Does a FET act like a resistor at low currents?

[uoficowboy]

Hi - in some battery powered applications I would like to put a voltage divider on a battery and measure the voltage on that divider with a microcontroller to monitor the battery. My question is this: I would like to not have the resistors burning power 100% of the time. So - I was thinking that it might be possible to switch the bottom side of the divider with a FET. My worry is that I'm not sure if a FET acts just like a resistor at low currents, or if there are additional effects that I need to be worried about.

For example, let's say I want to measure a 9V battery with a microcontroller that has an ADC reference voltage of 5V. I might make my resistive divider be two 100K resistors, with a BSS138 (RDSon = ~4 ohms). Can I assume that the FET will essentially not effect the circuit at all? I mean, would the voltage drop across the FET will be something like 9V * 4/(100K + 100K + 4)? Or are there other effects I need to be worried about?

Thanks!

 

[ALERTLINKS]

This application is using FET as a switch, which is most common use of FET. In mcu when you set
reference voltage for a comparator,it is same type of application. Detail of this module can be seen in datasheets. Digital ics use FETs for logic level switching. Fet can also be driven in analoge region. It is used as variable resistor in feedback control circuits replacing LDRs, thermistors etc.

 

[FvM]

Switching the bottom resistor will hardly have the intended effect, because the top resistor continues to draw current from the battery with the divider output clamped at the ADC input. Disconnecting the high side is the only reasonable way, usually with a PMOSFET. Besides some minimal leakage currents, there should be no problems involved. Particularly no offset voltages as BJT switches do.

 

[kak111]

Fets as voltage controlled resistors

https://www.qsl.net/n4xy/PDFs/Semic...otes/Vishay-Silcnx_FET_VoltCntrlRES_AN105.pdf

https://arxiv.org/ftp/arxiv/papers/0708/0708.3498.pdf

 

[poorchava]

Generally it depends on wheter you want to measure dc or ac signal. For DC it is common practice to use JFET as voltage modulated resistance. For AC signals (or signals swinging from below and above ground potential) it is possible to use JFET as described in AN105, althrough it is far more convenient to use a photoresistive optocoupler (diode + photoresistor optically coupled in one package). On the other hand photoresistor turn-off resistance can't get anywhere close to that of a FET transistor which disqualifies it as switching element.

 

[FvM]

Although many correct points have been mentioned about FETs as controllable resistors, they don't apply to the original problem which is about using a FET as an analog switch. I suggest to read more than the (actually misleading) thread title.

 

[alexan_e]

An alternative solution would be to use a high value resistive divider so that the current consumption is minimum, for example a 100K divider (or more) will have a consumption of just 120uA for 12v battery.
The problem in this case is that the impedance presented to the ADC input will be fairly high compared to the 10K that is usually specified as the max source impedance for mcu ADC, you can easily solve this problem by using a small capacitor in parallel with the ADC input.
Since you measure DC this capacitor will not be a problem and will be able to effectively present a low impedance to the ADC input.

Alex

 

[FvM]

The high resistance divider is a solution. The original post was assuming however a 9V battery, which has usually about 500 mAh capacity. So the the battery will be exhausted after 8 month. Rather annoying.

 

[uoficowboy]

Switching the bottom resistor will hardly have the intended effect, because the top resistor continues to draw current from the battery with the divider output clamped at the ADC input. Disconnecting the high side is the only reasonable way, usually with a PMOSFET. Besides some minimal leakage currents, there should be no problems involved. Particularly no offset voltages as BJT switches do.

Hi FvM - that is a good point. But in some situations this circuit will still work, I believe. For example, I often want to measure the voltage of a single LiPo battery on a 3.3V MCU. Many MCUs have 5V tolerant inputs - so there shouldn't be any worry about clamping diodes. In this situation - would the FET work how I described? The reason I ask is that my understanding of BJTs is that they would not function this way (that there is sort of an offset voltage to a BJT) - but I don't remember encountering any time where a FET had anything but a fairly constant resistance.

Thanks!

 

[alexan_e]

The project involves an mcu which will consume several mA, if we assume two 100K resistors in the divider then the consumption will be 9v/200K=45uA, this current compared with the mcu consumption would probably be about 1% (if we assume 4.5mA for the mcu) so the battery will last 1% less time, doesn't seem a big deal to me.
The OP said that this is a battery powered application so I guess that the mcu circuit consumption will be far worse than the above assumption of 4.5mA so the difference that the divider will make will be even less.

Alex

 

[uoficowboy]

The project involves an mcu which will consume several mA, if we assume two 100K resistors in the divider then the consumption will be 9v/200K=45uA, this current compared with the mcu consumption would probably be about 1% (if we assume 4.5mA for the mcu) so the battery will last 1% less time, doesn't seem a big deal to me.
The OP said that this is a battery powered application so I guess that the mcu circuit consumption will be far worse than the above assumption of 4.5mA so the difference that the divider will make will be even less.

Alex

Hi Alex - most of my professional work is on battery powered devices. Ultra low power battery powered devices. In sleep mode (where they spend most of their lives) their current draw is typically less than 1ua. So you can understand why I don't want a voltage divider always on my battery

 

[alexan_e]

In that case you have to use the Pmosfet as FvM has suggested but this mosfet will be connected to 9v (mosfet source) so you will need an additional control device to turn the mosfet on/off.
If you use a pullup resistor for the gate to have 9v to the gate so that the mosfet is off by default then you will need a device (like a Nmosfet) controlled by the 3.3v of the mcu that will be able to ground the gate of the Pmosfet to turn it on.

Alex

 

[poorchava]

I guess you could go with voltage divider, but with much higher resistance (in order of 10 M?) and then buffered with some low power opamp. The opamp should have sleep/disable function. Further, the inputs should be of FET type to so that the input resistance of opamp doesn't alter the divider ratio. I.e. using 10M divider introduces additional 500 nA current, which isn't really that much. I guess you could even go with 100M but getting precision resistor in that order of magnitude may be hard, not to mention insulation resistances and other parasitic effects. On the other hand that would introduce only 50 nA of additional current which is really small (at least IMO)

Another matter is the frequency of signal. Precision, jfet input, low power amplifiers with high bandwidth tend to cost truckloads of money.

 

[uoficowboy]

In that case you have to use the Pmosfet as FvM has suggested but this mosfet will be connected to 9v (mosfet source) so you will need an additional control device to turn the mosfet on/off.
If you use a pullup resistor for the gate to have 9v to the gate so that the mosfet is off by default then you will need a device (like a Nmosfet) controlled by the 3.3v of the mcu that will be able to ground the gate of the Pmosfet to turn it on.

Alex

Hi Alex - I recognize that the P-FET on top would be best in the original situation I described - but my original question remains somewhat: would there be no voltage drop across the FET? (assuming the on resistance of the FET is negligible in comparison to the voltage divider resistance)

Thanks!

 

[alexan_e]

The voltage drop will be the same as a 4ohm resistor which will be negligible in a 100K divider.

Alex