The best engineers know when to buy, not build.

The layout of SMPS, plus the cost of parts almost all of which will be SMD, is not something I'd recommend for a beginner. However a switcher is arguably more reliable than a linear regulator as its more efficient and generates less heat. At 300mA on 5V your 7805 is going to have to lose >2W (nearly 70% of your input power) as heat into your enclosure compared to <400mW for a pretty mediocre switcher. But if you insist, the 34063 is showing its age, its a nearly 20y old design. There are better, more efficient chips.. I'd go for an LT1076-5 in a TO220-5 lead case. Almost as easy to use and mount on PCB as a 7805. Datasheet attached.

But my recommendation is still to buy a module, solder some header pins on it and mount it on your PCB, just like any other component. You've got more than enough learning to do getting this up and running.

Re transformers, the first will give you about 700mA on load, the second about 1.8A.

 

The best engineers know when to buy, not build.

The layout of SMPS, plus the cost of parts almost all of which will be SMD, is not something I'd recommend for a beginner. However a switcher is arguably more reliable than a linear regulator as its more efficient and generates less heat. At 300mA on 5V your 7805 is going to have to lose >2W (nearly 70% of your input power) as heat into your enclosure compared to <400mW for a pretty mediocre switcher. But if you insist, the 34063 is showing its age, its a nearly 20y old design. There are better, more efficient chips.. I'd go for an LT1076-5 in a TO220-5 lead case. Almost as easy to use and mount on PCB as a 7805. Datasheet attached.

But my recommendation is still to buy a module, solder some header pins on it and mount it on your PCB, just like any other component. You've got more than enough learning to do getting this up and running.

Re transformers, the first will give you about 700mA on load, the second about 1.8A.

Understood, I will look for modules as well.
About the transformers are you talking about the 9V or 12V ones?

Thanks

 

Understood, I will look for modules as well.
About the transformers are you talking about the 9V or 12V ones?

Thanks

The TRIAD 12v ones you linked to... I think the first is more than enough as you're no longer powering the relays from 5v.

 

The TRIAD 12v ones you linked to... I think the first is more than enough as you're no longer powering the relays from 5v.

Ok, Thanks!

 

Designing a reliable and cost-effective switch-mode power supply is certainly a big deal, even for an experienced engineer. The challenges are not only the design of the circuit itself, but also the physical implementation of that circuit.
But for this project the question becomes one of how important is it to have the absolute maximum efficiency. 2 watts of wasted heat in a device with 200 watts of UV lights is not going to be a big deal.

 

Designing a reliable and cost-effective switch-mode power supply is certainly a big deal, even for an experienced engineer. The challenges are not only the design of the circuit itself, but also the physical implementation of that circuit.
But for this project the question becomes one of how important is it to have the absolute maximum efficiency. 2 watts of wasted heat in a device with 200 watts of UV lights is not going to be a big deal.

Yea, wasting heat is not too much of a concern but the heating and reliability, if the LM78xx are reliable enough, I can go back to my first iteration use a 7812 for relay and 7805 for mcu.

Thanks

 

Re transformers, the first will give you about 700mA on load, the second about 1.8A.

How did you calculate it? I want to be able to do it myself if I find another one
Also, the characteristics are @ 25 deg, but I guess my cabinet's temperature will be much higher than that, no problem huh? I remember you telling about MTBF, but have got no clue how to calculate it..

 

You probably don't need a regulator for the relays, they aren't critical for voltage. They'll be happy with anything from 10 - 15v.

The 7805 will get hot, but nothing a small clip-on heatsink can't handle.

Transformer calculations are fairly straightforward.

Your DC volts, off load, will be
AC volts x 1.4 - 2 x diode drop (use 0.7)

So DC v = 12 x 1.4 - 2 x. 7 =15.3v

Your available DC current is approx 0.7 x AC current

AC current isn't often stated, but the VA of the transformer usually is. AC current = VA/ACVolts

So 10VA transformer gives 10/12 = 0.83A AC = 0.56A DC.

The DC voltage and current depends on the size of the smoothing capacitor.

It gains charge every time the AC voltage exceeds the capacitor voltage + 2 diode drops. It loses charge while the AC volts is lower and the diode are off. To a first approximation the rate is DC current * [diode off time] ie I x dT and the voltage drop dV is roughly dV = I x dT/C, or the capacitance based on peak ripple is C = I x dt/(V * 2r) where r is % ripple, so for 0.5A, 10mS and 10% (=0.1),
C = .5 x 01/(12 * 0.2) = 2000uF.

 

You probably don't need a regulator for the relays, they aren't critical for voltage. They'll be happy with anything from 10 - 15v.

The 7805 will get hot, but nothing a small clip-on heatsink can't handle.

Transformer calculations are fairly straightforward.

Your DC volts, off load, will be
AC volts x 1.4 - 2 x diode drop (use 0.7)

So DC v = 12 x 1.4 - 2 x. 7 =15.3v

Your available DC current is approx 0.7 x AC current

AC current isn't often stated, but the VA of the transformer usually is. AC current = VA/ACVolts

So 10VA transformer gives 10/12 = 0.83A AC = 0.56A DC.

I'm clear with this part.

The DC vintage and current depends on the smoothing capacitor.

I don't understand this.

Also, the transformers characteristics are given @ 25 degs. How do I find the recommended operating range of it?
How to find out the output regulation i.e, output voltage on full load and no load?
There is no data in the datasheet about that.

 

Yea, wasting heat is not too much of a concern but the heating and reliability, if the LM78xx are reliable enough, I can go back to my first iteration use a 7812 for relay and 7805 for mcu.

Thanks

A supply onlyn used to operate relays and a pilot light really does not need to be well regulated. And if it only operates one relay the regulation is even less critical. A relay requires a higher voltage to pull in than to stay leld in, and so poor regulation can actually be a bit of a benefit.

 

A supply onlyn used to operate relays and a pilot light really does not need to be well regulated. And if it only operates one relay the regulation is even less critical. A relay requires a higher voltage to pull in than to stay leld in, and so poor regulation can actually be a bit of a benefit.

But the output can vary anywhere between 14-22v after rectification from the 12v transformer. But the max permissible voltage for OMRON G7L series is 110% of rate voltage, that's like 14v.

 

The voltage at the input capacitor of a full wave rectified supply will approach the peak value of the input voltage. Any current drawn will reduce that voltage towards the RMS value of the input volts. So just adding a pilot light will reduce the voltage quite a bit.
And the "max permissible" relays is a constant voltage, not a momentary input. So the small filter capacitor at the input will only need to keep the voltage at the minimum required by the regulator..
It sounds complicated but it really is not so complex.

 

The voltage at the input capacitor of a full wave rectified supply will approach the peak value of the input voltage. Any current drawn will reduce that voltage towards the RMS value of the input volts. So just adding a pilot light will reduce the voltage quite a bit.
And the "max permissible" relays is a constant voltage, not a momentary input. So the small filter capacitor at the input will only need to keep the voltage at the minimum required by the regulator..
It sounds complicated but it really is not so complex.

Thanks.
So, I should size the filter capacitor so that they keep the voltage at 9V(min for the 12v relay), am I correct?

 

I have been reading about using transistor as a switch. Can I use a high power Darlington Transistor as a switch to control the load?
They are very cheap compared to relays.

 

IGBT or Mosfet.
Max.

 

Thanks.
So, I should size the filter capacitor so that they keep the voltage at 9V(min for the 12v relay), am I correct?

An FET or a MOSFET work quite well for switching DC loads, but to switch an AC load with them becomes a bit more complicated. I am not sure how it could be simply done and still have isolation from the mains on the control side. There must be ways to do it, but I have not explored them.

And for the power supply capacitor, the actual value of capacitance needed also depends on the effective series resistance of the capacitor used, so while the math is not bad the unknowns are not known.

 

An FET or a MOSFET work quite well for switching DC loads, but to switch an AC load with them becomes a bit more complicated. I am not sure how it could be simply done and still have isolation from the mains on the control side. There must be ways to do it, but I have not explored them.

And for the power supply capacitor, the actual value of capacitance needed also depends on the effective series resistance of the capacitor used, so while the math is not bad the unknowns are not known.

You could use an opto isolated triac driver to drive a triac. I’ve done that before.

 

Given the reactive load, sizing a Triac and it's associated snubber network, is not something I'd recommend. And an SSR to meet this requirement is considerably more expensive than the relays. Plus triacs and ssr are not able to fully isolate the load from the supply. Stay with the relay, it's a simple and safe solution to the need.

 

I have been reading about using transistor as a switch. Can I use a high power Darlington Transistor as a switch to control the load?
They are very cheap compared to relays.

Not on AC at this level, only DC. There are ways, but only for low-level audio and RF, not power circuits.

 

Thanks.
So, I should size the filter capacitor so that they keep the voltage at 9V(min for the 12v relay), am I correct?

No, it needs to be above 9v by a margin because this is a first approximation. The value of 2000uF gives 10% ripple, or 2.4v p-p so approx 9.6 to 14.4v at the full 500mA load. It might even be a bit on the small side.