Design Critique: Using PIC16F690

Motardo

Joined Sep 21, 2011
22
With 2k2 base resistors, you've only got 2 mA or so base current into the 2222s, so you should keep the collector current less than 20 mA or so to keep the transistors in saturation. But you've got 17 mA through the coils, plus about 24 mA through the LED's? That seems like a lot for the LED's. I would drop it to about 5 mA (4k7 resistors in series) if you don't need them super bright. Or else keep the LED's how they are and use 1k base resistors to increase the drive.
 

chimera

Joined Oct 21, 2010
122
Hello,

I was wondering if someone could critique my design if they can see any major flaws or a better solution.

Thanks!!!
okay.. the micro part looks okay. reduce the LED current limiting resistors,
but I am more concerned with your power supply.

1- You are using a 24Vrms transformer.
2- You have no smoothing cap on it (other than the one in your LM7805 circuit)
3- The input voltage to the LM7805 is going to around 33V (rectified and smoothed)
4- That means that your power dissipation across the poor regulator is going to be huge. You are going to be needing an incredibly huge heat sink.
5- Recommendation: Use a switching buck regulator IC to drop the voltage down to 10 or so volts (LM2576-adj) and then use a 7805. This way, no heat sink (or a small one). http://www.ebay.com/itm/LM2576S-Adj...t=LH_DefaultDomain_0&var=&hash=item231b959833

Enjoy!
 

nerdegutta

Joined Dec 15, 2009
2,684
Hi,

All of the above and:

  • Add a capacitor near the IC's VDD pin, to GND.
  • Consider switch bouncing in you code.
  • Sink the LED instead of source it. (RC3).

A good short book is "Microcontroller Interfacing Circuits". Nearly all kinds og interfacing is talked/explained/showed there.
 

Thread Starter

tigerapna

Joined Dec 14, 2011
73
okay.. the micro part looks okay. reduce the LED current limiting resistors,
but I am more concerned with your power supply.

1- You are using a 24Vrms transformer.
2- You have no smoothing cap on it (other than the one in your LM7805 circuit)
3- The input voltage to the LM7805 is going to around 33V (rectified and smoothed)
4- That means that your power dissipation across the poor regulator is going to be huge. You are going to be needing an incredibly huge heat sink.
5- Recommendation: Use a switching buck regulator IC to drop the voltage down to 10 or so volts (LM2576-adj) and then use a 7805. This way, no heat sink (or a small one).
Enjoy!
Will making the PCB layout for this power supply be difficult with the LM2576 and 7805? Will it be easier to just use a 6V step down transformer with the 7805?
 

Motardo

Joined Sep 21, 2011
22
1- You are using a 24Vrms transformer.
2- You have no smoothing cap on it (other than the one in your LM7805 circuit)
3- The input voltage to the LM7805 is going to around 33V (rectified and smoothed)
4- That means that your power dissipation across the poor regulator is going to be huge. You are going to be needing an incredibly huge heat sink.
The worst case current draw through the 7805 is only 8 mA or so (three transistor bases drawing 2.13 mA each, plus two input switches drawing 500 uA each, plus say 1 mA for the uC clock (probably much less). So if the 7805 drops 28 V at 8.4 mA, that's less than a quarter Watt. It doesn't really need a heatsink, but a small one wouldn't hurt.
 
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