What do shunts in a microwave oven transformer do?

  • #1
gary350
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I removed the high voltage secondary winding of a microwave oven transformer and replaced it with 9 turns of wire I need 2.1 vac 50 amps. Should I remove the shunts?

I have a 100a bridge rectifier I need 3 vdc 50a for about 5 to 7 days. I don't think a MOT is 100% duty cycle so I plan to use a 1500 w variac to power the MOT with 60 vac or less to be able to run for 7 days continuous.

150 watts is not much but I still want built in safety shut down.

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  • #2
I think that's a Ferroresonant transformer design (AKA Constant Voltage Transformer). The shunts are designed to saturate after exposure to a certain amount of flux (volt-seconds from the primary). These designs will normally have a resonating capacitor on an output winding. The idea is to regulate voltage at the secondary.

I would remove them unless you want to do that sort of design. Since your repurposing this, I would look at it as a mostly new design and only put in what you want and understand.
 
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  • #3
gary350 said:
Should I remove the shunts?
Yes.
When I make a 1:1 isolation transformer from a pair of microwave oven transformers, I keep the primary windings and replace the magnetic shunts with wooden wedge blocks to hold the windings in place.

I believe the magnetic shunts are there to prevent one-sided saturation of the transformer when used with the half-wave rectifier that supplies the magnetron HV DC.
 
  • #4
I read online the shunts make the HV transformer come up to full power very fast for the magnetron start. There are also several other answers online.

I did the math for that EI core 100 turn primary needs to be 220 turns for continuous duty at 120 vac 60Hz. I'm pretty sure the 100 turn primary will melt down if I leave it run 7 days. I just don't want to rewind another EI core I am looking for an easy way out like 60v or less primary. I have a 1500w variac I can make 60v or less.
 
  • #5
I removed the shunts 9 turn secondary is now 9.36v. 2 turns is 2.08v. Now it works like a transformer should work.
 
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  • #6
It's probably easier at this point to just do some measurements an see how hot it gets than for us to philosophize about magnetizing inductance etc. You are kind of stuck with the core/primary you've got without doing a new design from scratch.
 
  • #7
DaveE said:
It's probably easier at this point to just do some measurements an see how hot it gets than for us to philosophize about magnetizing inductance etc. You are kind of stuck with the core/primary you've got without doing a new design from scratch.
That is what I will do. I know the primary winding should have 2 times more turns to be 100% duty cycle but it will run at 150w instead of 1200w that = 8 times less power. I will attach a thermo switch to the transformer if the temperature gets above 120°f it will turn off the 120vac.
 
  • #8
I made a 3 turn 60a secondary winding voltage is too high so I changed it to 2½ turns secondary is 3.05 vdc. Perfect. Primary is 120vac the transformer has warmed up 10 degrees after a few minutes of testing. Next I need a thermo switch on the 120vac primary then let it run for several hours to see how warm it gets. When I finish the circuit and put a 50a load on the transformer it might get hotter. I need to check DC output with a scope. I used these 2 caps because they were right in front of me. I have 100s of caps I might be able to find 50,000. uf for 3 vdc. I have a heat sink fan also it might be needed when the circuit is running 50 amps.


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  • #9
What is the application of the rectified low-voltage supply?

A low-voltage full-wave rectifier is inefficient because the two diode voltage drops are similar to the transformer voltage. Efficiency may be about 50%.

So why do you use a transformer to drive a rectifier bridge, when low voltage switching regulator modules are so cheap ?

A low-voltage switching supply, will have synchronous MOSFETs to rectify the output. MOSFETs have a lower voltage drop than diodes, so can often get better than 90% efficiency.
 
  • #10
Baluncore said:
What is the application of the rectified low-voltage supply?

A low-voltage full-wave rectifier is inefficient because the two diode voltage drops are similar to the transformer voltage. Efficiency may be about 50%.

So why do you use a transformer to drive a rectifier bridge, when low voltage switching regulator modules are so cheap ?

A low-voltage switching supply, will have synchronous MOSFETs to rectify the output. MOSFETs have a lower voltage drop than diodes, so can often get better than 90% efficiency.
Show me your circuit drawing that produces 3 vdc 50 amp none stop for 7 days. Power source is 120 vac.
 
  • #12
Baluncore said:
Buy it assembled and tested.
Search eBay: Regulated Switching Power Supply

https://www.ebay.com.au/itm/386972623426
3.0V, 40A, 120 W = AU

What is the power source for that???
Baluncore said:
What is the power source for the 3v 40a device.
 
  • #14
It's a bit difficult to make a good high current DC rectifier with just a big capacitor. Either you get very small conduction angles, hence high RMS currents, or you get pretty poor load regulation. If you aren't already, use schottky diodes. Also mind the ripple current ratings on your capacitors.

One of my favorite references for this stuff is the appendix of an ancient National Semiconductor Audio handbook. Before we all had Spice on our desktops.
 
  • #16
gary350 said:
it can also do 460a also
Probably not for long. That's 1.4KW output and probably something similar dissipated in the diodes.
 
  • #17
Also, it should be easy to wind a center tapped secondary which will get rid of half of your diodes. That's important at very low voltages.
 
  • #18
After 30 minutes the MOT transformer is almost too hot to touch. With 120 vac on the primary the amp meter shows almost no amps. With load on secondary meter is 2.80 vdc 2 amps = 5.6 watts. 40 carbon rods in salt water should have .2 amps on each rod but they don't. I need a transformer 30w 120vac to 3vac.

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  • #19
gary350 said:
With 120 vac on the primary the amp meter shows almost no amps.
Where? Which current?

I'm not sure exactly what you're doing, but why do you need those big capacitors? They're going to cause more heat in the wires and diodes because of the reduced conduction angle (high peak currents). My ancient car battery charger doesn't have them.
 
  • #20
DaveE said:
Where? Which current?

I'm not sure exactly what you're doing, but why do you need those big capacitors? They're going to cause more heat in the wires and diodes because of the reduced conduction angle (high peak currents). My ancient car battery charger doesn't have them.
This project is suppose to be 50a at 3v. Your suppose to use 1000uf per amp = 50,000. uf. Carbon rods have a surface of about 1 sq inch. Carbon rods are suppose to be .2a per sq inch. 40 carbon rods are 40 sq inches. 40 x .2 = 8a. I don't have enough carbon rods to do 50a. I tried a different DC power supply it is 12vdc and volt meter show the voltage across the salt water is up to 3.4 vdc. Voltage has come up from 2.8v to 3.4v. According to my test a 24v power supply should read 6.8v on the salt water. Tomorrow I replace power supply with a 1500w variac on 120 vac then dial in 8a on the salt water that should work. Data I have says to keep carbon rod at 3.5v .2a to prevent the rods from self districting. Data I have tells the voltage and amp but not how to build the power supply to get that. Variac should make it possible to dial in 3.5v 8a. I tested the resistance across the salt water cell and the carbon rods my meter shows 1.6 ohms. Tomorrow I do more tests and take more voltage and amp reading. I have 4 lbs of pure salt dissolved in 10 lbs of distilled water.
 
  • #21
gary350 said:
Your suppose to use 1000uf per amp
You are? Sorry, I still don't understand, which is OK, since it's your problem, not mine.
 
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  • #22
Problem solved 1 of the blue caps was bad. I learned 3.5vdc works but 3vdc does not, a 2kw variac connected to 120vac is working good. Amp meter reads 12a at 3.5v = 42 watts. All wires had to be changed to 15a wire. 40 carbon rod electrodes at the anode. Stainless steel pot is the cathode.

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  • #23
DaveE said:
You are? Sorry, I still don't understand, which is OK, since it's your problem, not mine.
There is a range of capacitor size that needs to be based on allowable ripple.
-
I did a quick Google for Astron power supplies which have their schematics published and available. I would say their rule of thumb for capacitor size 1500 and 2000 uF per amp. These are 12 volt regulated power supplies.
 
  • #24
Averagesupernova said:
There is a range of capacitor size that needs to be based on allowable ripple.
-
I did a quick Google for Astron power supplies which have their schematics published and available. I would say their rule of thumb for capacitor size 1500 and 2000 uF per amp. These are 12 volt regulated power supplies.
I can not use a constant voltage power supply. I found some good engineering data that says not to worry about holding voltage a 3v often voltage needs to go up to get the current to go up also.

Carbon rods should have .3a per square inch of surface area. Surface area on my carbon rods in the salt water is slightly more than 1". 1x40=40 sq in. 40x.3=12amps. Data says my circuit needs to run at 12a if amps are low turn up the voltage until I get 12a. Circuit runs good at 12a 3.5 volts it produces too much chlorine gas to do this inside. This needs to be outside on the covered cement patio to run for 24 hours. I still need to note if variac over heats at 12a its rated 2.2kw. This circuit is 12a x 3.5v=42 watts. I need to learn amp rating on variac.

The microwave oven transformer will go on a shelf to wait for a different project.
 
  • #25
Averagesupernova said:
There is a range of capacitor size that needs to be based on allowable ripple.
-
I did a quick Google for Astron power supplies which have their schematics published and available. I would say their rule of thumb for capacitor size 1500 and 2000 uF per amp. These are 12 volt regulated power supplies.
Check out the link I gave in post #14. Which includes this, among other useful stuff:

1719638169730.png

These values are best designed, not chosen by someone else's rule.
Case in point: I'm still not sure why he needs a capacitor at all.
 
  • #26
DaveE said:
Check out the link I gave in post #14. Which includes this, among other useful stuff:

View attachment 347538
These values are best designed, not chosen by someone else's rule.
Case in point: I'm still not sure why he needs a capacitor at all.
This project needs DC current if I do nothing about ripple efficiency is 70.7%. If this project needs to run 2 days on flat line DC it will need to run 3 days on ripple DC. No flat line DC is not required but it saves time. Good example is, assume your driving along the highway at 70 mph and you keep turning the engine off 3 seconds of every 10 seconds.
 
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  • #27
DaveE said:
These values are best designed, not chosen by someone else's rule.
Case in point: I'm still not sure why he needs a capacitor at all.
DaveE said:
You are? Sorry, I still don't understand, which is OK, since it's your problem, not mine.
DaveE said:
Where? Which current?

I'm not sure exactly what you're doing, but why do you need those big capacitors? They're going to cause more heat in the wires and diodes because of the reduced conduction angle (high peak currents). My ancient car battery charger doesn't have them.
Cap the attitude. I don't understand the project either. Current through an electrolyte. Ok, fine, whatever the end goal is, that's what it is. I gave an example of what someone else has done. Technically so did you. Just about all design guidelines are 'someone else's rule'. We may need caps, we may not, we may need them big, maybe not.
 
  • #28
The main purpose of the shunts is to limit the output current of the transformer. They create a bypass route for the magnetic flux so that part of it does not go through both windings. This effectively adds inductance in series with the transformer (leakage inductance) which limits the output current.

Because part of the primary flux does not go through the secondary, open circuit output voltage should be reduced a little bit too. How much, not sure. Perhaps I should actually measure it.

The magnetron would take plenty of current if it were supplied from a "stiff" transformer with low leakage inductance. Enough to damage the magnetron or the transformer, or to blow the supply fuse.

If you remove the shunts, the output voltage will fluctuate less when loaded, and you can draw more current from it but there's a higher risk of damaging the transformer which is not dimensioned for full-time full-load use, an of blowing the fuse if you mess up something.



The output circuit of a classic microwave oven power supply is not a half-wave rectifier but a level shifter. See:

https://www.quora.com/What-is-the-p...he-capacitor-and-the-body-in-a-microwave-oven
 
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