Hi all,
I need your help for the following problem:
I'm developing a hand-held device that contains an analog switch to monitor the various voltages inside by a single A2D for Built In Test (BIT). One of the voltages is the RTC back-up battery. The problem is that during power off (When main batteries outside of device) the analog switch is powered off as well and it causes to high current consumption from the back-up battery.
The backup battery voltage is 1.6v to 1.2v.
I can't use a relay because I don't have space for it. (very small device)
What kind of protection can I use in this case?

Thanks

 

What is this current-gobbling switch, and what does the circuit look like?

 

I'm using the MC74HC4051A analog switch.
The RTC backup battery is connected directly to one of the inputs.

 

There are two immediate possibilities.

1. Your circuit draws more than 25 mills, and the IC is blown up and leaky.

2. The select pins are low when off, which satisfies the chip enable and selects channel 0 to conduct - possibly leaving the battery connected.

 

Any suggestion how to solve this problem?
How to make this input disconnected from backup battery during power off?

 

Got a meter? Check for leakage current. Check for normal current draw to see if it exceeds 25 ma. Pull up the chip select line so the chip can't conduct while off.

 

I pulled down the Chip select line but the leakage current is still high, about 1 mA.
any more suggestions?

Thanks

 

The problem is that during power off (When main batteries outside of device) the analog switch is powered off as well and it causes to high current consumption from the back-up battery.
The backup battery voltage is 1.6v to 1.2v.

If you removes the main batteries, then the 74HC4051 chip and other connected circuits are not powered. However, the backup battery is still connected to one pin of the CMOS gate.

One thing that you obviously not aware of is most if not all CMOS gates has internal protection diodes to VDD and VSS/VEE. These diodes are normally not conducting, as long as power is provided to the chip and the gate voltage is between these rail limits. If you removes the main power, then the chip supply voltage collapse to below that of backup battery and the gate protection diodes become conducting, limited only by the gate input series resistor of some 100 to 200Ω.

The current you saw is the conduction current via the protection diode to the rest of the main circuit.

I pulled down the Chip select line but the leakage current is still high, about 1 mA.
any more suggestions?

Would you expect the chip select to work if the chip has no power supply to it?

 

O.K, I understand that.
Do you have any idea how to avoid that conduction current via the protection diode?

Thanks

 

If you removes the main power, then the chip supply voltage collapse to below that of backup battery and the gate protection diodes become conducting, limited only by the gate input series resistor of some 100 to 200Ω.

I am not sure if that's the problem, with some exceptions. if the power is truly cut off via a switch to the rest of the circuit, obviously the battery leakage via the upper protection diode wouldn't be a problem: it couldn't form a loop to ground.

if the switch is on the primary side, a few possibilities:

a) the leakage in the filter caps. shouldn't be that big of a deal but it may.
b) a bleeder on the cap can be shorting the battery.

most rtcs require very little current. so the cheapest solution would be to put a large resistor in serial with the battery to power the rtc. the resistor doesn't really solve the problem but it will help contain the current drain.

 

Do you have any idea how to avoid that conduction current via the protection diode?

Sure.

You would need to "disconnect" the gate from the backup battery if there is no main supply.

Here I have used a small MOSFET to do that. The added MOSFET will not consume any current. The CMOS gate protection diodes are simulated. A quick simulation run of "Before & After" shows that the scheme is working.

You can use any small n-ch mosfet, like 2N700x or the BS170. The simulation also shows why there is a current flow(in R2) if the MOSFET is not in circuit, the situation you have to begin with.

 

eblc1388- Thanks for your effort.
Few question about your circuit:

1. Isn't the N-channel source should be connected to ground? (Low side switch), Otherwise the mosfet won't switch ON and OFF correctly. The analog switch output is connected to 100K resistor to ground.
2. What will be the switch status when the gate is floating and not 0v like what you simulated?

Thanks

 

Isn't the N-channel source should be connected to ground? (Low side switch), Otherwise the mosfet won't switch ON and OFF correctly. The analog switch output is connected to 100K resistor to ground.

The MOSFET is connected as a "high side switch", switching 1.6V into 100K resistor. Its gate is at (5-1.6)V higher than its source pin so it will conduct.

What will be the switch status when the gate is floating and not 0v like what you simulated?

What "gate" are you referring to? Why would it be floating?

 

The MOSFET is connected as a "high side switch", switching 1.6V into 100K resistor. Its gate is at (5-1.6)V higher than its source pin so it will conduct.

According your circuit R1 is connected to drain not to source. That's why I don't understand how you determine the Vgs.

I reffer to the n-channel "gate" that is connected to Vsupply. The Vsupply is floating during power OFF when there're no batteries inside the device.

 

According your circuit R1 is connected to drain not to source. That's why I don't understand how you determine the Vgs.

It does not matter. There is no current through R1.

In your final circuit, you don't need to use R1, I placed it there to let me "measure" the current in simulation only.

I reffer to the n-channel "gate" that is connected to Vsupply. The Vsupply is floating during power OFF when there're no batteries inside the device.

No, it will not float.

The MOSFET gate will be connected to the Vsupply, shared along with other components/devices. When there is no batteries inside the device, these connected devices will help to pull down the Vsupply to 0.6 volt. I have simulated this load by the 1KΩ at the left of the voltage source. Its value does not matter. 1K or 100K or 1M, it will pull down the Vsupply to 0.6V so the MOSFET gate pin will not float.

To see this happen in real life, you can place a voltmeter across the Vsupply and ground while the main batteries is there and then removes the batteries. See how the Vsupply voltage drops to about 0.6V.

 

It does not matter. There is no current through R1.

In your final circuit, you don't need to use R1, I placed it there to let me "measure" the current in simulation only..

Probably I don't understnad some mosfet basic.
Without R1 the 1.6v backup battery is connected to the mosfet Drain.
The Mosfet gate is connected to 5v.
So what is the source voltage to determine the Vgs?

No, it will not float.

The MOSFET gate will be connected to the Vsupply, shared along with other components/devices. When there is no batteries inside the device, these connected devices will help to pull down the Vsupply to 0.6 volt. I have simulated this load by the 1KΩ at the left of the voltage source. Its value does not matter. 1K or 100K or 1M, it will pull down the Vsupply to 0.6V so the MOSFET gate pin will not float.

To see this happen in real life, you can place a voltmeter across the Vsupply and ground while the main batteries is there and then removes the batteries. See how the Vsupply voltage drops to about 0.6V.

The device shelf life should be at least 2 years with backup battery and RTC. during that time I belive that the Vsupply voltage will be discharged completely. Does it matter?

 

Probably I don't understnad some mosfet basic.
Without R1 the 1.6v backup battery is connected to the mosfet Drain.
The Mosfet gate is connected to 5v.
So what is the source voltage to determine the Vgs?

The source voltage is that of the drain because the MOSFET is conducting. i.e. 1.6V battery terminal voltage.

Vgs is then 5V -1.6V = 3.4V, when the +5V main batteries is present.

The device shelf life should be at least 2 years with backup battery and RTC. during that time I belive that the Vsupply voltage will be discharged completely. Does it matter?

No. It won't matter.

Sorry we are on different wavelengths at this moment. Why mention the 2 year timing thing? The voltage decay of Vsupply will happen in seconds or a minute.

Just measure(as I have suggested in previous post) how long the Vsupply decays from +5V down to 0.6V in your existing circuit after you have removed the batteries and post back the result.

We will then pick up the discussion again.

 

The source voltage is that of the drain because the MOSFET is conducting. i.e. 1.6V battery terminal voltage...

But why it's conducting in the first place? The initial state of the mosfet is Non conducting. Than the 1.6v isn't present on the mosfet source.

 

But why it's conducting in the first place? The initial state of the mosfet is Non conducting. Than the 1.6v isn't present on the mosfet source.

Would the MOSFET be conducting if its source pin is at 0V?

If you are not sure, then its time to revise the MOSFET circuit theory.

 

Would the MOSFET be conducting if its source pin is at 0V?

If you are not sure, then its time to revise the MOSFET circuit theory.

Why do you think that the source pin is at 0V????
Doesn't it depend on the common output from the analog switch?