# power supply issue

Discussion in 'Electronic Basics' started by Anders Nesheim Vinje, May 3, 2005.

1. ### Anders Nesheim VinjeGuest

Hi.
I have built this dual power supply.
Transformer rating: 2 x 24 V 60VA
Pretty standard with a bridge rectifier and ground connected to the center
tap. So i got + 24 and -24.
The i have two 4700uF 50v caps, one connected at each powerline to provide
filtering. Then there is a 7815 and and a 7915 as regulators.
Ok here is the issue. When i measure the voltage over the filtercaps i see
+37,5V and - 37,5V! How is this remotly possible?. Even the peak to peak
voltage isent that high. The 7815 and 7915 is only rated up to 35V so i am
scared they might blow or something.

Another problem is that since there is a voltage drop 37.5 -15 = 22 V over
the regulators. It cant provide much current before they shut down because
of the heat. I know from the datasheet they can handle up to 10W with a
decent heatsink and a max current of 1,5A. I dont need that much current
anyway since the transformer can only deliver 30VA / 24 = 1,25A per
secondery anyway.
I have another question too. How much current goes through the primary side
of the transformer with max load? I was thinking of having a fuse there

Anders Vinje

2. ### Anders Nesheim VinjeGuest

I took a new measurement and found out the the ac value was acutally 27V not
24. So when i muliply with the square of two i roughly get 37,5V peak to
peak.
But why do i see the peak to peak voltage after it has been rectified???

Anders

3. ### John FieldsGuest

---
With good reason.

Since each of the supplies is connected to the transformer as a
full-wave center-tapped supply, the rectifiers will only drop about
0.7V in front of the filters, so with 27VRMS into the rectifiers the
peak voltage (the voltage across each filter cap) will be:

Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V

Which is pretty close to what you got.
---

---
Since a transformer transfers power and the secondary can deliver:

P = IE = 1.25A * 24V = 30 watts

into a resistive load, then the primary will need to take 30 watts
from the mains in order to transfer that into the load.

Then, since:

P= IE

we can rearrange to solve for the primary current:

P 30W
I = --- = ------ = 0.25A
E 120V

for 120VRMS mains.

Assuming your transformer is 80% efficient means that the mains will
have to supply extra current to compensate for the transformer's
losses, so that means you'll need to draw about

0.25A
It = ------- ~ 0.31 amperes
0.8

from 120V mains.
---

---
Unless you're measuring the difference in voltage between the filter
caps you're not measuring peak-to-peak, you're measuring peak.

Since each of the supplies is connected to the transformer as a
full-wave center-tapped supply, the rectifiers will only drop about
0.7V in front of the filters, so with 27VRMS into the rectifiers the
peak voltage (the voltage across each filter cap) will be:

Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V

Which is pretty close to what you got.

You don't say whether the load is across the filter cap or across the
output of the regulator but, assuming it's across the filter cap, the
current through the resistor will be

E 37.7V
I = --- = ------- ~ 0.94 amperes
R 40R

This will cause the voltage across the filter to vary as it charges up
through the rectifier and discharge when the voltage out of the
rectifier falls below the voltage on the cap. This voltage change is
called 'ripple', and its amplitude varies with the load current, like
this:

I dT
dV = ------
C

where dV is the ripple voltage in volts,
I is the load current in amperes,
dt is the ripple period in seconds, and
C is the capacitance in farads

So,

0.94A * 8.3E-3s
dV = ---------------- = 1.66 volts
4.7E-3F

That means that that the caps could charge up to 37.7V, but that that
voltage would fall to about 36V when the rectifier output fell to
of those two voltages, about 36.8V, so you probably wouldn't even
intent on noticing it.

If the 40 ohm load was connected across the output of a regulator, the
current would be:

15V
I = ----- = 0.375A
40R

and the ripple voltage would fall to:

0.375A * 8.3E-3s
dV = ------------------ ~ 0.66 volts
4.7E-3F

so you'd probably be even less likely to notice that difference.

There are a few more problems to consider, but in order to do that we
need to know what your maximum load current is going to look like.

4. ### jgreimerGuest

Why are you seeing 27 V instead of 24 V at the output of the transformer?
Is your line voltage 27/24 * 120 = 135 V?

5. ### Anders Nesheim VinjeGuest

I see i need another transformer. Something like 2X 15V instead.
The reason i see 27 V is under no load. I think its 24 V when under max
I live in Norway and we got 220V 50HZ mains here.

Anders

6. ### BanGuest

That is strange, I think Norway has 230V like almost all west-European
countries. If the transformer is rated for 220V this will explain half the
overvoltage. The secondary voltage is indeed rated at max. AC-current. This
will explain the other half. Bigger transformers(M85b) have only 106% output  