I need to build a circuit that will take 0-5 VDC input, multiply it by 2, add 10, and provide that value as the output VDC. I plan on using this with an air-fuel ratio measuring unit that I have. It provides the 0-5 VDC and I want to scale that to use with a automotive digital voltmeter that fits in the dashboard. Thanks.

 

If I understand correctly, your digital voltmeter measures 20V full scale so you want the signal scaled accordingly.

It can be done but you will need a 20V minimum power supply. A boost switching supply can do this but it's not trivial or cheap.

Seems like it would be easier to modify the voltmeter to read 0-5V or use a different voltmeter that does.

 

You are correct about the 20V full scale. I won't have a 20V power supply available, nor do I want to get into anything exotic or expensive to build. I have no idea how easily such a task might be accomplished, but it sounds like it wouldn't be very practical.

Would it be any easier to build a circuit if all I'm looking for is to take the input 0-5VDC and double it for the output? That would be a compromise I could accept as the reading on the voltmeter would only be off by a consistent 10 volts, i.e. 2.2 VDC input would read 4.4 on the output, meaning an air-fuel ratio of 14.4.

Thanks.

 

Since the current is very small, possibly a voltage pump such as the LT1054 would easily double the automobile voltage of 12 to 24.

__________________
The great thing about electronics is unlimited ways to do the job. The only limit is one\'s imagination. I generally think my way is best.
Show me a different way. I have an open mind.

 

Good idea. I had looked at Linear Tech's switched-cap IC's but didn't see that the LT1054 could be used to double 12V. Should work.

Then an op amp can be used to add the gain and offset to the signal.

 

The LT1054CP is made by TI and is a little cheaper than the Linear Tech's

__________________
The great thing about electronics is unlimited ways to do the job. The only limit is one\'s imagination. I generally think my way is best.
Show me a different way. I have an open mind.

 

If you could get into the voltmeter and force the tens digit to a "1", then you could get by with a gain of 2 (0-10V). You probably don't have enough experience to do this.

__________________
Ron

 

The voltmeter might be any one of several scales (such as 0-5V) internally. If you can get into the voltmeter, find the voltage divider and rework it.

 

The problem is that he wants to read out fuel-air ratio which requires a 10V offset and a gain of two. That can't be achieved solely by changing the internal voltmeter sensitivity.

 

If he could scale the meter to 0-5V on the internal attenuator (most digital voltmeter modules are 0-200mV without the attenuator), and force the tens digit to "1", he wouldn't need any gain, offset, or higher voltage supply.

__________________
Ron

 

Thanks for all the suggestions. It doesn't sound like there's an easy solution.

 

This is not a difficult circuit to build! You described a linear system.
Your output equation is simply:

Vout=[2Vin+10] volts, where Vin=0-5 Vdc

Dagwood a 24 volt or higher IC DC-DC converter could be used for your V+ inputs to (dual or quad) single supply (24 volt or higher), rail to rail op-amp circuits. Your 0-5 volt input would be feed to an inverting amp (gain=2) with this output feed to an inverting summing amp (gain=10), the output:

Vout=2Vin+10 volts

You may or may not, depending on the sensor and digital volt meter specs need to design an appropriate interface between your designed circuit and these input/output devices-for proper loading, buffer etc?
Some suppliers give free samples ( ex. microchip ), I recieved 5 free 22 uH inductors for a dc-dc converter, just by asking and without ordering anything. Though the prices for the dc-dc converter and op-amp IC along with any support components needed for this project are reasonable.