Tailcap current

[Jufran88]

So, I was experimenting with my laser and decided to measure the current at the tail cap. I set my DMM leads for the 10A setting and this is what I got:

Host: Solarforce L2 Gunmetal
Driver: Micro FlexDrive V5 @ 1.5A
Battery: LG 18650 unprotected @ current voltage of 4.10V

At the tail cap I am getting 2.25A. What is the relation of the tail cap current to the current flowing through the LD driven by a boost circuit?

 

[Toke]

The boost circuit acts like a DC "transformer", as in turning low voltage high current, into higher voltage but lower current.
A buck driver does the opposite.

What is the relation of the tail cap current to the current flowing through the LD driven by a boost circuit?

The tail cap current will be higher, you can expect it to be voltage difference*driver efficiency. The later is hard to predict.

 

[Cyparagon]

What is the relation of the tail cap current to the current flowing through the LD driven by a boost circuit?

It depends on the supply voltage. The flexdrive supplies and draws (approximately) constant wattage. As the battery voltage falls, the driver will draw more current.

 

[Jufran88]

So, this is what I'm thinking:

driver power out = LD power in
Vbatt x Ibatt x Efficiency = Vf x If
(equation used at CPF)

Where,
Vin = Vbatt = voltage fed to the driver
Iin = Ibatt = current fed to the driver (tail cap measurement)
Efficiency = efficiency of the driver (could include other factors)
Vf = forward voltage of LD
If = current outputted from driver

In terms of efficiency of the driver (drlava states 80-95% efficiency for the flexdrives) we get:

Efficiency = (Vf x If) / (Vbatt x Ibatt) = (4.2*1.5)/(4.1*2.25) = .682 or 68.2%

Now, this is pretty low so I think the other factors that create loss in efficiency would be wire resistance, test probe resistance, battery resistance, but would that matter since the current values aren't very high?

And can we suggest that we can measure power dissipated from the LD or driver by using P=VI

So power dissipated by the driver:
P=VI = (4.1)(2.25) = 9.225W

and power dissipated by the LD:
P=VI = (4.2)(1.5) = 6.3W

Efficiency = 6.3/9.225 = .682 or 68.2%


Now I notice I get the same efficiency when I calculate it both ways. Coincidence? Can anyone else test this out and see what they get?

And I thought as the voltage falls that current will not go up but down, where if it hits minimum voltage the LD will just turn off.

 

[Cyparagon]

Are you calculating using open circuit voltage, or load voltage?

I can assure you it won't be 4.2V when you place a 2A load on it. You need to use the voltage while under load.

 

[Jufran88]

Well, it would be load voltage and since there's no datasheet for A140 LD's I wouldn't know exactly the Vf at 1.5A. I'm just using the standard 4.2V for forward voltage of the LD.

Would anyone else like to participate and see what values they get?

 

[Cyparagon]

No no no... the battery voltage under load.

 

[Jufran88]

Oh, got confused because you said 4.2V. My battery was 4.1V at the time and that's open circuit voltage.

Now that I have calculated it under load its 3.65V

So, now calculating for efficiency:

Efficiency = (Vf x If) / (Vbatt x Ibatt) = (4.2*1.5)/(3.65*2.25) = .767 or 76.7%

 

[Benm]

Seems reasonable enough

Don't rely too much on current measurements when using switchmode drivers though. The will have input capacitors, but there is still an alternating component to the current they draw, and most multimeters cannot measure that properly.