• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

R + C what are they doing?

Status
Not open for further replies.

Scarr

Member
Hi Guys,

Came across this circuit and was not 100% sure (actually 10%!) what the 1u cap and 180ohm resistors are actually doing, I am sure you guys will know straight away.

Capture.JPG

L + N are 240v UK mains.

Thanks

Steve
 

schmitt trigger

Well-Known Member
This is the classic capacitive (transformerless) supply.
They have split the series cap in two branches. Good safety feature, in case one capacitor fails.
The series 180 ohm resistor is to limit inrush current.
 

tomizett

Active Member
Just to expand on that, in case you're not familiar, the principle is to produce a potential divider which drops the mains voltage to the required output voltage.
In a "normal" potential divider Vout = Vin * R2 / (R1+R2) where R2 is the resistor that is connected across the load - R1 and R2 appear in series across the supply.

In this design, R2 is substituted for the combined reactance of R39, R42, ZD6 and the load on the DC side of the bridge.... the clever bit is that R1 is substituted for a capacitor. Actually, in this case, it's substituted for the series combination of C32, C39, R36 and R47.

But why is a capacitor substituted for a resistor? Since it is an AC circuit with a known frequency, the capacitor has a known reactance, Xc=1/(2*pi*f*C) and so, as far as the "potential divider" action of the circuit is concerned, it behaves as a resistor of this value. Unlike a resistor, however, it does not dissipate any power. And that's why this circuit is often used - it is a simple way to drop voltage without dissipating an impractical amount of power.

Hope this helps!
 

audioguru

Well-Known Member
Most Helpful Member
A resistor voltage divider heats the resistors, they might need to be huge and need a fan to cool them. A capacitor voltage divider does not heat the capacitors.
 

Scarr

Member
Thanks whilst the math does go over my head a bit I understand the principle, it's basically I voltage divider using a cap to prevent massive power loss due to heat.

A little bit more history, the PCB's that matches the circuit are scrap PCB's I was salvaging parts from, I noticed what looks a common fault on several PCB's hence me getting interested in this deisgn, R36 or R37 have basically exploded, not both together, but on one board it will be R36 and on others it will be R37, I know this is a hard question that might be impossible to answer but what would be some common causes of these resistors exploding?

Steve
 

Scarr

Member
How strange, I'm am no guru (as you can tell) but it is exactly what I thought, so you guys will be able to say, based on what you see there is no inrush current protection on the circuit you can see? also what sort of current would cause this?

Steve
 

Colin

Active Member
That's what the resistors are doing. They are absorbing the inrush current. Maybe the device is being switched on and off a lot of times and the resistors have gradually taken the "hit" many times and gradually failed.
The actual working current is very low 35mA so they have not overheated over a period of time. So, the only conclusion is the inrush current. Maybe the capacitors are still charged when the device is turned on again and this will create excessive current. That's one reason why the caps have 100k across them - to discharge the cap. Not so you don't get a bite from the pins of the plug.
 

MikeMl

Well-Known Member
Most Helpful Member
...L + N are 240v UK mains.
Question about how the UK power distribution works: What is the voltage between L and the Ground symbol?

What is the voltage between N and the Ground symbol?
 

schmitt trigger

Well-Known Member
How strange, I'm am no guru (as you can tell) but it is exactly what I thought, so you guys will be able to say, based on what you see there is no inrush current protection on the circuit you can see? also what sort of current would cause this?

Steve
One question;
would you take a picture of those resistors?
 

tomizett

Active Member
Yea, would be interesting. They may just be undersize.... possibly replacing them with something beefier would get these boards working again and make them more reliable.
 

Scarr

Member
Sorry for the delay but I wanted to get the best shots for you and this took some doing :)

There was a comment of 35ma from Colin, is this the maximum safe current to draw from this design?
 

Attachments

audioguru

Well-Known Member
Most Helpful Member
Maybe you had lightning damage or maybe a capacitor is shorted (maybe from lightning)?
 

Scarr

Member
Hi Audioguru, I know it looks like very severe damage but I don't think these would all blow in the same way due to a strike would they? I think I favor the capacitor theory, so if a cap shorted would this cause this type of damage however I have tested the caps using ESR70 and they read 1u as expected. so if it's highly likely not lightning and not a cap short any other ideas?

Thanks

Steve
 

audioguru

Well-Known Member
Most Helpful Member
The peak of a 240V sinewave is 340V. When first turned on the capacitors are discharged then the current at 340V peak is 340V/(180 + 180)= 944mA. Then the momentary power in each 1W resistor is (944mA squared) x 180 ohms= 160.5W. Of course the little resistor will soon blow up if it gets 160.5W frequently.
 

Scarr

Member
Hi again,

Been a little busy but also been thinking, if this is inrush what is the best way to protect from this, a varistor? another replacement larger wattage resistor?

Thanks

[Edit]

Dam I'm now getting myself confused how can it be inrush if the resistor array including caps limit the current :-(

[Edit]
Looking at the pics and not being a expert by far, these pics look like a sudden shock not overheating, so with all your experience what causes a resistor to explode, voltage or current?
 
Last edited:

schmitt trigger

Well-Known Member
When the caps are fully discharged, and you switch the power at peak sinewave, the inrush current might be 10 or 20 times larger than average.
I believe you are correct, that a NTC thermistor, rated for inrush limiting duty, is the best way to go.

Vishay and many other vendors make them.
 

Scarr

Member
This is turning into a project :)

OK got a ERZV05D471 varistor, (PDF Here) the PCB is rated at 5A but I think this is just for fusing, I think the average consumption is more like 2.5A, anyway, does it look like the right varistor for the job?

Thanks
 

schmitt trigger

Well-Known Member
No
You are selecting a Varistor
You require an Inrush limiter NTC thermistor.

Something like these: https://www.digikey.com/products/en/circuit-protection/inrush-current-limiters-icl/151

Quote from a Cantherm datasheet:
Power NTC Thermistor. The MF72 series Power NTC Thermistors provide inrush current suppression for sensitive electronics. Connecting a MF72 in series with the power source will limit the current surges typically created at turn on. Once the circuit is energized the resistance of the MF72 will decrease rapidly to a very low value, power consumption can be ignored and there will be no effect on normal operating current. Using the MF72 Power NTC Thermistor is a most cost-effective way to curb surge current and protect sensitive electronics from damage.
 

Pommie

Well-Known Member
Most Helpful Member
Question about how the UK power distribution works: What is the voltage between L and the Ground symbol?

What is the voltage between N and the Ground symbol?
Power in the UK is 440V 3 phase which gives 240V phase to neutral. Actually, they have gradually reduced that - probably around 220V phase to neutral now.

Mike.
Edit, for clarity, phase to phase = 440V.
 
Status
Not open for further replies.

EE World Online Articles

Loading
Top