LM317 Output

[AnalogKid]

This might have gone by already, but how big is the heatsink/fan?

ak

 

[Suraj143]

Hi Crutshow

Thank you very much for the circuit design. That's the one I was looking for.I redrawn the circuit with the transistors.Is my drawing correct?All emitters of the transistors are going to -1.25V point..!!

Can I also use a negative voltage regulator to reduce parts count in the red area...

Thanks.

 

[ronsimpson]

Can I also use a negative voltage regulator to reduce parts count in the red area

(R9 & D3) could be a negative adjustable regulator. The rest of the red area should be the same.

 

[Suraj143]

A quick question regarding the TL431A.

How the -1.25V being generated in crutshows schematic?I cannot see the divider resisters in TL431 to apply the formula Vref *( R1/R2) :-(

 

[crutschow]

How the -1.25V being generated in crutshows schematic?

With the TL431 sense input connected to its anode, the voltage across it is 2.5V (its minimum voltage)
When the voltage control pot, U3, is at zero resistance then the voltage divider action of R2 and R5 generates -1.25V at the ADJ pin.

I redrawn the circuit with the transistors.Is my drawing correct?All emitters of the transistors are going to -1.25V point..!

Not quite.
For R1=120Ω in your schematic, then R4 also must be 120Ω going from the transistors emitters to the -2.5V from the TL431.
Also R4 must be changed to 2.5kΩ, 1/2W.
And the base resistors should be more like 3kΩ, not 1kΩ.

But the bigger problem is that the transistors won't completely shut off now with 0V on their inputs because their emitters are going to -1.25V.
You will need to add another transistor as a level translator such as this:

Can I also use a negative voltage regulator to reduce parts count in the red area...

Using a negative regulator instead of the TL431 would eliminate two parts, resistors R4 and R5.
Here's the LTspice simulation of that using an LM337 negative regulator.

 

[Suraj143]

Very useful information Everything understood clearly.

But the bigger problem is that the transistors won't completely shut off now with 0V on their inputs because their emitters are going to -1.25V.
You will need to add another transistor as a level translator such as this:

That will be too much complex to the PCB.Its better to add a relay to shutdown condition

 

[ronsimpson]

Depending on tolerance, you could add resistors (pot) to adjust the negative regulator. (-1.25 to -1.4V) If you worry about shutting off to 0.00 volts.

 

[crutschow]

That will be too much complex to the PCB.Its better to add a relay to shutdown condition

You could also add a P-MOSFET in series with the regulator input to act as a switch and cutoff the power such as this:

 

[Suraj143]

Hi crutshow thank you for the circuits. That's the one I'm looking for.Adding a current source is the easiest way of doing I guess.

My final choice have being done.I also added a short circuit protection.Want to know whats the best place to connect the "X" point.Is it ok to connect it to "Y" point or send the "X" signal to micro controller & make shutdown condition?

 

[crutschow]

Want to know whats the best place to connect the "X" point.Is it ok to connect it to "Y" point or send the "X" signal to micro controller & make shutdown condition?

If you connect to "Y" then the minimum output is 1.25V, which is the problem we've been discussing.
Connect "X" to the the base of T5 or to the μC.

Note that your "Shutdown" signal is 0V to shut off the regulator.
Thus with no connection to the "Shutdown" node, the regulator is off.

But I see a problem.
10kΩ for R7 is way to large a base resistor to allow T6 to conduct 1.5A or more.
R7 should be such as to provide a base current of at least 1/20 of the collector current or 75mA for 1.5A.
R7 should thus be about 400Ω for 1.5A output.
(This large base current requirement for bipolar transistors is why MOSFETs are often used at higher currents).

 

[Suraj143]

Hi I got a small problem.

I lowered the R7 to 470R. When the shutdown signal is on (logic 5V) the uC brownout occurs :-(

I cannot replace the TIP2955 with MOSFETs , this must do with TIP2955

 

[audioguru]

The value of R7 does not affect the current from the uC, R8 affects it instead.
Use a PNP darlington (TIP125, TIP126 or TIP127) that has much more current gain than a TIP2955, then the value of R7 and R8 can be much higher. A darlington can also be used for T5 then the value of R8 can be much higher.

Isn't your logic backwards? When the "shutdown" signal is +5V then the transistors and the LM350 are turned on, not turned off.

 

[Suraj143]

Hi I changed my shutdown circuits transistor to a darlington one.Now the circuit works, but R6 is getting too much heat

Yes my logic is bacwards the driver transistor is always on to power the regulator

 

[ronsimpson]

R6 is getting too much heat

If R6 is too hot then why is R7 not too hot?
You could make R6=22k and the heat will be 1/2.

Because the TIP125 has Base to Collector resistors of 8120 ohms R6 is not needed.

 

[Suraj143]

Hi Ron

Yes you are correct.R7 is also too hot.I removed the R6.But the R7 is still 10K & its being hot.

 

[ronsimpson]

Series two 4.7K resistors will give the same current at 1/2 hear each. 3mA
Parallel two 22k resistors will........

The transistor probably has a gain much higher then the 1000 minimum (shown on data sheet).
You could change the single resistor to 22k and 1/2 the heat and no the current will be 1.5mA.

 

[audioguru]

When the supply is 33V and T5 is turned on, when R7 is 10k ohms then its current is about 31.2V/10k= 3.1mA. Then since R7 has 31.2V across it and a current of 3.1mA in it the heating is 31.2V x 3.1mA= 0.097W which is fairly low and is fine for a 1/4W resistor.

 

[Diver300]

That circuit won't work if there is a load connected between the +ve output and the virtual ground point.

 

[Ian Rogers]

Sorry Diver300 This thread was reopened by a newby.