Hello Everyone,

Background on this question: I am currently working on restoring a car. I'm working on designing a circuit to replicate the old fuel sending units output so I can upgrade the sending unit and still use the stock gauges. The old sending unit was a resistive type. It is wired in such a way that it makes up the lower half of a voltage divider circuit. The upper half is embedded in the gauge unit.

Here is my circuit to spoof the gauge.


The 130 ohm resistor is embedded in the gauge cluster. The input of the transistor will be driven with a DAC (MCP4725). So what I want is Vout to swing as close to 0V as possible. From testing I know that the Empty state will register if Vout is equal to approximately 0.46V. However, with the current transistor I can only get Vout to go down to around 0.8V ish.

Is there a way of getting Vout all the way down to 0.46 V without using a germanium type transistor? I have tried simulating with MOSFETs instead of the PNP transistor and the lowest I could get Vout was around 0.6V.

Thanks!!

 

PNP is hooked up backwards, try NPN. Also assuming 12 volts and 130 homes implies 12/130. -> 92 mA or so.
To have an output voltage of say .4 volts means a maximum resistance if using a MOSFET of .4/.092 -> about
4 ohms. What MOSFET did you try?

 

Guess I should clarify. This is not a simple switch circuit. Vout needs to swing from around 0.46V to 5V according to the voltage on its base (I think this configuration is also referred to as an emitter follower). ~0.46V is what the gauge sees as empty, ~5V is what the gauge sees as full.

The voltage at the top of the divider is also 5V not 12V.

As far as the MOSFET I tried, I actually tried just about every one that comes with LTSpice... So I guess what you are saying though is I would need a Rds MAX of 11 ohms? (5V/130Ohms = 38mA. 0.4V/0.038A ~= 11ohms)

 

You are using a PNP transistor so it is acting as an inverted emitter-follower circuit.
Even if you correct the transistor correctly (emitter to R1, collector to ground) the minimum output will still be about 0.7V.

MOSFETs won't work as source followers here, since there gate-source voltage is generally greater than the base-emitter voltage of a BJT.

Alternatives to get a lower voltage are to use a germanium transistor or add an op amp circuit.

Another alternate shows the LTspice sim of a circuit below, using two NPN and one PNP silicon transistor in a current-mirror configuration that goes down to about 170mV minimum:
You may have to tweak the value of R3 to get the desired output span.

Edit: For best operation, you can use dual matched transistors in one case for the 2N3904's, such as the DMMT3904
Edit2: Changed sim to use proper 130Ω value for R1.

 

Is the 130Ω resistor the resistance of the gauge itself? Or is it a different resistor in the cluster?
What is the full-scale-deflection current of the new gauge? What is its resistance?

 

Another alternate shows the LTspice sim of a circuit below, using two NPN and one PNP silicon transistor in a current-mirror configuration that goes down to about 155mV minimum:

Ah man that is Perfect! Exactly what I was looking for! Thanks so much!

 

See my revised sim in post #4.

 

This should work great for what I wanted to do! However, just for curiosity's sake, what kind of configuration would I need to make this work with an op amp? If there is a name for it anyway..

 

what kind of configuration would I need to make this work with an op amp?

Below is an example circuit:
It could be called an op amp non-inverting circuit with a buffer sink-transistor output.

The circuit output follows the input nearly exactly.
The minimum output is <70mV.
The capacitors are needed to stabilize the circuit due to the added loop gain from the common-emitter transistor.
The op amp is an inexpensive rail-rail type.