Making prototype PSU for Microcontroller board[SOLVED]

Thread Starter

Pushkar1

Joined Apr 5, 2021
416
@Irving , @bertus I came back with new ideas after doing some research.

I've designed below power supply unit. if I get good feedback from experts. I will reserve this PSU for future work. I will buy all the components and make a prototype circuit board as soon as possible

1626257294678.png

List of materials

  • 5mm Dc power jack
  • Small Power supply on-off Switch SPDT
  • W04m Single bridge rectifier 400v, 1.5A
  • 1000uf 25v Electrolytic Capacitor
  • 0.33 uf ceramic capacitor
  • 5V LM7805 Voltage Regulator
  • TO-220 Heatsink
  • 0.1 uf ceramic capacitor
  • 2K2 resistor
  1. Power indicator LED

I've tried to keep clean diagram as I can. Waiting for the feedbacks
 
Last edited:

Irving

Joined Jan 30, 2016
3,843
  • Switch in diagram is DPDT, in BOM is SPDT. Needs to be rated 30vDC/3A if DC input, AC rating irrelevant.
  • 0.33u capacitor in BOM is not on schematic, though C2 1u on schematic not in BOM. C2, C3 on schematic shown as polarised when not if ceramic caps.
  • 2k2 resistor gives 3.5mA in LED on 12v DC input allowing for drop in bridge rect. Will probably be OK for brightness.
  • TO220 heatsink - what type/rating? Is it sized correctly?
  • >9vAC will severely stress the 7805.
 

Thread Starter

Pushkar1

Joined Apr 5, 2021
416
What heatsink do you have? Can you post a picture?
I've attached picture of heatsink. Nothing is written on the heatsink like model number ratings. So It is very hard to tell exactly it's ratings?

Update : Image size now fixed
 

Attachments

Last edited:

ericgibbs

Joined Jan 29, 2010
18,766
hi P1,
When posting images of objects, it helps if you have a known size object on the photo as well as the object.
Some use a Coin others a 12inch rule.
We can then estimate the size of the object.
E
 

Thread Starter

Pushkar1

Joined Apr 5, 2021
416
hi P1,
When posting images of objects, it helps if you have a known size object on the photo as well as the object.
Some use a Coin others a 12inch rule.
We can then estimate the size of the object.
E
sorry I had uploaded image from my smart phone now fixed
 

Thread Starter

Pushkar1

Joined Apr 5, 2021
416
  • Switch in diagram is DPDT, in BOM is SPDT. Needs to be rated 30vDC/3A if DC input, AC rating irrelevant.
  • 0.33u capacitor in BOM is not on schematic, though C2 1u on schematic not in BOM. C2, C3 on schematic shown as polarised when not if ceramic caps.
  • 2k2 resistor gives 3.5mA in LED on 12v DC input allowing for drop in bridge rect. Will probably be OK for brightness.
  • TO220 heatsink - what type/rating? Is it sized correctly?
  • >9vAC will severely stress the 7805.
Here is revised circuit diagram

1626265406642.png
 

AnalogKid

Joined Aug 1, 2013
10,986
With a 9 Vac input, at any current greater than approx. 0.5 A the output will pull out of regulation, sagging below 5 V 120 time per second.

Because the power supply load is driven by a constant-voltage regulator, the combination of the load and regulator appear as a constant-current load to the filter capacitor. For example, if the 5 V load current is 1 A, it stays at 1 A whether the DC input voltage to the 7805 is 7 V or 24 V. Thus, the current through the regulator is a constant 1 A regardless of the supply voltage, the definition of a constant-current load. This makes calculating the bulk capacitor ripple voltage much easier, because there are no exponential terms in the equation.

This is the full-blown equation that relates the voltage across a capacitor to the current through it:

1626272055987.png
The good news is that if the current (I) is a constant and the initial voltage (V0) is 0 V, it reduces to:

EC=it

E x C = i x t

Change in capacitor voltage (in volts) times the capacitor value (in farads) equals the constant current (source or load) value (in amperes) times the (charge or discharge) time period (in seconds)

In the US, the power line frequency is 60 Hz, so the time period for a full-wave rectified power source is 8.33 ms.

The capacitor is 0.001 F (1000 uF)

A 9 Vac source has a peak value of 12.72 V.
Minus two diode Vf drops equals 11.3 V.
Minus the 7805 minimum input voltage value of 7 V equals 4.3 V

Solving for i:

i = (E x C) / t

= (4.3 x 0.001) / 0.00833

= 0.516 A

Thus, the negative peaks of the capacitor ripple voltage will drop below 7 V when the output current is greater than 0.5 A. Because of the sinusoidal shape of the input voltage waveform, the actual time period is a bit less than 8.33 ms. This increases the actual max current value a small amount.

ak
 
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Irving

Joined Jan 30, 2016
3,843
hi P1,
When posting images of objects, it helps if you have a known size object on the photo as well as the object.
Some use a Coin others a 12inch rule.
We can then estimate the size of the object.
E
This is true, however, knowing this is a TO-220 heatsink its not hard to guess the dimensions well enough to find a catalog entry that matches...

@Pushkar1 That heatsink, mounted vertically has a thermal resistance of about 18degC/W. I'm guessing you don't know much about heatsinks so here's a quick tutorial:

The purpose of a heatsink is to help move heat from the device's active junction internally to the surrounding air. On each step of that journey the heat encounters some resistance, just like current does in a circuit. In an electrical circuit the flow of energy is current, driven by the voltage drop across the circuit elements. A thermal circuit is similar, the current becomes heat flow and voltage drop becomes temperature drop. So here is a model of a device on a heatsink:

1626269840594.png

Each 'resistor' represents a change in temperature (in Kelvin K) depending on the amount of 'heat' to be transferred (in joules per second or Watts W).
  • R1 represents the thermal resistance from junction to case, as given in the datasheet;
  • R2 the thermal resistance of the interface between the case and the heatsink surface using some thermal paste (without paste it will be a much bigger number); and
  • R3 the thermal resistance of the heatsink to the surrounding air through natural convection/radiation.

So, for example, if we start at the bottom, with worst case air temperature of 40degC, then the heatsink surface will be 18K above ambient for each Watt of power loss in the device.

Just like resistors in series in an electrical circuit we can add the thermal resistances together to get the total thermal resistance between the air and the junction. So adding all together the resistance is 20.2K/W and the junction temperature T(j) will therefore be 20.2degC above ambient for each Watt of power loss in the device.

The maximum temperature for the junction is given in the datasheet. For this device its 125degC, but you should never run a device at the maximum. For safety 80% is considered worst case, so 100degC is what we'll use. Since we know the maximum junction temperature is 100degC we can calculate the maximum wattage as:

\( W_{max} = (T_{(jmax)} - T_{(amb)})/R = (100 - 40)/20.2 = 2.97W \)

That means that at 1A the maximum voltage drop across the 7805 is 3v, since 3v * 1A = 3W. Alternately, at 12v DC input, assuming 1v drop across each diode in the bridge rectifier, the drop is (12 - 2 * 1 - 5) = 5v so the maximum current draw with that heatsink is 3W/5v = 600mA. And with 12v AC input it will be approx (15 - 5) = 10v, so max current of 3W/10v = 300mA.

So, your options:
1. fan cool the existing heatsink, though given its small size its unlikely a fan will improve it much, maybe reduce it from 18K/W to 12K/W which isn't much help...

2. A much bigger heatsink; To be safe under all conditions it would need to be better than (100 - 40)/10 = 6K/W total, or
6 - 2.2 = 3.8K/W for the heatsink alone.
 
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Thread Starter

Pushkar1

Joined Apr 5, 2021
416
Wouldn't it be a good idea to set a variable current limit option in the PSU. It will not supply more than the safe limit. so as not to harm system. The PSU will shut down the output if the current draw goes above the limit and therefore protect system.

So for the variable current limit, should I add a potentiometer to the output of the voltage regulator?
 

Irving

Joined Jan 30, 2016
3,843
On the face of it, yes, but the datasheet tells a different story... it says its 2.6K/W for a 75degC rise with natural convection. That implies a 29W input (75/2.6), so on a 10W input the rise should be 2.6*10 = 26degC... but look at the curves... for a 10W input the temp rise is given as 40degC ie 4K/W. That suggests the fin design is not so efficient at lower wattages.
1626276968154.png
 

Irving

Joined Jan 30, 2016
3,843
Wouldn't it be a good idea to set a variable current limit option in the PSU. It will not supply more than the safe limit. so as not to harm system. The PSU will shut down the output if the current draw goes above the limit and therefore protect system.

So for the variable current limit, should I add a potentiometer to the output of the voltage regulator?
I've said it 3 times... you can't do current limiting with just a passive device. Where are you going to put it? In series? What happens to the voltage?

1626279698576.png
 

Irving

Joined Jan 30, 2016
3,843
Here's one way to do current limiting at the input to the 7805. The box on the output of the 7805 is an active-load that allows me to sweep the output current from 0 - 1A.

I'm not suggesting you build this, but you might find it instructive to work out how it works.


1626289323621.png
 
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Thread Starter

Pushkar1

Joined Apr 5, 2021
416
I'll have to stop here for now. I went to the store to buy components. But I am not getting the components shown in the circuit.

I can order components online but it will be very expensive which includes delivery charges for each components. Maybe I can buy two buses for that much money.

I have two options , I can buy a new PSU, I can work with what is available with me. If I buy new PSU I won't be able to learn much about PSU circuits. If I work with what I have available, then I need to compromise on the specification. The advantage of this would be that I would get practical experience for the power supply and if a problem occurs in real time, I could also learn to solve it. I don't even waste my money in this, if I do any mistake.

look forward to hearing suggestions from you @Irving
 
There is a really nice power supply on ebay that you can set voltage and current with a potentiometer. I used it to make a universal" wall wart and it became permanent for a 6V one. It can be set into a translucent box, so the display is visible and, I think, they added a blanking input now.

So, I put 5.5/2.1 and 5.5/2.5 connectors on the input and the radio-shack type of cord on the output. I just wish the polarity was locking. A 12V wall wart can get me anything less than 9V. I need 19V for a router, so a 24 V supply works for that.
 

Irving

Joined Jan 30, 2016
3,843
As I said previously, 500mA at 5v will comfortably drive most simple MCU projects - its the same output as a USB port. Live with that compromise, learn what you can and look out for other equipment you can harvest parts from... it need not cost a lot.

Buying from Mouser, etc. can be expensive because of the delivery charges. eBay can be a good source of cheap parts...
 

MrChips

Joined Oct 2, 2009
30,706
I would not worry about current limiting an LM7805 voltage regulator.
It already has built-in short circuit protection. You can short the outputs and there will be no harm done.
 
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