Boost converter drops to 4.5V when trying to achieve 6V with an 18650 cell

I don't have a lot of experience with this but I'm working on a prototype device which I want to run off a 18650 cell. I need to boost the voltage from 3.7V to 6V using a boost converter that uses a LM2587 chip that I bought on amazon.

When I connect the load, the voltage drops from the 6V that I need to 4.5V.

The cell is rated for 6.5A continuous and the boost converter is rated at 3A continuous.

At 4.5V, the circuit is drawing around 1.8A. What am I missing?

• Welcome to EE.SE. We love schematics. There's a button on the editor toolbar and its easy to use. Please provide datasheet links. – Transistor Aug 15 '16 at 22:29
• Updated my answer with the links. Looking into the schematics editor – david2391 Aug 15 '16 at 22:43
• You seem to be working with bare Lithium-ion cells. What precautions are you taking? – Nick Alexeev Aug 15 '16 at 23:05
• Does the battery voltage also see a drop? Does anything get warm? – Marcus Müller Aug 15 '16 at 23:06
• @MarcusMüller the battery voltage drops from 3.8 to 3.4 when the load is connected. The boost converter board does get warm when the load is connected but the battery does not. – david2391 Aug 15 '16 at 23:16

Specification sheet claims are usually "best case" with optimum vin vout and power. The conditions you require do not match the converter's optimum operating conditions.

You are seeking to achieve 6V across 3 Ohms. Power = V^2/R = 36/3 = 12 Watts.

At eg 12V in you you would need eg ~= 1.3 A average input at 75% efficiency to get 12 W out - and you could easily achieve 12 Watts output.
At 6V in you'd need 2.6A average and still doable.
At 4V in you need 3.9A average in and with duty cycle considerations you are near or above the limit for 12W out depending on overall achieved efficiency.
A single LiIon 18650 cell has a max of 4.2V, average operating voltage of 3.6V and useful voltage range of about 3V to 4V.

Let's see what we can expect.
The LM2587 has a 5A peak internal switch.

Duty cycle Toff:Ton ~~~= Vin: Vout/efficiency
Say 3.5: 6/75% = 3.5 :8 ~= 30% off, 70% on.
Max switch current = 5A.
Available Iin avg ~= Imax /2 x ton/tcycle
= 5/2 x 70% = 1.75A
Power in max = 3.3V x 1.75A = 5.8W
Power out max = Power in x efficiency
Say 5.8 x 75% = 4.3 W
Available V into 3 Ohms . Power = V^2/R
Or V = sqrt(Power x R) = = (4.3 x 3)^0.5 = 3.6V
You are getting somewhat better - converter is presumably operating at better efficiency than the 75% I used.
BUT, while the converter is capable of providing more power under the best Vin, Vout and load combinations, it falls somewhat short in this case.

E&OE. YMMV.

• Thanks for the detailed breakdown Russell. I'm going to have to do learn some more about the calculations made in the second half but it's clear that it's not possible to get what I need. Now my question is, is there a chip/board that can handle this or will I need to stack 2 LiIons in series to get the higher voltage? The reason I didn't want to go this route is because the device needs to be recharged without removing the batteries. – david2391 Aug 16 '16 at 21:33
• @user1530141 Factors of relevance include (not limited to) Vin, Vout/Vin ratio, Switch Imax, Connection losses, battery Ioperate current, ... || battery should be OK at 6A continuous rating. | Check Vdrop in Vin cct. Measure Vbattery AT battery and on PCB tracks of Vin connector (not just at connector terminals). Your baattery is flat top and made to have SPOT WELDED leads attached (NOT SOLDERED) and if using a holder the holder and wires and connector can add significant losses. | You can: buy boost converters with more input current capability./ parallel two of these with a little care. – Russell McMahon Aug 17 '16 at 1:59
• This module MAY do what you want. There are 3 on the site - the dearest (of course :-) ) of the 3 is rated at "enough" - see specs at bottom of page. Heatsink suggests it is "more serious"m- IC unspecified. seller MAY advise if asked (easy for him to look and see). – Russell McMahon Aug 17 '16 at 2:23
• The name LTC1781 doe NOT appearto be a 'western' part number and I could not find a data sheet. It MAY be their product number. Worth a look but be sure the spec is able to be as claimed. – Russell McMahon Aug 17 '16 at 2:23
• @user1530141 Aha! LTC1871 datasheet (ie not 1781) is a superb performance IC that can do anything desired as it uses an external MOSFET switch. That makes total sense IF that is what ios used here. Vin is >= 2.5V. || Here is a "real" one. Maybe for \$8. Buyer beware as always. Search: LTC1871 boost converter – Russell McMahon Aug 17 '16 at 2:33