Because for example, I would I of known this to be happening without the simulation?
Yes, it's a general rule-of-thumb for a full-wave rectifier circuit with capacitor input (don't understand dannyf's comment about that :confused:).
It will vary some depending upon the transformer and diode impedances, as well as the size of the capacitor but for typical designs you should derate the DC output current to no more than about 50-60% of the transformer's RMS rating.
It can be determined mathematically if you generate a mathematical equation for the current pulses out of the transformer and use that to calculate the RMS current value.
This derating was known to be required well before simulators and computers were invented. Simulators just make it easier to determine the value.
 

Thread Starter

FurryLemon

Joined Aug 14, 2016
20
You are misunderstanding the word "loading" as used by engineers in the business. You are taking the phrase "loading is high" to mean a high resistance load that draws a small current from the supply, and the phrase "loading is low", or "low load" to mean a low resistance load. Your understanding is just backwards compared to common usage among power supply engineers. When we (engineers) say "loading is low", we mean a load that doesn't take much current from the supply.

Did you read the link I gave you: http://forum.allaboutcircuits.com/threads/derating-transformers-for-rectifier-use.68839/
Right, thank you.

In dannyf's post, he says that when the load is low (low current is being drawn), the peak current is high. But in the simulation it isn't. When the current being drawn is low the peak current is low. It is opposite when there is a large load. There is a larger peak current on the transformer side.
 

The Electrician

Joined Oct 9, 2007
2,971
Right, thank you.

In dannyf's post, he says that when the load is low (low current is being drawn), the peak current is high. But in the simulation it isn't. When the current being drawn is low the peak current is low. It is opposite when there is a large load. There is a larger peak current on the transformer side.
dannyf is mistaken.

Do you understand why there is a very peaky current drawn from the transformer secondary? It's because once the capacitor is charged up to a voltage near the peak of the applied sine wave, further current can only come from the transformer when the secondary voltage is more than the voltage already on the capacitor, and that only occurs for a short time. Since the DC current into the load must be supplied from the transformer during short time intervals, the average current occurring during those short intervals has to be the same as the total DC load current. The DC current is flowing all the time; it's steady. Mathematically speaking, the integral of the current pulses from the secondary has to equal the integral of the DC load current. The only way this can happen is if the peaky currents have larger peak amplitude during their short time than the DC current.

And, it's also a mathematical fact that given two currents having the same DC component, but different waveshapes, the more peaky current will have the greater RMS value. It's the RMS value of the current that determines how much heating the copper wire of the secondary will experience, and that heating is what determines the rating. In capacitor input circuits like yours, the peaky current out of the transformer secondary will always have a larger RMS value than the DC current into your load, so you have to derate the transformer. I fully explain this in the long thread I linked to in the earlier post.

What I mean by "DC component" is this: if you use a meter that responds to DC and use that meter to measure the current out of the rectifier (but before the capacitor; the peaky currents in other words), and compare that to the DC current out of the capacitor, the two measurements will be the same. The time integral of the peaky current (this is the average current) is the same as the DC current out of the capacitor; it has to be that way.

The manufacturer's rating for a transformer assumes a resistive load. So your transformer is rated for .6 amps only if you connect a resistor that would draw .6 amps directly to the secondary, without a rectifier/capacitor. With a resistor load, the current drawn from the secondary has the same wave shape as the grid voltage, a sine wave, not a peaky waveshape.
 

Thread Starter

FurryLemon

Joined Aug 14, 2016
20
dannyf is mistaken.

Do you understand why there is a very peaky current drawn from the transformer secondary? It's because once the capacitor is charged up to a voltage near the peak of the applied sine wave, further current can only come from the transformer when the secondary voltage is more than the voltage already on the capacitor, and that only occurs for a short time. Since the DC current into the load must be supplied from the transformer during short time intervals, the average current occurring during those short intervals has to be the same as the total DC load current. The DC current is flowing all the time; it's steady. Mathematically speaking, the integral of the current pulses from the secondary has to equal the integral of the DC load current. The only way this can happen is if the peaky currents have larger peak amplitude during their short time than the DC current.

And, it's also a mathematical fact that given two currents having the same DC component, but different waveshapes, the more peaky current will have the greater RMS value. It's the RMS value of the current that determines how much heating the copper wire of the secondary will experience, and that heating is what determines the rating. In capacitor input circuits like yours, the peaky current out of the transformer secondary will always have a larger RMS value than the DC current into your load, so you have to derate the transformer. I fully explain this in the long thread I linked to in the earlier post.

What I mean by "DC component" is this: if you use a meter that responds to DC and use that meter to measure the current out of the rectifier (but before the capacitor; the peaky currents in other words), and compare that to the DC current out of the capacitor, the two measurements will be the same. The time integral of the peaky current (this is the average current) is the same as the DC current out of the capacitor; it has to be that way.

The manufacturer's rating for a transformer assumes a resistive load. So your transformer is rated for .6 amps only if you connect a resistor that would draw .6 amps directly to the secondary, without a rectifier/capacitor. With a resistor load, the current drawn from the secondary has the same wave shape as the grid voltage, a sine wave, not a peaky waveshape.

ahhh, yes!, It's just clicked. Sorry, its sometimes hard to interpret what people are saying. But it makes sense now! "Mathematically speaking, the integral of the current pulses from the secondary has to equal the integral of the DC load current." Yep makes sense.

I've read your post on derating transformers. It's a very nice practical guide to derating it. I'll have to do it.

Thank you.
 
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