Electric Bikes - A question about voltage fall

Dear people: I noticed that when I twist the throttle rapidly the 4-leds digital voltmeter indicates the voltage lows, even with full batteries. Before disassembly the batteries case I like to ask you about this.

I weld all the main connections from the battery to the controller and replaced that original weak fuse holder with a round one and a 15A tubular fuse.

I built a resistive load of 1.2 ohms with some Ralco resistances and I connect it at the output of the batteries case. The voltage is 38.4 V without load and 35.5 V with it. The intensity is 11.5 A.

Could be the internal resistance of batteries or any parasitic resistance of the key switch or fuse holder?

PD: They're CSB gel lead-acid, 3 x 12 V 12 AH
http://www.csb-battery.com/english/01_product/02_detail.php?fid=4&pid=45

Voltage drop across a device is equal to current X resistance, a lot depends on where in circuit the LED voltmeter is connected. So if it's connected across the battery, you may be seeing instantaneous current draw or inrush, that may not show up in your steady state simulation.

Dear people: I noticed that when I twist the throttle rapidly the 4-leds digital voltmeter indicates the voltage lows, even with full batteries. Before disassembly the batteries case I like to ask you about this.

I weld all the main connections from the battery to the controller and replaced that original weak fuse holder with a round one and a 15A tubular fuse.

I built a resistive load of 1.2 ohms with some Ralco resistances and I connect it at the output of the batteries case. The voltage is 38.4 V without load and 35.5 V with it. The intensity is 11.5 A.

Could be the internal resistance of batteries or any parasitic resistance of the key switch or fuse holder?

PD: They're CSB gel lead-acid, 3 x 12 V 12 AH
http://www.csb-battery.com/english/01_product/02_detail.php?fid=4&pid=45

Not sure what you mean by "twist the throttle rapidly ", and "... resistive load of 1.2 ohms ...The voltage is 38.4 V without load and 35.5 V with it. The intensity is 11.5 A." doesn't properly compute because I=V/R, I=35volts/1.2ohms=29Amps (not 11.5). Every cell has internal resistance which will cause a voltage drop when the load is applied. That's normal. Every other conductor in the circuit to the load contains resistance that will cause additional voltage drop at the load so minimizing circuit resistances is the right thing to do if the voltage drop is a intollerable problem in your design. But circuit resistance cannot be totally eliminated without super cooling.

38.4 V / 11.5 A = 3.39 ohms and the load has 1.2, so there are (3.39 - 1.2) = 2.19 ohms in the circuit somewhere ...

I built a resistive load of 1.2 ohms with some Ralco resistances and I connect it at the output of the batteries case. The voltage is 38.4 V without load and 35.5 V with it. The intensity is 11.5 A.

IF
1. you are measuring correctly
2. the voltage at the battery leads is 35.5V
3. the amperage the battery is putting out is 11.5A
4. the load consists of resistance only (as opposed to pulse-width-modulation etc)

THEN
the load is ~3.1 Ohms

...which says nothing bad about the battery, but seems to say that the contacts you are using and/or the wires you are using during this measurement are producing ~1.9 ohms of resistance.

I think there's nothing at all wrong with your battery if, after a full charge, it drops 3 volts due to a 11 amp load.

Maybe it's the problem of ur battery.I guess Ur battery include 3 cells.The volt of each cell is 12V.One get worse the whole done.It's not base on ur OHM.

anytime you put the voltmeter across the battery terminals and see a voltage sag, it is due to internal resistances of the battery. This is either due to a weak battery or excessive current draw.

I hope this helps

anytime you put the voltmeter across the battery terminals and see a voltage sag, it is due to internal resistances of the battery. This is either due to a weak battery or excessive current draw.

Even when fully charged and in very good condition, e-bike batteries will have a measurable voltage sag when putting out ~10 amps. For most batteries, a 1 amp current should be enough to cause a voltage drop that you can see on a normal multi-meter or volt meter.

The rating of the battery set you have is exceeded. And when you do that the internal resistance of lead acid battery's goes up. They get hot and the resistance goes up even more. The nominal current draw for a 12AH should be around 3 Amps. Now you exceeded that by a factor of almost 4.

And you don't subtract the load resistance. You have an internal battery resistance of 3.39 ohms period. That's like 1.13 ohms per battery, for a 12V lead acid battery that is about right, for the current draw you have.

Bottom line is: lead just isn't that good of a conductor of electrons.

Later OM