# LTC4065L connection to active circuit

We're curious on how the LTC4065L performs in the following situation, and what the recommended design is.

There's a battery connected to LTC4065L, a 5V input to the LTC4065 and a circuit connected to the battery as well. See the attached diagram.

If I have understood this correctly, during charging, some of the charge current will go to BAT1 and some to U2. However, LTC4065L will have no way to determine this, and if U2 constantly draws a current >1.9mA (1/10 of the charge current), LTC4065L will never detect a fully charged battery. Further, the expected time to charge BAT1 will depend on how much of the programmed current U2 draws. Is this correct?

Further question; how would one design a circuit with the LTC4065L where the battery is in integrated into the product and therefore cannot be disconnected during charging? It is important that U2 receives no power dips as this can cause a restart (I figure some ripple that can be suppressed with capacitors is OK). Any pointers or reference designs would be much appreciated!

• oscar.e,

• The LTC4065L current-regulating circuit does, in fact, regulate the total current passing through it.  This includes the cell and any load on the cell.  You are correct that this directly affects the cell charge current.
• Note that the LTC4065L is a 5V part.  You will not be able to use it to charge a 9V battery.
• What is your input voltage, and battery voltage?
• What kind of battery do you want to use?
• What is your load voltage?  Does it need to be well-regulated, or can it be derived from the battery (which will have a range of voltage)?
• Hi WATaylor,

The 9V was a mistake! Its a LiPo-battery. Further details:

• The input is 5V and the battery voltage is 4.2V.
• The battery is a Lithium ion polymer battery: 3.7V (nominal), 115mAh (nominal), 1.0C max discharge current, 1.0C max charge current
• We're power-regulating the output with one buck-boost (3.3V output) and one boost (5V output). The boost drives a high-intensity LED and the 3.3V drives MCU, radio and sensors. Additionally, there is a battery gauge to measure battery level and a voltage supervisor to disable the power regulators when the LTC4065L is trickle-charging (it has a cut-off voltage of 3.08V). I've attached a more detailed schematic. Note that I've updated the component designators and replaced U2 with a resistive load (but in reality its a mix of processor, radio and sensors).

• Any update on this?

• I apologize for the delay in responding.  I did set up the LTC4065L-4.1 (I did not have a LTC4065-L on hand) on the bench, and tried running it with a load and cell emulator (adjustable power supply with a resistive load to absorb charging currents).  The total output (LTC4065L BAT pin) current is limited according to the selection of the resistor at the LTC4065L PROG pin.  In this case, RPROG=1.54kΩ, and Icharge=135mA.  I added an additional 50mA load to emulate an additional load on the BAT pin.  Net charge to the battery emulator then was 135mA-50mA=85mA as expected.

In a typical application with an LTC4065L, initially. the cell would charge through ~4.1V while the load would see the cell voltage as it charges at the rate of (programmed current - load current).  At around 4.1V, the charging current will begin to decrease as the cell voltage approaches 4.2V.  At some point between 4.1V and 4.2V cell voltage, the programmed charging current will drop below the system load current.  Seamlessly, the additional current will be provided by the cell, and it will begin to discharge.  As the cell voltage drops, the charger output current will increase.  Again, the output voltage follows the cell voltage, and there is no interruption of power to the load.  At some point in this process, the timer in the LTC4065L will time out, and charging will cease.  All power will be provided by the cell.  When the cell discharges below ΔVRECHRG, a recharge cycle will begin.  Again, this will happen seamlessly with the output following the cell voltage.

Is this what you are looking for?  Let me know if you need more information or have more questions.

• Hi WATaylor,