Two similar LED constant current sources, which is better and why?

Thread Starter

dcbingaman

Joined Jun 30, 2021
799
I have the following constant current sources for LEDs side by side in the same circuit for comparison:

1668477859377.png
R1, Q1, R2, D3 and D4 form a constant current source for D1 LED.
R3, Q2, Q3 and R4 form another constant current source for D2 LED.

I am interested in the ability of each to regulate the current in the respective LED over the V1 range of 3.3V to 15V.

Both work similarly but still have subtle differences.

Here is a DC sweep of the current in D1 and the current in D2:

1668478095248.png

The current in D1 is the green trace.
The current in D2 is the blue trace.
The horizontal axis is the V1 voltage being swept from 3.3V up to 15V.
The vertical axis is the current in the D1 (green trace) and D2 (blue trace)

The LTSpice circuit is attached.

It appears the the second circuit using Q2 and Q3 is superior at maintaining constant current over the range specified. The question I have is why. Which circuit would be recommended and why.
 

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WBahn

Joined Mar 31, 2012
27,903
The second circuit has a higher output impedance (which is what you want for a current source).

Notice that the first one is open more loop. The two diodes establish a base voltage that is very insensitive to the LED current. As you increase the supply voltage, you increase Vce and that results in increased collector current due to the Early effect. Now as the LED current increases, that does increase the voltage across the emitter resistor, which reduces the Vbe which partially tamps down the increase in collector current. But with the greater Vce voltage, less base current is needed to produce the same collector current, which reduces the current in the base-bias resistor, which increases the base voltage, which increase Vbe, resulting in more collector current. So this effect is at odds with the negative feedback just described.

In the second circuit, the negative feedback is much stronger. An increase in collector current due to increased Vce again causes an increase in the emitter voltage, but this same increase is applied to the base shunt transistor which increases the current that it draws away from the base, which reduces the base current and reduces the collector current back toward where it is supposed to be. This effect is very strong because of the current gain of Q3.
 

crutschow

Joined Mar 14, 2008
31,129
In short, the reason that the two transistor circuit makes a better (more stiff) current source is that the Q3 current-sense transistor has gain which reduces the current change versus voltage roughly by the gain of the transistor, so that would be the preferred circuit in my opinion.

(Incidentally, the dark blue trace in your posted simulation is barely visible on my computer screen.
Using red or yellow for the trace will show up much better against a black background.)
 

MrSalts

Joined Apr 2, 2020
2,621
The left version will vary with temperature more than the right version. In the left circuit, you have two pn junction drops for the two diodes in series for about -4mV/°C vs 2mV/°C for the base-emitter pn junction on the npn.

In the right circuit, both NPN base-emitter drops are matched at 2mV/°C which I assume changes the current less than the left version with changing temp. Whether the temp is caused by changes in ambient temp or by self-heating of the circuit.
 

LowQCab

Joined Nov 6, 2012
2,650
Here's a couple of other ways to do it that are very stable, super-simple, and cheap.
( 1-TO-92 Package, a Resistor, and a Bypass-Capacitor ).
( the Resistors can be ~1/4-Watt in your case ).
.
.
.
Current Regulator 1 .png
 

crutschow

Joined Mar 14, 2008
31,129
Here's a couple of other ways to do it that are very stable, super-simple, and cheap.
That does make a very stiff current-source, but the main disadvantage of the LM317 is that it requires about 2V across it for proper operation plus the 1.25V drop of the reference voltage across the current sense resistor.
For battery applications that 3V plus drop may be undesirable.
 
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LowQCab

Joined Nov 6, 2012
2,650
Voltage-Drop across a Resistor, or across a Semi-Conductor-Device, or both at the same time,
is all the same to the LED.

The amount of Heat which must be dissipated is exactly the same no matter what You do,
and there is enough Voltage, ( 3.3V ),
to make the Regulator-Voltage-Drop almost irrelevant to the equation.

Also, the Regulator-Voltage-Drop is slightly less at the lower Current-Levels, ( ~10mA ), being used here.

Greater Efficiency would require a "Switch-Mode-Current-Regulator",
which wouldn't be significantly more efficient when working at ~3.3V.
It might not even have enough Voltage-Overhead to operate properly.

I think a single "Rail-to-Rail" Op-Amp would be the most elegant solution.
.
.
.
 

Ian0

Joined Aug 7, 2020
6,713
Bob Pease had an interesting variant. If you add a Schottky diode to the second circuit, cathode to the top of R4 anode to the base of Q3, it lowers the dropout voltage, and temperature compensates it. It also needs a bias resistor from the base of Q3 to a stable supply.
I’m intrigued how well it might work with @Danko ’s circuit - when I get back to a computer with SPICE on it, I’ll try it.
 

crutschow

Joined Mar 14, 2008
31,129
Voltage-Drop across a Resistor, or across a Semi-Conductor-Device, or both at the same time,
is all the same to the LED.

The amount of Heat which must be dissipated is exactly the same no matter what You do,
and there is enough Voltage, ( 3.3V ),
to make the Regulator-Voltage-Drop almost irrelevant to the equation.
I don't follow your reasoning (?).
Any effect on the LED is not what I am referring to.
The regulator voltage drop is added to the LED drop.
The minimum supply voltage must then be the regulator drop plus the LED drop.
Thus the large LM317 drop is a disadvantage if you want to operate down to a minimum battery voltage.
The transistor circuit can operate down to about a volt above the LED voltage.
 

Thread Starter

dcbingaman

Joined Jun 30, 2021
799
In short, the reason that the two transistor circuit makes a better (more stiff) current source is that the Q3 current-sense transistor has gain which reduces the current change versus voltage roughly by the gain of the transistor, so that would be the preferred circuit in my opinion.

(Incidentally, the dark blue trace in your posted simulation is barely visible on my computer screen.
Using red or yellow for the trace will show up much better against a black background.)
Good way to look at it. Thanks, that does help in understanding it.
For the circuit on the left, any increase in current in the emitter resistor causes an increase in emitter voltage that results in a drop in base emitter voltage that then counteracts the current increase by decreasing base emitter current.
In the second circuit you still have that effect but in addition you have the increase in emitter current causing an increase in base emitter voltage of Q3, thus increasing base emitter current of Q3 multiplied by Q3 current gain steals current from the base of Q2 resulting in a drop again of emitter resistor current.
It would be interesting to see an analysis using Ebers-Moll equations but I imagine it would be very involved.
 

Thread Starter

dcbingaman

Joined Jun 30, 2021
799
Being there appears to be plenty of ways to do this, I placed the ones people have come up with on one circuit to compare:
Being there is not much difference in behavior above 6V I limited the plot to 3.3V to 6V. Thanks for everyone's input.

I will probably go with D2 method. Component count is reasonable and the cost of the parts is low. I like the Rail to Rail op amp solution but those chips tend to be more expensive the BJT's.

I tend to agree with
crutschow
Changes in current for the LED would not be noticed much by a human.
 

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Ian0

Joined Aug 7, 2020
6,713
Here's the Pease circuit (in green). Interesting to note that 33k is an optimum value for R2, above or below and the current varies with the supply voltage. It works better if the 33k goes to the (variable) supply voltage than to a fixed reference.
Screenshot at 2022-11-15 21-27-05.png
 

Thread Starter

dcbingaman

Joined Jun 30, 2021
799
Here's the Pease circuit (in green). Interesting to note that 33k is an optimum value for R2, above or below and the current varies with the supply voltage. It works better if the 33k goes to the (variable) supply voltage than to a fixed reference.
View attachment 280787
What has changed? In the circuit with D3 configuration I was getting around 10mA at 3.3V. The circuit here has less than 2mA at 3.3V? I noticed the LED's and transistor types have changed?
 

Ian0

Joined Aug 7, 2020
6,713
What has changed? In the circuit with D3 configuration I was getting around 10mA at 3.3V. The circuit here has less than 2mA at 3.3V? I noticed the LED's and transistor types have changed?
The "step" command varies the temperature from 0°C to 100°C.
I thought it might be more visible in the picture, but it is rather tiny.
 

MisterBill2

Joined Jan 23, 2018
13,805
One more consideration is that the forward voltage drop of the two diodes used as a reference is temperature dependent and varies in the wrong direction to be helpful. (Referencing to the circuit in post #1.)
 

Ian0

Joined Aug 7, 2020
6,713
One more consideration is that the forward voltage drop of the two diodes used as a reference is temperature dependent and varies in the wrong direction to be helpful. (Referencing to the circuit in post #1.)
One diode cancels out the Vbe variation with temperature. The other reduces the reference voltage, so that the current will fall with increasing temperature. Helpful in that it reduces LED power dissipation. Unhelpful in that the LED light output also falls with increasing temperature, so makes it worse.
 
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