Alice; Ah! you want to drive a pressure transducer. Is it a bridge style? And you hav e a 1.5V rail? Why not connect the + on the transducer to the rail and generate a negative on the - side? If you use an op-amp in a servo setup, you could drive the - terminal until one of the return terminals is regulated to 0VDC, and the other terminal would then be a ground referenced signal. Linear Technology (ref Jim Williams, bridge transducer circuits) is fond of these methods. If you step up to 3V, you would have to deal with a 1.5V common mode with a 1.5V rail, which is ugly. With the ground reference, you can then avoid an instrumentation amplifier. If you need to measure bipolar signals, you could be in the thick of it again, as your PIC HATES any negative voltages on an analog pin. Whacks the other analog readings REAL fast. Would this circuit have poor performance in light of the total V applied to the sensor varying? Do you have to use a bridge with four varying elements to avoid this? Hmm... Chris Eddy Pioneer Microsystems, Inc. Alice Campbell wrote: > > Date: Mon, 8 Nov 1999 17:26:08 -0500 > > Reply-to: pic microcontroller discussion list > > From: Wagner Lipnharski > > Organization: UST Research Inc. > > Subject: Re: [OT] Ringing Chokes and batteries > > To: PICLIST@MITVMA.MIT.EDU > > Well, i'm about to replace the 9v battery in my datalogger with two > AA 1.5v and a stepper to 5v, but i was curious how they worked, and > wanted to see for mysef. I've been messing with switched capacitor > circuits, and saw the inductor circuit, but couldnt get it to work. > Then i wondered if i could run it at 3v, stepped up from 1.5v, and > let the pressure transducer work ratiometrically. The pressure sensor > only pulls a few mA, the pic and adc and memory chip only a few mA. > The problem with the switched cap design is the oscillator part, the > self-oscillation of the coil looked nicer. > > alice > > > Hello dear Alice, looking for a small step up conversion circuit? > > What are you trying to power up? > > What is the input and output Voltage x Current? > > :) > > > .