PIC Power Supply - Modifications

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Suraj143

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I built the below power supply & I got those problems.

1) R2 1K resister heats too much.
2)When relay is in OFF position voltage across capacitor is 25V. When relay is ON the voltage is 6V across the capacitor.
3)Relay won't turn ON with R5.

Help me to solve those issues.
 

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Please tell me about your relay. What is the coil resistance, voltage, current, part number? I need more information.

There are two problems.
The relay is pulling too much power.
The 25V across C2.

Where did you get this circuit?

You understand this is on the power line! You can get hurt.
C1 MUST be a high voltage capacitor! What are you using?
 
My relay is 12V & it takes 30mA current.(I tested the relay from a bench power supply & measured current using a DMM).

Yes C1 is 400V capacitor.
 
Problem 1:
The circuit was designed for 110 vac and you are using 220 vac.
Is this true?
 
I tried the circuit at 110Vac and the voltage on C1 is 13V with the relay open.
With 220Vac I get 26V on C1 with the relay open.

I think it was designed for 110Vac.

Where did you find this circuit?
 
Do you have a Zener? From 12V to 20V?
----edited------
I am using 400 ohms resistor for the relay.
Do you have more 1N4007 diodes? (1N4001 is also OK)
 
Last edited:
Oops sorry you are measuring some other capacitor (current limiter).

I am talking about C2 (220uF) capacitor.You need to measure that voltage.
 
My circuit is different!
1)It does not matter where R2 is. I changed.
2)You need more power to drive the relay! I changed from 2 diodes to 4. (half wave to full wave) This will give 2x more power to the relay.
3)More power makes the problem of too much voltage on C2 worst!
4)I added a 16V Zener on the relay supply to keep the voltage down when the relay is open. This might get hot.
5)R4 is the relay 400 ohms when closed and 400k when open. (my test)
6)You need a 0.1uf on the micro controller! I also added a 22uF.
7)I did not have a 5.1V Zener so I used 4.7V. Keep your 5.1V Zener.
8)When the relay is on there will be no current in the 16V Zener. When the relay is open the Zener will use up this power.
9)You can use low voltage diodes for the first 4 Diodes. The voltage is low. It is about the same is the relay supply.
10) I used 4k for the micro computer load.
 
I changed from 16V Zener to 12V Zener.
With relay on there is about 0 watts in Zener.
With relay off there is 1/2 watt in the Zener. (so the relay must use 1/2 watt)
So two 6V Zeners (in series) will each have 1/4 watt loss.
OR
You could use two of you 5.1V Zeners and a little resistance. (47 ohm to 100 ohms)
 
More simple.
I used two 12V Zeners inside the full wave bridge. (less parts)
The relay supply is running 12.3V
 
Xc(f) for 0.68uF @100Hz =~2k2 ohms
for 220uF = @ 100Hz = ~ 7 Ohms / duty cycle
your Relay using 30mA @12V = 400 Ohms

When Vout> 5.1V the load series R of 1K is now in parallel.

thus your overall impedance divider gives poor ripple and insufficent current.

With this arrangement with 230Vac you probably need to pump 300mA into the zener to get 30mA average out at 12V.
Not very well designed.

The Diode bridge will attempt to charge up 140% of average voltage before decaying.
The decay time constant , T=RC= 400*220uF= 7.3ms which at 100Hz is only 70% of the pulse interval when you should have at least 2T to 3T ( 5T=95%Vpeak)
Thus your e-cap is too small for the relay.
Since your decay T = 7.3ms /10ms interval the droop voltage will be more than (1-1/e)=62% if it were T=10ms thus your cap impedance will active most of the time (i.e. 80%) and . 7 Ohms /80% ~<10 OHms

Try to understand why you are wasting current with your Zener.

COnsider a Buck regulator offline or a better design as possibly suggested already.

Back to the drawing board.
 
Tony Stewart said:
Xc(f) for 0.68uF @100Hz =~2k2 ohms
Click to expand...
What is 100Hz?
The cap in the relay supply dose not need to have small ripple.
The cap on the micro supply does need to last 2T or 3T. I did not look at that.
Tony Stewart said:
Try to understand why you are wasting current with your Zener.
Click to expand...
I don't know what you are saying. This type of power supply needs a load. When the relay is open there needs to be a load to keep the voltages down.
I don't care about 1/2 watt loss.
A 220VAC buck converter will not be small/low cost.
 
Here is the power up and down with the relay on. The relay, if on, will hold to probably 6 volts. So that about 50mS. The micro will work to (I don't remember) 3 or 2.5V. If the relay is off this is much much longer.
 

Hi Ron,
100Hz is the rectified 230Vac/50Hz input
The relay will chatter with 80% ripple, but the 1K on the Zener with 7 Ohm cap only makes it worse as the 1K discharges the relay cap at a rate of 220uF*1k

Converting this into a buck regulator only means adding an LDO, comparator , FET & choke. expensive? I dont think so.
 
What 80% ripple? Look at post #15.
Xc(f) for 0.68uF @100Hz =~2k2 ohms
Click to expand...
The current through the 0.68 cap is 50/60hz not 100hz.
The LDO will not fix the voltage getting too high on the 220uF cap.
The Buck will make it even worse.
1k discharging cap
Click to expand...
Yes but that is what keeps the micro alive.
 
Last edited:
KMoffett said:
"7 Ohm cap" ???????

Ken
Click to expand...
Ken,
EDIT ( I was right in first place)

I mis-estimated impedance at 2f in my head using
Xc=1/2pifC , it should be 1/0.130 =7.2Ω , ESR ought to be ~1 Ohm typ. unless low ESR. type.
This 220uF forms an impedance divider with the plastic cap @0.68uF from the line when a diode conducts and determines partly the DC voltage on the cap. The sag is controlled by the load when the diode is off.

I was referring to Original unregulated Voltage from rectified 50Hz on the original schematic when the Vdc dropped to 6V with relay on. 2.6% of the AC RMS input. This is a "full bridge"

Of course the rectified AC is 100Hz with harmonics that reduce Xc by 2 for each harmonic from 200,300,400 etc.

Correction:
The 2 diodes in the OP's design is a voltage doubler with the shunt diode across the cap & (if) with no load R's achieves 650Vdc

But Ron's 4 diode bridge is not a voltage doubler but doubles the current vs above., i.e. no load R's achieves 325Vdc from 230Vac.
 
Last edited:
ronsimpson said:
What 80% ripple? Look at post #15.

The current through the 0.68 cap is 50/60hz not 100hz.
The LDO will not fix the voltage getting too high on the 220uF cap.
The Buck will make it even worse.

Yes but that is what keeps the micro alive.
Click to expand...
All full-wave rectifiers on AC double the fundamental frequency, without exception ( unless you have active PFC) .... that goes for Voltage and Cap Current here.

The LDO gives a bandgap ref for a simple buck regulator.
A Buck regulator does not use a 75Ohm series resistor which , of course , might make it worse
 
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