# How to record the current draining out of a battery?

Discussion in 'The Projects Forum' started by Harnee15, Jan 19, 2016.

1. ### Harnee15 Thread Starter New Member

Sep 14, 2015
24
0
I have a circuit for discharging a battery (details below) which is working fine but it needs some additions so that actual current being discharged from the battery can be recorded. Can anyone please help me with this, please?

About the circuit : The circuit shown in figure below is being used for discharging a battery (which is connected at the point VBATT on the top right corner.).The battery is drained depending on the value written from the DAC. The value written on the DAC goes through two OPAMPS (A &B) and then to a Darlington transistor which is further connected to a 10 ohm resistor. The resistor is enabled or disabled through a switch. There is another resistor of 680k ohm which can also be used to discharge the battery if DAC is not being used. So, either the value written from the DAC discharges the battery or the 680K.The voltage of the battery (VBATT) is recorded on dsPIC (AD41) and also the emitter voltage of the darlington is recorded (AD42).This system is working as desired. However, I want to record the value of the actual current that is being drained out of the battery. The actual current will be different from the demanded current (Current values written through DAC) when the voltage of the battery is significantly low and it cant provide the current demanded.
Question: How can the actual current be determined from the given circuit? Shall I use a current mirror (or follower), if yes, then how can it be added to this circuit?

File size:
122.7 KB
Views:
17
2. ### ronv AAC Fanatic!

Nov 12, 2008
3,292
1,255
What is the desired current drain, and battery voltage? What is the power supply for the op amps and the micro? What is the minimum battery voltage that you care about?
The way the circuit is set up there is a slight error because the base current for the darlington also flows thru the 10 ohm resistor.

3. ### Harnee15 Thread Starter New Member

Sep 14, 2015
24
0
Hello,
Following are the details of the parameters:

Battery Voltage : Can be 1.5-3.6V
Desired Current Drain: 20mA
Power supply for the Opamp : 3.3V
Power supply for the micro 3.3V
Minimum battery voltage: Till the end of the battery (or when it stops generating 20mA)

Yes, you are right that the base current of the darlington flows through the resistor. So, there is a bit of error there.
Can you please suggest some means for recording the actual current draining out of battery?

4. ### wayneh Expert

Sep 9, 2010
12,151
3,058
Guys, that base current causes less than a 1% error, right? Given the other variables such as precision of the resistor, measuring devices, temperature fluctuations and so on, that 1% is nothing to worry about. Measured voltage across the 10Ω resistor, minus 1% if you wish, would be a very good estimate.

Or you could just add another low ohms shunt resistor at the battery.

5. ### ronv AAC Fanatic!

Nov 12, 2008
3,292
1,255
The circuit as it is should work down to a battery voltage of about 1 volt depending on which darlington transistor you are using.
You could use just a single high gain transistor like a BC547C. Then it would work down to about 0.25 volts.
The voltage across the 10 ohm represents the current thru it. In this case 0.2 volts = 20 ma.

6. ### Harnee15 Thread Starter New Member

Sep 14, 2015
24
0
Hi guys,

@wayneh I'm recording only two values for my calculations i.e. battery voltage and emitter voltage of the darlington transistor. So, can I estimate the current flowing through the 10 ohm resistor with the emitter readings accurately?

@ronv Apologies for my lack of knowledge here but do you mean that voltage will not go beyond 1 V for any battery? Also, as I don't have any means of recording the value of current flowing through the 10 ohm resistor. Is it possible to estimate it with the emitter voltage accurately?

7. ### ronv AAC Fanatic!

Nov 12, 2008
3,292
1,255
With the darlington transistor you cannot discharge the battery below about 1 volt.
You know you are measuring the voltage across a 10 ohm resistor, so you can calculate the current using ohms law I = E/R. Or in this case 10 volts per amp. So if you made your amplifier with a gain of 10, at 20 ma it would have an output of 2 volts or 0.1 ma per volt.

8. ### Harnee15 Thread Starter New Member

Sep 14, 2015
24
0
Hi @ronv I have one more concern please, , there is a switch (MOSFET) connected to the 10 ohm and the 680k resistor such that either of one is switched on at a time. So, when the 10 ohm resistor is off, there will be still some emitter voltage recorded although no current will be actually flowing through the 10 ohm resistor. So , if we go by using the formula I=E/R we might think that current is flowing through the resistor even when it is switched off.
(The on/off between the 10 ohm & 680k resistor is determined by a simple scheduling algorithm written on dsPIC.)

How can we then determine the accurate value of the actual current flowing through the 10ohm resistor? Could you please pour in your thoughts on this , please?

Thanks much!

9. ### ronv AAC Fanatic!

Nov 12, 2008
3,292
1,255
I think you can do it with code. If the micro told it to turn off don't let it look at the a to d.

10. ### wayneh Expert

Sep 9, 2010
12,151
3,058
Can you not add another voltage, the bottom of the 10 ohm? I suppose that once you verify that the MOSFET is truly switched on to an Rdson in the milliohm range, you know there is very little voltage across it.