LT1001 precision current source : what voltage ? - Page 1
Forum section: electronics / projects

author: Overspeed
author_rank: Frequent Contributor
timestamp: January 25, 2023, 11:18:45 am
content: HelloI am not a electronic engineering but I try to understand and progressI try to built a precision current source from 25 mA to 500 mA base on LT1001 circuit from LT1001 specs doc , a similar subject have already been open with :LT1001 precision current sink: Why the JFET?on this forumMy questions are :1) does that possible to power the LT1001 in +/- 9V ? as I have a already available circuit ?2) does the +/- Vcc shall be be non symmetrical as usually Op Amp have symmetrical Vcc ?3) does the voltage Vin is negative as the formula is not clear V- + 1V ? or positive ?4) as the 2N3685 drive the 2N2219 , does that possible to use a bigger transistor than the 2N2219 as collector current is only 800mA and that a small TO39 , something in TO-220 with 1.5 amp ?Thanks in advanceRegardsOS

author: Kleinstein
author_rank: Super Contributor
timestamp: January 25, 2023, 11:32:15 am
content: The supply does not have to be symmetric. In most cased the OP-amp does not even see the ground / 0.A +-9 V supply is perfectly OK for the LT1001 or similar OP07.For the given circuit V_in should be negative. So with a +-9V supply between 0 and -8 V. As the positive voltage range is not used much an asymmetric supply, like +5 V and -12 V would perfectly make sense for this circuit.One can use a bigger transistor (e.g. TO220 case). However the FET still has to drive the base current of output current divided by the gain factor. The maximum current of the JFET is somewhat limited and for really high current one may need a darlington transistor.For the transistor it is not just the current rating that matters. Often the power dissipation is the more limiting factor and one usually does not have perfectly cooling and thus can use only something like half of the Ptot rating.

author: David Hess
author_rank: Super Contributor
timestamp: January 25, 2023, 03:16:31 pm
content: Quote from: Overspeed on January 25, 2023, 11:18:45 amThe JFET removes the error contribution from the base current of the bipolar transistor, which would otherwise add to the output current. It effectively makes a bipolar transistor with zero base current.The JFET also increases the load resistance on the operational amplifier which improves open loop gain and thereby precision.QuoteThe LT1001 will operate fine on lower voltages. Lower supply voltages create an input offset voltage error based on the power supply rejection, but this error is small and can be ignored.QuoteAsymmetrical supply voltages create an input offset error based on the common mode rejection but this error is small and can be ignored.QuoteSure, that will work, and would also be desirable because of lower cost and increased power dissipation capability.

author: mawyatt
author_rank: Super Contributor
timestamp: January 25, 2023, 03:25:19 pm
content: Quote from: Overspeed on January 25, 2023, 11:18:45 amOne could replace both with a NMOS Power device, just make sure you have enough Vcc to support the NMOS Vgs at desired max current.Best