Question: Accurately altering circuit resistance

[Mhorton91]

Hoping this is a simple question for people in the EE field... I study math, so I need some help in the real world (haha).

I'm working on a project car that I'm trying to build over the summer, and I'm putting aftermarket gauges in it. Here is my issue.

My factory fuel sending unit has a resistance range from 0 ohms empty, to 280 ohms full... and that's a fairly uncommon range, so the gauges that offer that are pretty pricey (pretty much only made by Autometer, for anyone interested)... I'm really hoping there is a way to alter the circuit coming out of my sending unit. The gauge I want to run has a range of 0 ohms empty, to 90 ohms full... I assume it would be easier to add resistance instead of decreasing, but I'm assuming (read: hoping) there is a way to do it.

Hoping someone can help out, with either an answer, or a method to figure it out myself.

Edit-- For anyone unfamiliar, Automotive wiring is 12 VDC. I assume that makes a difference in a calculation such as this.Thanks!
Marshall

 

[Mhorton91]

Not one to just look for answers from others... I've done some reading, and it seems the place to start is I = V/R... Which is good, except when the gauge reads empty, I end up with division by zero... so I'm trying to decide where to go from there.

Edit- Now I'm seeing that Ohm's law is sort of like Hooke's law... it only holds in very specific situations...

2nd edit... I found that R₂ = (R₁ * R_T)/(R₁ - R_T)

Which works out to 132.6315... ohms, a calculation I found for power dissipation (?) is P=V²/R... Which looks like said resistor would need to be rated for something like 1 watt...

Am I getting close to something?

 

[jim hardy]

Post a link to how your gage works?

We don't know if your car uses 5V for instruments or another voltage

or what the gage does with resistance

also what are ohms on your sender at 3/4, 12, 1/4 full?
You'll want to maintain linearity i would think

 

[Baluncore]

With this type of gauge circuit you can use a single resistor in parallel to get the full scale correct, but you will get very poor linearity. The simple circuit here will give a much better linearity. It works by scaling the effective resistance over the full range. Less than one volt will be dropped across the transistor, that is why I have reduced the parallel R2 from your 132.63R to about 125R. Adjust R2 to fix any full scale error. The transistor can be low power because it drops so little voltage in this configuration.

[Edited, see later post].

 

[Mhorton91]

Post a link to how your gage works?

We don't know if your car uses 5V for instruments or another voltage

or what the gage does with resistance

also what are ohms on your sender at 3/4, 12, 1/4 full?
You'll want to maintain linearity i would think

I will [attempt] to get some accurate measurements when I go back out to my dad's house tomorrow morning. I haven't been able to find the information online.

With this type of gauge circuit you can use a single resistor in parallel to get the full scale correct, but you will get very poor linearity. The simple circuit here will give a much better linearity. It works by scaling the effective resistance over the full range. Less than one volt will be dropped across the transistor, that is why I have reduced the parallel R2 from your 132.63R to about 125R. Adjust R2 to fix any full scale error. The transistor can be low power because it drops so little voltage in this configuration.
https://www.physicsforums.com/attachments/104135

Thank you for this! This electronics stuff is incredibly interesting! I'm actually going to Radioshack to buy a book that has been recommended called Getting Started in Electronics.. I will pick up some npn transistors, and various resistors to attempt to build this circuit and test it out!

In your diagram, does the arrow at the bottom represent the direction heading towards the gauge?

 

[Baluncore]

In your diagram, does the arrow at the bottom represent the direction heading towards the gauge?

No. The sender is usually connected to the chassis = ground at one end. The arrow at the bottom is that ground.
The gauge usually goes between the top terminal of my diagram and a regulated positive voltage in the instrument panel.

 

[Mhorton91]

No. The sender is usually connected to the chassis = ground at one end. The arrow at the bottom is that ground.
The gauge usually goes between the top terminal of my diagram and a regulated positive voltage in the instrument panel.

Thank you for the clarification! I feel 100% in over my head as far as actually getting your circuit to work, but I'm absolutely still going to try.

 

[Baluncore]

Linearising a bi-metalic gauge with a mismatched resistance sender is a bit more involved than expected.

In this circuit the regulated instrument supply is the top of the diagram, the chassis is at the bottom.
The old 0 to 280 ohm sender is Rsender with one side connected to ground. The new gauge is Rgauge.

A resistor, Rcalib is used that has the same resistance as the old gauge series resistance. This can be selected to calibrate the system. It will need to be a power resistor as it replaces a resistor heated bimetallic strip. Do not use a variable resistor as it could be adjusted to zero at the same time as Rsender was zero. If possible measure the old gauge resistance to find the appropriate value.

How it works. Rcalib and Rsender makes a voltage divider. The op-amp controls the gate of M1, the N channel MOSFET, so as to make the drain voltage the same as the potential divider. So M1 is pretending to be a lower resistance version of the sender.

 

[Baluncore]

This unusual circuit makes a 0 to 280 ohm sender look like a 0 to 93.3 ohm sender.
How does it work? It is all done with current mirrors.

It uses PNP transistors because one side of the sender is grounded.
The current through the old sender flows down through Q1 with the Q1 base current being provided by Q4.
Q1, Q2 and Q3 all have the same base emitter voltages so all have the same collector current in parallel. The total current through the three upper transistors is three times that through the sender, so the sender appears like it has one third the resistance. Zero to 280 / 3 = zero to 93.3 ohms, which is probably close enough to 90 ohms.

Q5 is different, it needs to be a higher power transistor. It keeps the voltages across the Q2 and Q3 transistor collectors the same as Q1. That keeps the power dissipation in all the upper transistors similar and so stops thermal errors or runaway.

 

[Mhorton91]

Thank you, Baluncore! I'm definitely going to try my hand at assembling this voltage mirroring circuit! It looks the cleanest, to me atleast. I stopped by Radioshack to look around today, they were completely out of PNP transistors, so I'm going to try the other location tomorrow, or I may try to order them online.

 

[Baluncore]

The op-amp and MOSFET circuit does an accurate job without any extra voltage drops.
The mirror circuit will drop about 1 volt of the available instrument voltage but that should be close enough for a fuel gauge.

If I was building the mirror circuit today in Europe or Australia I would build it using BC557 transistors for Q1,2,3 & 4. I would select BC640, a 1A, 1W audio output transistor for Q5. The mirror circuit could be built entirely out of BC640 but they are more expensive.

I am not sure of the PNP equivalent transistors commonly available in the USA.
It would be easier if one side of the sender was not connected to the tank and chassis, as then the common NPN 2N2222A might be used.

 

[Mhorton91]

The op-amp and MOSFET circuit does an accurate job without any extra voltage drops.
The mirror circuit will drop about 1 volt of the available instrument voltage but that should be close enough for a fuel gauge.

If I was building the mirror circuit today in Europe or Australia I would build it using BC557 transistors for Q1,2,3 & 4. I would select BC640, a 1A, 1W audio output transistor for Q5. The mirror circuit could be built entirely out of BC640 but they are more expensive.

I am not sure of the PNP equivalent transistors commonly available in the USA.
It would be easier if one side of the sender was not connected to the tank and chassis, as then the common NPN 2N2222A might be used.

I'm not sure what is meant by the mirror circuit dropping 1 volt, is this a potentially dangerous situation?

If it makes a difference, my factory instrument panel voltage supply wire (that I plan to reuse to supply power to all the gauges) gets a full 12.4v. Does the mirror circuit dropping a volt mean that the fuel gauge would only receive 11.4v?

I located the transistors you recommended online, however, if the voltage drop is a potential issue, and the op-amp and MOSFET circuit could be viewed as a better option, how would I go about deciding on a specific op-amp, and n channel MOSFET.. I scrolled through several pages of both on ebay and saw several different specs that as of now I know nothing about.

Thank you.
Marshall Off topic, but this very short conversation has me signed up for an info session in the engineering department, this is amazing stuff.

 

[Baluncore]

I'm not sure what is meant by the mirror circuit dropping 1 volt, is this a potentially dangerous situation?
If it makes a difference, my factory instrument panel voltage supply wire (that I plan to reuse to supply power to all the gauges) gets a full 12.4v. Does the mirror circuit dropping a volt mean that the fuel gauge would only receive 11.4v?

It is not dangerous. The thing about gauges is that the voltage needs to be regulated better than the +/–1 volt. It is important that the voltage does not jump around as the regulator switches in and out at idle. With bi-metallic gauges the ambient temperature often has more effect than instrument voltage.

The resistance of the gauge is in series above the mirror so the full 12.4V will never appear across the sender or mirror transistors. The mirror circuit only inserts two transistor Vbe voltage drops into the sender to gauge path. That totals about 1.2 volt and will not be a problem. The advantage of the mirror circuit is that it is very simple and does not need a power supply.

Students get taught to look at voltages across resistors with an oscilloscope, they do not get taught to think of currents flowing through PN junctions in current mirrors. The problem comes when those who try to modify the mirror circuit do not understand how it actually works.
https://en.wikipedia.org/wiki/Wilson_current_mirror
https://en.wikipedia.org/wiki/Current_mirror

See if you can get the mirror working first. It will certainly be accurate when the tank is near empty. The error at the full end will be the width of the pointer or less, which does not matter since you know when you have filled the tank. Getting the ends of the scale correct is relatively easy, the linearity problem that this circuit fixes it to keep the pointer close to the half full mark when the tank is actually half full.

 

[Mhorton91]

It is not dangerous. The thing about gauges is that the voltage needs to be regulated better than the +/–1 volt. It is important that the voltage does not jump around as the regulator switches in and out at idle. With bi-metallic gauges the ambient temperature often has more effect than instrument voltage.

The resistance of the gauge is in series above the mirror so the full 12.4V will never appear across the sender or mirror transistors. The mirror circuit only inserts two transistor Vbe voltage drops into the sender to gauge path. That totals about 1.2 volt and will not be a problem. The advantage of the mirror circuit is that it is very simple and does not need a power supply.

Students get taught to look at voltages across resistors with an oscilloscope, they do not get taught to think of currents flowing through PN junctions in current mirrors. The problem comes when those who try to modify the mirror circuit do not understand how it actually works.
https://en.wikipedia.org/wiki/Wilson_current_mirror
https://en.wikipedia.org/wiki/Current_mirror

See if you can get the mirror working first. It will certainly be accurate when the tank is near empty. The error at the full end will be the width of the pointer or less, which does not matter since you know when you have filled the tank. Getting the ends of the scale correct is relatively easy, the linearity problem that this circuit fixes it to keep the pointer close to the half full mark when the tank is actually half full.

I will read those articles! I will also plan on sticking with the voltage mirror. I searched online and found an this site that does transistor cross reference (http://alltransistors.com/crsearch....&ueb=0&ic=0&tj=0&ft=0&cc=0&hfe=0&caps=&page=0)

I'm assuming any equivalent transistor to the BC557 will work for the Q1, Q2, Q3, and Q4... and any equivalent to BC640 for Q5?

 

[Baluncore]

I'm assuming any equivalent transistor to the BC557 will work for the Q1, Q2, Q3, and Q4... and any equivalent to BC640 for Q5?

I selected those because they have sufficient voltage to survive in an automotive environment. Keep Q1,2,3 thermally close together. Keep Q5 away from the others to prevent it differentially heating the others.
We do not yet know the new gauge resistance so we do not know the maximum gauge current. Use BC640 for all until you can find the maximum gauge current.