How to start LM317 regulator from 0 Volts

[Mohammed Ayaz Quadri]

TL;DR

I want to make power supply from LM317 from 0 volts instead of 1.25 volts. I have seen method of using diodes and zeners but I think there maybe a better way.

I want to make power supply from LM317 which starts from 0 volts instead of 1.25 volts. I have seen method of using diodes and zeners but I think there maybe a better way.

We know Voltage across R1 (see image 1) is 1.25V. So if I take the output across R2 rather than at OUT of IC, won't I get variable voltage from zero volts? Why not people do this? Does this effect regulation in any manner?
Refer image2. We can add addition capacitor and two diodes for short protection and further stabilization? How do they ger changed if I use my method of taking output across R2?How can I increase the current output above 1.5A? I've seen ways of adding a transistor. How would that work here in case of taking output across R2?
Thank you.

 

[Mohammed Ayaz Quadri]

I started to think if I apply load across R2 the current it needs should come from R1 but ideally Curreny thru R1 is constant. Thus regulatuon will be destroyed. Isnt there another way to getting from 0V output where I can apply those additional capacitor and diodes for protection and additional stabilization and also I can increase current output much over 1.5A.

 

[Baluncore]

We know Voltage across R1 (see image 1) is 1.25V. So if I take the output across R2 rather than at OUT of IC, won't I get variable voltage from zero volts? Why not people do this? Does this effect regulation in any manner?

Regulated output is from LM317 out terminal. If you load the potential divider Adj terminal it will no longer be regulated.
To regulate at zero volts you need a 1.25 V negative supply or a 1.25 V drop after the output.

 

[Baluncore]

To boost output current you can use a pass transistor. When current through the LM317 rises towards one amp, the voltage across R1 turns the transistor on and provides more current.

 

[alan123hk]

I want to make power supply from LM317 which starts from 0 volts instead of 1.25 volts

Please refer to the page 11 of the TI LM317 datasheet in the following link:-
http://www.ti.com/lit/ds/slvs044x/slvs044x.pdf

It is easy to derive the formula for the output voltage (Vo).

Because Vo [(R2+R3)/(R1+R2+R3)] - 10[R1/(R1+R2+R3)] + Vref = Vo
By solving this equation, we obtain Vo = Vref [1+(R2+R3)/R1] -10

Remark : A negative supply voltage of -10V was applied to the design

 

[Baluncore]

That circuit uses a negative voltage rail and a voltage reference diode to offset the output voltage to include zero.

 

[Tom.G]

...voltage reference diode...

Uhmm... It at least appears to be an ordinary diode symbol. The formula in the data sheet,
V_OUT= V_REFx (1+(R₂+(R₃)/R₁) -10V,
indicates the lower end of the divider is at -10V.
Must be a protection diode. Maybe to avoid the output from feeding back to the -10V supply during the shut-off transient?

That still leaves the -10V supply requirement though. Oh well. At least it needs only 45mA.

Another possibility is to put two Silicon diodes in series with the output before the load. This will drop about 1.4V from the regulator output.

The two drawbacks of using series diodes are:

• the voltage regulation at the load will be worse because of the diode resistance
• and you will need a minimum load current of a few mA to handle the diode leakage current.

Cheers,
Tom

 

[alan123hk]

I speculate that even if a negative voltage far below -1.25V is applied to the ADJ pin, the output voltage of this TI example circuit will never become negative.

Therefore, I performed an accurate circuit simulation using TI's LM317 SPICE model and confirmed this assumption .

Negative voltage changed from -10V to -20V in order to produce a negative voltage much lower than -1.25V for the ADJ pin.

 

[Baluncore]

The problem with a resistor to the negative supply is that regulation is unreliable and supply noise appears on the output.
Here are a couple of quieter solutions.
The first uses the constant reference current and an op-amp to generate a negative reference voltage.
The second uses a voltage reference diode, biased with sufficient current to a negative supply.

 

[DaveE]

Do you have to use an LM317? It's not the best starting point for this design since you will need a negative PS and the 1.25V reference is sort of buried in the device. There are other issues with this older design that can cause problems in non-standard applications, primarily the bias currents required.
I haven't looked at it recently, but you might consider the LM10 (0.2V reference and op-amp) and a power transistor of some sort. I believe the input CMR includes ground in this IC.
Or, just go to the TI or Analog Devices website and search for a more modern IC that does what you need.

 

[Mohammed Ayaz Quadri]

Uhmm... It at least appears to be an ordinary diode symbol. The formula in the data sheet,
V_OUT= V_REFx (1+(R₂+(R₃)/R₁) -10V,
indicates the lower end of the divider is at -10V.
Must be a protection diode. Maybe to avoid the output from feeding back to the -10V supply during the shut-off transient?

That still leaves the -10V supply requirement though. Oh well. At least it needs only 45mA.

Another possibility is to put two Silicon diodes in series with the output before the load. This will drop about 1.4V from the regulator output.

The two drawbacks of using series diodes are:

• the voltage regulation at the load will be worse because of the diode resistance
• and you will need a minimum load current of a few mA to handle the diode leakage current.

Cheers,
Tom

I am actually making a dual power supply with lm317 and lm337. So I do have negative supply. So I can use another regulator to generate the -10V or -1.25V. But problem is that I want to boost the current using transistor and I want to use the protection diodes as well. But I don't have enough knowledge so I can understand how the circuit will change. Plus if I usea regulator to generate the negative voltagr for adj terminal I need to add protection diodes there. Do I need to add booster transister there as well? It seems people here are giving awesome solutions which I can't understand right now. Thank You fr your efforts.

 

[Mohammed Ayaz Quadri]

Do you have to use an LM317? It's not the best starting point for this design since you will need a negative PS and the 1.25V reference is sort of buried in the device. There are other issues with this older design that can cause problems in non-standard applications, primarily the bias currents required.
I haven't looked at it recently, but you might consider the LM10 (0.2V reference and op-amp) and a power transistor of some sort. I believe the input CMR includes ground in this IC.
Or, just go to the TI or Analog Devices website and search for a more modern IC that does what you need.

Thank You I will try considering that. Can you tell me a negative compliment to Lm10?

 

[Baluncore]

Linear regulators like the LM317 generate most heat when high current flows with a low output voltage. Keep both the maximum voltage and current as low as possible. That will keep the cost and heat down.

I want to make power supply from LM317 which starts from 0 volts instead of 1.25 volts.

You must specify a maximum output voltage.

How can I increase the current output above 1.5A?

You must also specify a maximum output current.

 

[Baluncore]

Can you tell me a negative compliment to Lm10?

There is no complementary LM10 because it is an op-amp with a low voltage reference in one package. For negative voltages the LM10 external circuit would need to be changed.

Managing power dissipation with heatsinks will be a problem unless you use switching regulators.
You need to specify the limits for output current and voltage if you want us to recommend a reliable circuit.