0
\$\begingroup\$

I found BCR421 driver and want to ask how to properly drive TSSF4500 LED. All the parameters seem to make sense if look separately, but where do I start to determine the necessary parameters?

It should be a continuous mode with voltage supply of say 3.3V. I would like to be able to adjust the current through the LED from 10mA to 100mA.

enter image description here

For that, I am planning to use the trimmer resistor in series with a 10 Ohm resistor in place of the Rext, where the 10 Ohm on its own will drive the Iout to about 90mA, and additional resistance from the trimmer will allow to drop the current to 10mA, according to the plot below:

enter image description here

  1. Should I care about the Vout value? It seems to me like a chicken and an egg problem. Vout depends on the voltage drop across the LED and Rext, and determines the current Iout, according to the plot. It the same time, the current through the LED depends on the voltage drop across the LED, which also dictates the Vout that determines the Iout. Well, I can not seem to get out of this loop. So, what determines all the other?
  2. How do I determine the current through the Rext, or the Vdrop? Is it correct to assume that for low Rext values the current through it is ~Iout?
  3. What is the advantage of this circuit as opposed to simple resistor-diode in series (except the EN pin that allows fast switching)?
|improve this question
\$\endgroup\$
  • \$\begingroup\$ Not sure if you should want 100mA current through the trimmer, that will likely damage it. Why not set the maximum current with a resistor and vary the current by changing I(EN) on pin 1? It my need a little fiddling to get a useful input signal, but at least it is much easier to control. \$\endgroup\$ – jippie Mar 25 '16 at 13:56
  • \$\begingroup\$ Also notice that the driver drops little over 1V before it starts to act as a current source. Then if you power supply is only 3V3, there is only 2V3 left for the LED's forward voltage which is on the low side for white LED's. \$\endgroup\$ – jippie Mar 25 '16 at 13:59
  • \$\begingroup\$ @jippie I agree that 100mA through a random trimmer needs to be checked against the data sheet- it's not likely a good idea for reliability. I don't think reducing the current through Ien is a good idea- this thing is not a current mirror - it depends on the two diodes conducting to stabilize it. \$\endgroup\$ – Spehro Pefhany Mar 25 '16 at 14:05
  • \$\begingroup\$ @SpehroPefhany you are probably right. On a different subject: notice that the datasheet mentions at least 4V5 for the enable voltage, over a volt more than the proposed power supply voltage of 3V3 (unless that is really intended to be the LED voltage, which would make more sense). (Although 3V3 is later mentioned for 1.2mA I(EN) typical) \$\endgroup\$ – jippie Mar 25 '16 at 14:27
  • \$\begingroup\$ @jippie I think it's okay. The current for the enable input of the 421 is specified at 3.3V. The curve is pretty mushy at 100mA out in fig. 3.22, so more voltage would be better. \$\endgroup\$ – Spehro Pefhany Mar 25 '16 at 17:54
2
\$\begingroup\$

  1. Yes, you should care. The BCR421 will try to maintain a (roughly) constant current at the collector terminal, however if Vout drops too low then it will not be able to maintain the constant current, as you can see from the graphs you posted. What will happen is that the output will saturate at a lower current than you want. To find out if this will happen, assume that the maximum desired current is flowing through the LED. Subtract the maximum possible LED voltage (which will occur at the maximum current) from the minimum possible supply voltage and compare with the dropout voltage of the BCR421 at that current.

    So the LED drops 1.35V typical at 100mA (1.6V maximum). Your supply voltage is 3.3V nominal, let's dock 5% for tolerance to give 3.13V. That gives the chip 1.54V to work with. The parameter VSmin indicates that typically 1.4V is enough at 25°C, and the graphs you posted indicate that above about 1.1V it's fairly constant current. So it's likely okay unless you have some extreme operating conditions.

  2. The current through the external resistor when it is regulating correctly will be approximately Iout - 10mA. The internal resistor conducts the 10mA. It will be slightly higher because of base current (maybe 0.5%), but that's immaterial in this situation, given the +/-10% tolerance on the 10mA current. As @jippie points out in a comment, 100mA through a pot wiper is probably not a good idea, depending on the pot construction (okay for a wire-wound, but they are expensive). Minimum resistance may also be a problem. It's generally recommended not to put much current through pot wipers- they work better as voltage dividers than rheostats.

  3. This circuit maintains a fairly constant current regardless of changes in the LED forward voltage (or differences from unit-to-unit when first assembled) and regardless of changes in supply voltage.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ I forgot to ask about the short circuit condition. In Absolute maximums, it says that Imax=200mA. What happens if Rext is shorted? Any ideas what would be the Imax and for how long? I just want to assess the light illumination hazard for the maximum LED radiation at maximum possible current provided by the driver. \$\endgroup\$ – Nazar Mar 25 '16 at 14:30
  • \$\begingroup\$ And with 10 Ohm and 100mA, the Rext power would be 100mW, is that so bad for the trimmer? \$\endgroup\$ – Nazar Mar 25 '16 at 14:40
  • \$\begingroup\$ @Naz most trimmers are happy with 100mW total spread over the entire element, but both concentrating it at one end and concentration at the wiper contact point(s) can be a problem. \$\endgroup\$ – Spehro Pefhany Mar 25 '16 at 17:41
  • \$\begingroup\$ If Rext is shorted you probably need to buy a new driver and a new LED so don't do that. \$\endgroup\$ – Spehro Pefhany Mar 26 '16 at 4:20
1
\$\begingroup\$

Should I care about the Vout value? It seems to me like a chicken and an egg problem. Vout depends on the voltage drop across the LED and Rext, and determines the current Iout, according to the plot.

What you want to do is make sure Vout is high enough (i.e. that your supply voltage is high enough) that you are on the flat part of the graph. Then the output current becomes dependent only on the resistor value.

How do I determine the current through the Rext, or the Vdrop? Is it correct to assume that for low Rext values the current through it is ~Iout?

You can assume the current through the two paralell resistors is approximately Iout. You are told that the internal resistor value is 100 ohm so working out the proportion that goes trhough each resistor shouldn't be too hard.

What is the advantage of this circuit as opposed to simple resistor-diode in series (except the EN pin that allows fast switching)?

The main advantage is that you can maintain a fixed current into the LEDs across variability in the characteristics of the LEDs and across variations in the supply voltage (which means you can use an unregulated supply).

|improve this answer
\$\endgroup\$
1
\$\begingroup\$

(1) Yes within limits. If V(Led) + Vout(minimum) > 3.3V this will not work at 3.3V supply.

The graph tells you that Iout is roughly independent of Vout for Vout > 1V so take Vout (min) = 1V and add Vf (=1.3 to 1.6V). So Vout(min) + Vled = 2.6V < 3.3V and we're OK. (But running this LED at 1.5A, or running a white LED, would not work).

(2) Current in Rext = Iout - (Iout for Rext=infinity).
Practically, Current in Rext = Iout

(3) It keeps Iout more stable as your supply voltage and LED temperature varies. For example if running off a battery, Iout would be roughly constant until Vout fell below 1V (Vsupply < 2.6V).

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.