Overheating regulator

[Robbie Banks]

Hello all,

Can anyone shed any light on the following. I have a 12V regulator after a
bridge rectifier and transformer from mains. The regulator supplies a 12V
supply to a pulsed source, pulling about 500mA. The regulator is rated to
2A, yet when operating the regulator is hot enough to cook your
marshmallows. I'm trying to understand why. Any ideas?

--
Robbie Banks

 

[Check It Out]

Uh, what is the input voltage to the Regulator? Probably exceeding the
wattage spec.

"Robbie Banks" <robbie_banks@nospamhotmail.com> wrote in message
news:c3elkh$8a9$1@titan.btinternet.com...

Hello all,

Can anyone shed any light on the following. I have a 12V regulator after a
bridge rectifier and transformer from mains. The regulator supplies a 12V
supply to a pulsed source, pulling about 500mA. The regulator is rated to
2A, yet when operating the regulator is hot enough to cook your
marshmallows. I'm trying to understand why. Any ideas?

--
Robbie Banks

 

[Richard Rasker]

On Fri, 19 Mar 2004 11:30:25 +0000, Robbie Banks wrote:

Hello all,

Can anyone shed any light on the following. I have a 12V regulator after
a bridge rectifier and transformer from mains. The regulator supplies a
12V supply to a pulsed source, pulling about 500mA. The regulator is
rated to 2A, yet when operating the regulator is hot enough to cook your
marshmallows. I'm trying to understand why. Any ideas?

Your information is rather sparse, but I gather that you failed to take
the power loss in the regulator into account. A regulator must not only be
capable of handling the output current, but also the power generated in
the regulator itself. Do you have any experience in the field of
electronics? If not, read on.


Your circuit probably looks somewhat like this:

regulator
.------------. +Vin .-------. +12V out .------.
| Transfomer |------| |------o----| |
| Rectifier | `-------' | Load |
| Capacitor | | gnd | |
| |---------------------o----| |
`------------' `------'

Now what happens, is that Vin > 12V (otherwise the regulator obviously
wouldn't work). Vout = 12V, thus there is a voltage of (Vin - 12) across
the input and output terminals of the regulator. This voltage difference
must be multiplied by the current through the regulator (500mA) to
calculate the power loss in the regulator (the power developed in any
element is equal to the voltage across is, multiplied by the current
through it (at least for DC that is).

Let's assume you have a 12V mains transformer and a sufficiently large
filter capacitor (say, 4700uF*). The secondary AC voltage is 12V; across
the filter cap, you get 12 * sqrt(2) = 17 volts (the amplitude value of
the AC voltage). No, wait, let's take the voltage drop across the
rectifier into account: 17 - (2 * 0.7) = 15.6 volts.

So, Vin = 15.6 volts, Vout = 12 volts, and the current = 0.5 A. The power
loss in the regulator then is (15.6 - 12) * 0.5 = 1.8 watts.
If you're using a TO220 housed regulator (e.g. the 78S12) without a heat
sink, this is indeed enough to heat it up significantly.

As you can see, the power loss increases rapidly with higher values of
Vin; also, you can't use a lower Vin, or the regulator won't work
properly (Vin must at any time be at least Vout + 2V). I recommend
mounting your regulator onto a heat sink.

*): Actually, the capacitor value also plays an important role in the
calculations for these circuits, most particular with relation to the
so-called ripple voltage. With 4700uF and 0.5A load, the ripple voltage is
about 1 volt, which I didn't take into condiration here - otherwise this
explanation would double again in length.


Regards,

Richard Rasker

--
Linetec Translation and Technology Services

http://www.linetec.nl/

 

[Boris Mohar]

On Fri, 19 Mar 2004 11:30:25 +0000 (UTC), "Robbie Banks"
<robbie_banks@nospamhotmail.com> wrote:

Hello all,

Can anyone shed any light on the following. I have a 12V regulator after a
bridge rectifier and transformer from mains. The regulator supplies a 12V
supply to a pulsed source, pulling about 500mA. The regulator is rated to
2A, yet when operating the regulator is hot enough to cook your
marshmallows. I'm trying to understand why. Any ideas?

Would that regulator have a part number by any chance? What is the input
voltage? Are you using a heat sink?

--

Boris Mohar

 

[Robbie Banks]

ichard Rasker" <spamtrap@linetec.nl> wrote in message
newsan.2004.03.19.12.11.19.437684@linetec.nl...

On Fri, 19 Mar 2004 11:30:25 +0000, Robbie Banks wrote:

Hello all,

Can anyone shed any light on the following. I have a 12V regulator after
a bridge rectifier and transformer from mains. The regulator supplies a
12V supply to a pulsed source, pulling about 500mA. The regulator is
rated to 2A, yet when operating the regulator is hot enough to cook your
marshmallows. I'm trying to understand why. Any ideas?

Your information is rather sparse, but I gather that you failed to take
the power loss in the regulator into account. A regulator must not only be
capable of handling the output current, but also the power generated in
the regulator itself. Do you have any experience in the field of
electronics? If not, read on.


Your circuit probably looks somewhat like this:

regulator
.------------. +Vin .-------. +12V out .------.
| Transfomer |------| |------o----| |
| Rectifier | `-------' | Load |
| Capacitor | | gnd | |
| |---------------------o----| |
`------------' `------'

snipped

Cheers,

Turns out it was two problems, 1) ripple from AC source and the load it was
driving
The load was an series of LED's that were being driven with a duty cycle.
Although I mentioned that the load current was only 500mA (this was pulsed),
the load would dissapate more current if driven continuous. So, basically
the load being drawn was spiking higher than 2A, whereas the 500mA was an
average load.

Don't want to use a heat sink if possible as all the electronics are to be
potted.

Solution, uprated the reg and changed the input capacitor.

No more problem

Cheers for the feedback