Slow Turn On Voltage Regulator - To Drive a DC Motor ?

[ItsMike]

Hello,

I'm driving a 12v DC motor using an LM317 voltage regulator.
I want the motor to start up slowly, so I was thinking of using the circuit attached.
Will it work ?
Does the transistor go Active ? or straight into Saturation ? Also, what is the Diode for ?

Does this circuit need any extra protection for driving a DC motor ?

Thanks in advance.

 

[MikeMl]

The transistor is a "capacitance multiplier"; the capacitor value is effectively multiplied by the transistor β, thereby lengthening the time constant.

The diode discharges the timing capacitor when the input power is removed.

I think you need one more diode (Cathode to LM317 input, Anode to LM317 output) to protect the LM317 when the input power supply is turned off.

 

[ItsMike]

Where would the simulated capacitance appear ? between the Collector and Emitter ?
How does the current flow through the capacitance multiplier exactly ?

 

[MikeMl]

Where would the simulated capacitance appear ? between the Collector and Emitter ?

No, the effective capacitance appears between the adj. pin on the 317 and ground.

How does the current flow through the capacitance multiplier exactly ?

The emitter current sinks current away from the adj.pin. The emitter current is controlled by the base current, which in turn is defined by the 50K and the timing capacitor. In other words, the timing capacitor is charged by the current through the 50K+the base current.

 

[ItsMike]

For some reason I had trouble simulating this with ISIS (too lazy to install LTSpice), ISIS graph shows the voltage getting reaching max after about 18 seconds.

How do I calculate, approximately, the time it takes to reach max voltage ?
is it 5RβC ?

P.S

I have a very large capacitor in the input of the voltage regulator, don't I need to have a way of discharging it at power-down too ?

 

[Jaguarjoe]

If you let the output of the 317 become more positive than its input, the big pass transistor inside of it will blow out because of the reverse current flow through it. The large capacitor on the input helps you in this regard. A LM317K in the TO-3 can costs around $3.00. Adding a $0.02 diode across the regulator is very cheap insurance. Some people double their investment by putting a diode in series with the regulator's input to protect it from accidental polarity reversal as well.

 

[Mr RB]

I think you could do it with less parts with a darlington NPN and 2 resistors and a cap on it's base. The darlington will also drop only about 1v Vce when at full voltage, less than the LM317/117.

However it won't regulate the voltage, but I wasn't sure if that was needed as you said "slow turn on" and "12v motor" so if you are running from a 12v supply or 12v battery the darlington may be the better option. Just a thought.

 

[ItsMike]

Thanks for all the reply's guys.

This is kinda off topic, but since we are talking about voltage regulator protection...

I have a 7805 with big caps on the input and output, is it critical to get a diode to discharge them when the circuit is off ?

 

[Jaguarjoe]

I don't think anyone can predict whether the cap on the input side will discharge faster than the cap on the output side but if the one on the inlet side does discharge faster you will most likely lose the 7805.

 

[ItsMike]

Thanks a lot guys.