Constant Current Supply to Voltage Regulator

Hi all,

In your opinion is it possible to use a constant current source (i.e. LED Driver) as a digital circuit power supply (like attached schematic)? current source specification: Pout=3W, Isc=200 mA, Voc=20v and the desired regulated voltage is 5V, 100mA.

Thank in Advance,
Reza

yes u can use led driver as constant current powersupply for digital purpose but the o/p voltage should be 5v

What you show is a constant voltage output regulator, not a constant current.

rezaxyz said:

Hi all,

In your opinion is it possible to use a constant current source (i.e. LED Driver) as a digital circuit power supply (like attached schematic)? current source specification: Pout=3W, Isc=200 mA, Voc=20v and the desired regulated voltage is 5V, 100mA.

Thank in Advance,
Reza

Click to expand...

It seems like a strange idea. Does the digital circuit consume a constant current? And how would you keep the voltage within the required range for the IC power pins?

And yeah, that's not a constant-current supply. The 5-volt regulator would try to keep the output at a constant 5 Volts. The input constant-current source makes no sense, because the regulator will not consume a constant 1 Amp.

Most circuits with ICs need the power supply voltage to stay about the same, but need the power supply to be an agile source of current that can change quickly, to supply current when needed, as needed. Digital IC's, in particular, often draw current as pulses. So they need to be able to have fast-changing transient current available. For that, you use decoupling capacitors at each power pin, since drawing fast-changing currents through the power supply conductors would cause voltage spikes to be induced across the parasitic inductances of the conductors themselves, and also the correct current amplitude might not arrive at the IC's power pin at the correct time. V = L di/dt so even small-amplitude transient currents could create large voltage spikes, if the current amplitude changed very fast. So put a small 0.01 to 0.1 uF X7R ceramic capacitor very close to each power pin (within 2 mm) and then connect it to a nearby ground, and also connect a 10 uF or larger electrolytic in parallel with the small cap, at each IC power pin. They will act as small point-of-load power supplies, to supply the fast transient currents that the ICs will need.

I don't know why you need a so strange circuit. However in principle it could work even if the power dissipation have to be verified to be inside the maximum allowed.

I see that you specified the current source as a 3W, 200 mA with a maximum voltage of 20V (that is inside the range of the 7805). But in the sketch you wrote 1A: which one is correct ?
Furthermore you specified the load as 5V, 100mA, but in the sketch I see a resistor of 470 ohm that is about 10 mA: which one is correct ?

@pradeep.g.belchada:
How can I do it? The circuit output should maintain constant voltage instead of current.

@crutschow:
“What you show is a constant voltage output regulator, not a constant current.”

Yes, the main source is constant current and desired output is regulated dc voltage.

@tgootee:
“It seems like a strange idea. Does the digital circuit consume a constant current? And how would you keep the voltage within the required range for the IC power pins?”

No. and the problem is how to do it.

“And yeah, that's not a constant-current supply. The 5-volt regulator would try to keep the output at a constant 5 Volts. The input constant-current source makes no sense, because the regulator will not consume a constant 1 Amp.”

What would happen is practice?

“Most circuits with ICs need the power supply voltage to stay about the same, but need the power supply to be an agile source of current that can change quickly, to supply current when needed, as needed. Digital IC's, in particular, often draw current as pulses. So they need to be able to have fast-changing transient current available. For that, you use decoupling capacitors at each power pin, since drawing fast-changing currents through the power supply conductors would cause voltage spikes to be induced across the parasitic inductances of the conductors themselves, and also the correct current amplitude might not arrive at the IC's power pin at the correct time. V = L di/dt so even small-amplitude transient currents could create large voltage spikes, if the current amplitude changed very fast. So put a small 0.01 to 0.1 uF X7R ceramic capacitor very close to each power pin (within 2 mm) and then connect it to a nearby ground, and also connect a 10 uF or larger electrolytic in parallel with the small cap, at each IC power pin. They will act as small point-of-load power supplies, to supply the fast transient currents that the ICs will need.”

You are right but decoupling cap applicable when the regulator, regulate output voltage. How can I do it at first?

@albbg:
“I don't know why you need a so strange circuit. However in principle it could work even if the power dissipation have to be verified to be inside the maximum allowed.”

I need it because I have a led driver in my system with no cost.

I see that you specified the current source as a 3W, 200 mA with a maximum voltage of 20V (that is inside the range of the 7805). But in the sketch you wrote 1A: which one is correct?
Furthermore you specified the load as 5V, 100mA, but in the sketch I see a resistor of 470 ohm that is about 10 mA: which one is correct?

Sorry, the component values in the schematic are not correct. I use it only for showing the idea and the written parameters are correct.

The output voltage of an ideal current source as shown in your schematic would rise to inifinity. In so far, the schematic can't represent a real circuit.

If the real "current source" is voltage limited to 20 V (or something in this range), it will work as a voltage source if the load current is below 200 mA and everything is OK. But the specification (Pout=3W, Isc=200 mA, Voc=20v) isn't consistent, because 20V*200 mA refers to 4W output power.

Alternatively, you can build a 5V shunt regular. Depending on the LED driver design, either the first or the second variant will result in lower overall power consumption. From the voltage regulator viewpoint, a shunt regulator will dissipate (200-100) mA*5V = 0.5W and the series regulator 100 mA*(20-5) V = 1.5W, both at nominal load of 100 mA.

Just some additional questions:

1. is the current source connected only to the voltage regulator or it's supplying other circuits ?
2. what about the expected range of the power consumption from the load point of view (I think 100 mA are the maximum, but what about the minimum) ?
3. is you current generator an IC ? In this case could you specify the model ?

With 20volt input to the 78L05 and 100mA demanded by the load the 78L05 will go into SOA/thermal protection mode and shutdown (~1.5W dissipated across it). Plastic 78Lxx have a maximum power dissipation of about 500-700mW (depending on package), metal can types will go up to about 1.4W with a heatsink. If you can fit a heatsink to your device please do it as YOU require the maximum current from this device so cooling almost certainly will be necessary for circuit stability.
As you state the current source is rated at 200mA, I would add a resistor of across C1 to sink the excess current. This does 2 things -
1. Drops the input voltage to the 78L05 and therefore drops device dissipation.
2. Gives your current source a more stable load - current sources usually do not react well to being unloaded.
The value of this resistor is the problem, I assume your schematic is wrong with its 470 ohm resistor load (that's just 10.6mA of current at 5 volts) and you truly require 100mA, then a resistor of 100 ohms (3watt) is required. In this configuration the 78L05 dissipates about 0.5 watts, the added resistor about 1 watt with load (RL) connected, and about 3 watt with the load (RL) removed.

Hope this helps K.

which type of led driver u use is it pi ic or some thing else? it is important