[BlazerFRS]

I have a basic electronics question I'd like to put by some of the resident guru's here.

Based on my desire to learm more about electronics design and tangent's review I recently purchased "The Art of Electronics" and I'm very impressed so far. I'm about halfway through chapter two, and I've hit a bit of a stumbling block. How do you calculate the dissipation of a resistor/transister/zener diode in a circuit?

I understand P=I^2*R and P=V^2/R, and where they come from. I have no problem calculating the dissipation of a lone component, but I don't know who to apply it to a circuit, like the voltage regulator in exercices 2.3 and 2.4 (pg 68 and 69 respectivly).

I feel I must be missing something simple.

Thanks for any help
-Blazerfrs

 

[guzzler]

It's essentially the same thing. You take the voltage drop across a component, say a regulator, and the current you intend to draw and proceed. For things like regulators, you can get a little complicated when factoring in the efficiency of the device, but the simple approach will give you a ball park figure. This doesn't give you the temperature rise of the device, that's in the thermal properties table for the device, but gives you an idea of heatsinking requirements etc

g

 

[jcx]

you want to look to applying the more fundamental relation:

P = I*V

since for resistors: V = I*R, I = V/R you can just substitute into P = I * V to get the resistor Power eqs

 

[BlazerFRS]

What's the best way to figure out the voltage drop across a component? Using the Voltage divider equation I suppose?

 

[nikongod]

Quote:

Originally Posted by BlazerFRS What's the best way to figure out the voltage drop across a component? Using the Voltage divider equation I suppose?

voltage in-voltage out.

a 5 V regulator on a 12V power line has a 7V drop.

 

[dsavitsk]

A sort of related question ...

Suppose I am using a 10V regulated PS for an amp. Further suppose that the amp uses 1A of current at 10V. So, in some sense, this requires a transformer that can supply ~ 10VAC.

Assume that I am using a 35V transformer (let's say 35V after the regulator to keep it simple). Since the regulator is dissipating ~ 25V * 1A as heat, does this mean that I am actually pulling 35 watts through the transformer? That is, rather than a 10VAC transformer, I would actually need a 35VAC?

 

[Garbz]

not quite.
It depends if the power remains constant. If you pull 1A from a 10V transformer you pull 10 watt. Now if you feed that 35V the same circuit only consumes .35A

That said you pull 1A through 35V you will defaintly need a 35VA transformer.

A practical example of everything in the power desipation in this thread is my upgrade to a balanced amp.

The balanced amp goes from 2 channels to 4 channels. Each channel is identical. So assuming that 35V p-p voltage draw on the powersupply at 200mA because P = IV there will be a 7Watt draw on the powersupply.

Now if I add a second board I look at what remains constant and what changes. There are now 4 channels powered by 35V in parallel. This means same voltage and change current draw. 2x200mA = 400mA. 400mA @ 35v = 14watt. That's the draw on the powersupply.

If I substitute the 35V transformer witha 50v transformer I still need 35v to be delievered to the boards. All of a sudden there's a massive voltage drop of 15v + whatever the PSU was originally doing. So an additional 15v * 400mA = 6W of power needs to be decipated accross the powersupply components.

That's basically how it works as per my understanding.

I think the best way to understand the formulars is to digest a physics textbook. http://www.lightandmatter.com/area1.html has downloadable physics texts and book 4 has 2 chapteres on circuits.

 

[dsavitsk]

Right, but what I am saying is that with a PS that uses a LM317 to regulate voltage to say 10V, a 35V/1A transformer will provide the same current to a circuit as a 10V/1A transformer, but will draw an extra 25 watts to be dissipated as heat. Maybe this was obvious to everyone else, but I had not thought about it before.

 

[Garbz]

ahhh now I understand. In this case you need to think of things in terms of power before the PSU. Current is constant after the regulators and additional voltage is burnt as heat. I'm actually not sure about the answer to the question but considering the extra power is burnt as heat... this is much easier when there is a passive resitive element. I can't commit to an answer to your question. But I THINK that given a 10V 10VA transformer gets changed to 35V the VA rating would have to go up to 35VA as well.

The easiest way around it is just buy a huge transformer the extra current won't go anywhere

 

[BlazerFRS]

For the regulator example, if I understand things correctly, to find watts dissipated in the regulator you'd take the drop across the regulator, and multiply by the current consumed by the regultor(Iadj in the feedback loop); not but the total draw of the amp it was powering.

At least that's what I always thought; I'm probably completely off base.

 

[mono]

Quote:

Originally Posted by dsavitsk Right, but what I am saying is that with a PS that uses a LM317 to regulate voltage to say 10V, a 35V/1A transformer will provide the same current to a circuit as a 10V/1A transformer, but will draw an extra 25 watts to be dissipated as heat. Maybe this was obvious to everyone else, but I had not thought about it before.

That's not entirely true. A transformer does derive it's voltage from a turns-ratio between primary and secondary, BUT a transformer also floats high depending on load, and the larger transformer has more core field strength. With the 35V transformer, you would be able to get a higher amperage at 10V than with the 10V transformer. That's not necessarily a reason to use a 35V transformer though, that's too far an overshot for most practical purposes even if you had it lying around instead of having to buy it.

 

[mono]

Quote:

Originally Posted by BlazerFRS For the regulator example, if I understand things correctly, to find watts dissipated in the regulator you'd take the drop across the regulator, and multiply by the current consumed by the regultor(Iadj in the feedback loop); not but the total draw of the amp it was powering. At least that's what I always thought; I'm probably completely off base.

Not current of feedback but of the powered device. Drop a 24V pre-regulator supply to 18V out post-regulated and 1A current used by powered device, it'll be near 6W. If it were only the current in the feedback loop we'd never need a heatsink.

HOWEVER, the pre-regulated supply has to be measured while the load (powered device) is operating to determine the pre-regulated voltage, since a leser current draw will be progressively unloading the transformer, so it's voltage would rise.

 

[BlazerFRS]

Ok- that makes sense. Thanks for clearing that one up, i'm just a bit on the slow side lately...