Low voltage drop 8 amps Ac to DC power converter

[Walter6540]

Hello everyone. I made Ac to Dc power converter and it looks work. I still have a couple of questions about it.
I uploaded the diagram (see attached file please) and my questions are:
1. Why there is 14 volts before and after resistor R1. R1=15 Ohm and if current is 0.5 amps, voltage after R1 should
be 7 Volts, the load is 12 Volts 5 Amps.
2. Three TIP2955 transistors are too hot 60-70 degrees Celsius for 1.7 Amps per transistor load.
Thank You for your help
Walter

 

[davenn]

Did you actually measure 14VDC across C1 ?

I would expect it to be more like in the range of 17 - 20VDC

 

[duke37]

Rough calculation -

The voltage across R1 will be the emitter/base voltage of the TIPs (0.7V) plus the voltage across R2 to R4.
Thus at 3A output the voltage across R1 will be 0.7V + 0.1V = 0.8V.
The current through R1 will be about 50mA.

With a 14V transformer you should get up to 19V across C1 so in the worse case, each TIP transitor will dissipate 7W at 3A output. If the transistors get too hot, then you need a bigger heat sink.

 

[AnalogKid]

Also, 3300 uF is not nearly big enough to filter 8 A. After the diode bridge you should have around 18 V peak at C1, and there will be at least 1 V drop across the regulator (Vbe will be greater than 0.7 V at 3 A collector current.). That leaves less than 5 V of headroom, so you need around 15,000 uF to keep the regulator happy.

ak

 

[Walter6540]

Thank You very much for your quick response. The voltage across C1 is 19V without any load and 14v with load 12 v and 4 A. I will increase C1 up to 15000 uF and watch heat sink temperature. Thank You very much for your time and detail explanation.

 

[AnalogKid]

Note that the issue with the filter capacitor size is that if the cap is too small for the power being moved, the ripple voltage at the input to the regulator can be so big that the negative peaks go below the minimum value needed for clean regulation. The best way to see this is with a scope. A DVM doesn't separate the average value from the peak-to-peak variations. In other words, that 14 V measurement does not tell the whole story.

ak

 

[Walter6540]

I increased C1 up to 16000 uF. Voltage output, under a load of 4 amps, drops to
9.9 V in a minute. LM2940 is not hot. What could be the the problem?

 

[rickselectricalprojects]

those transistors do get hot. i remember i had a cheap ebay special 0-15v 1a power supply that had a 2n3055 and that got very, very hot but 74 degrees Celsius does seem very hot. the tip2955 is rated for 150 degrees Celsius.

 

[Walter6540]

Thank You. Yes, TIP 2955 are hot, LM 2940 is not hot. I hanged a transformer to 15v
output, but still have 9.5 V output and my load is 6 Amps

 

[Fredx-3gtvi]

Hi , I think I see your problem. You have used an LM 2940 as the regulator, and this is an LDO type regulator.
They are a good regulator and work well, but do NOT work like a normal 78xx type regulator, and so are not able to work with pass
transistors for higher currents. The reason for this is that when the regulator 'thinks' its about to drop out of regulation, it changes things internally and more current is drawn by the by the regulator. ie, more current flows into the input, and more current flows to ground
via it's common lead. This is fine with no external pass transistors, but with pass transistors, it changes the voltage across the
B-E of the transistors, which changes the current in the transistors and hence the output voltage.

There is a general "rule of thumb" for working out the capacitor after the rectifier and it is (for 50/60 hz supply) 1000uf per amp.
so for a 8 amp supply, the requirement would be 8,000uf minimum !

The TIP 2955 are hot because they have more voltage across them. Try using a 7812, and lots of uf for C1, but you may need
to replace the transformer with one with about 17-18 volts ac out.

Cheers.