# Electrical Circuit Help needed

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#1
A customer has requested a recommendation for the lowest possible cost power supply circuit with the following specification, and in the format of a written report:

1. Vin = 15 Vdc

2. Vout = 3.3 Vdc at 300 mA

3. An LED to indicate that the power supply is supplying 3.3 V to the load

Produce a report that details your recommended solution.

The report should include the following information:

- An introduction and specification

- A fully labelled circuit diagram, including showing all component values and part numbers

- Full working out to show the Ohms Law calculations for all resistors, the currents through all of the components, and the power dissipated by all of the components

- A diagram to show how you would test and measure the circuit, to check that it is working to the customer specification.

- A parts list including basic component spec, RS Components part number, component prices and total price for all circuit components

Imagine that you want to impress your customer so that they give you the contract to manufacture the circuits.

I don't know how to make the circuit. Would be helpful if you could tell me step by step how the circuit would look, as im using online circuit makers to test it and mine is not working.

Thanks
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4 months ago
#2
Sure I've seen this exact same question 3 times, Where's it from?

The problem within the problem is shown by a fag-packet calculation... the required circuit has to drop 11.7 V and deliver 300mA, if it does that in a naive way it's going to need to dissipate 3.5W as heat in order to deliver 1W of electrical power to the load.

That's energy inefficient and could be an issue for a mass market product when the green lobby get their teeth into it and demand regulation... it's also enough power to turn small components into smoke so you might have to consider heatsinking the component(s) dissipating the power to keep them in their safe temperature range... and heatsinking is an additional upfront cost.

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#3
(Original post by Joinedup)
Sure I've seen this exact same question 3 times, Where's it from?

The problem within the problem is shown by a fag-packet calculation... the required circuit has to drop 11.7 V and deliver 300mA, if it does that in a naive way it's going to need to dissipate 3.5W as heat in order to deliver 1W of electrical power to the load.

That's energy inefficient and could be an issue for a mass market product when the green lobby get their teeth into it and demand regulation... it's also enough power to turn small components into smoke so you might have to consider heatsinking the component(s) dissipating the power to keep them in their safe temperature range... and heatsinking is an additional upfront cost.

Thank you for the reply. It is from university, from the electronics module.
I thought the circuit was meant to look like this (in the picture) but its just the number of the resistors and all the calculations im not understanding.
Thanks
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#4
Hi,

If i had a total resistance of 39ohms in a circuit, with 300ma current, 3.3v output for LED, 15v input. There is also a zener diode and regular diode in circuit.

How would i figure out how to split the value of 39ohms in between the amount of resistors i may need?

Thanks
0
4 months ago
#5
Start by sketching something (anything) with those components and a couple of unknown resistors in it by way of scene setting, and then define where you apply a voltage and where you measure it. That will give you some context for a proper discussion and / or some simple calculations to determine how the resistors should be arranged.
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#6
Thanks for the reply. This is the sketch i have done.
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4 months ago
#7
(Original post by UniHelpNeeded)
Thanks for the reply. This is the sketch i have done.
R1, R2, R3 are in series in your sketch, that doesn't really help you, it just over complicates things, so replace them with a single resistor. You also said that there is a “regular diode” in the circuit as well as the LED and Zener. Is the forward bias voltage drop across the diode or LED specified as that can help set up the next bit of the problem.
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#8
I thought i couldnt replace it with one resistor as it would burn out?
And im not too sure. This is the most information i know- the values and that circuit ive tried to draw.

Thanks
0
4 months ago
#9
(Original post by UniHelpNeeded)
Thank you for the reply. It is from university, from the electronics module.
I thought the circuit was meant to look like this (in the picture) but its just the number of the resistors and all the calculations im not understanding.
Thanks
What degree course are you doing?
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4 months ago
#10
(Original post by UniHelpNeeded)
I thought i couldnt replace it with one resistor as it would burn out?
And im not too sure. This is the most information i know- the values and that circuit ive tried to draw.

Thanks

The idea behind using several series resistors here is to spread the power dissipation across multiple components, for example by using 3 x 1W resistors of equal resistance in series to dissipate 3W total.
to get the individual resistance values you will have to calculate the total resistance you require and divide it by the number of series resistors you need to cope with the power.
in the real world resistors come in standard resistance values if you can't reach the resistance you require using 3 series resistors you'll probably have to think about using 4 resistors of 1/4 the required total resistance each. e.g. 39 ohms total = 4x10 ohm resistors (only out by 2.5% which probably doesn't matter - cheap 1W resistors are 5% tolerance anyway)

Resistors all have a maximum power rating which is a separate parameter to their resistance value - but tbh if you run a 1W resistor at the full 1W constantly it'll get really damn hot (hot enough to burn any skin that touches it) and possibly start to char the circuit board over time too shortening the life of your product.

you can get high power resistors but they start to get really expensive

but you should really look at integrated circuit linear voltage regulators - they are not expensive and have overcurrent protection built in. if someone shorted the output of your circuit with a screwdriver, those resistors would go up in smoke but a linear regulator would just shut down safely.
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#11
(Original post by uberteknik)
What degree course are you doing?
General engineering, and then the next year you get to choose your specific choice in engineering. This is for the electronics module.
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#12
(Original post by Joinedup)
The idea behind using several series resistors here is to spread the power dissipation across multiple components, for example by using 3 x 1W resistors of equal resistance in series to dissipate 3W total.
to get the individual resistance values you will have to calculate the total resistance you require and divide it by the number of series resistors you need to cope with the power.
in the real world resistors come in standard resistance values if you can't reach the resistance you require using 3 series resistors you'll probably have to think about using 4 resistors of 1/4 the required total resistance each. e.g. 39 ohms total = 4x10 ohm resistors (only out by 2.5% which probably doesn't matter - cheap 1W resistors are 5% tolerance anyway)

Resistors all have a maximum power rating which is a separate parameter to their resistance value - but tbh if you run a 1W resistor at the full 1W constantly it'll get really damn hot (hot enough to burn any skin that touches it) and possibly start to char the circuit board over time too shortening the life of your product.

you can get high power resistors but they start to get really expensive

but you should really look at integrated circuit linear voltage regulators - they are not expensive and have overcurrent protection built in. if someone shorted the output of your circuit with a screwdriver, those resistors would go up in smoke but a linear regulator would just shut down safely.
so i did these calculations altogether.

total resistance = 15- 3.3. = 11.7 11.7/0.3= 39 ohms

power= 11.7 * 0.3 = 3.51W

Load= 3.3 * 0.3 = 0.99W

Efficiency = (3.3 * 0.3) / (15 * 0.3) = 0.22 = 22%

Are you suggesting to use 4 resistors and split the resistance equally between them? And how about the placement in the circuit? As there is also a diode and zener diode.

Thank you
0
4 months ago
#13
(Original post by UniHelpNeeded)
so i did these calculations altogether.

total resistance = 15- 3.3. = 11.7 11.7/0.3= 39 ohms

power= 11.7 * 0.3 = 3.51W

Load= 3.3 * 0.3 = 0.99W

Efficiency = (3.3 * 0.3) / (15 * 0.3) = 0.22 = 22%

Are you suggesting to use 4 resistors and split the resistance equally between them? And how about the placement in the circuit? As there is also a diode and zener diode.

Thank you
Well you could use one resistor if it had an adequate power rating - you will see on the RS website that resistors are available in a wide range of power ratings.

as I've said on the other thread I think that circuit is horrible in this application and you'd be better off borrowing a circuit from the data sheet of a linear voltage regulator like a LM317. The LM317 and similar ICs were designed by geniuses and there's no way anyone could possibly build a better voltage regulator out of discrete components for anything like the same cost, is there some reason why you need to use that zener circuit?

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