All text, can't draw a schematic where I am right now.
Without an opamp, differential pair, or some other zero-headroom (zero as in way less than Vbe) gain device, a constant current regulator has limitations. When using a transistor's Vbe as both the reference voltage and the setpoint, first to go is the idea of regulating down to zero or very low output current. For example, if you use the standard 2-transistor CC circuit, you can put a low power pot across the shunt resistor for the kind of adjustability I think the OP wants. But the minimum regulated output current always will be Vbe / Rshunt. For example, a 6.5 ohm shunt would yield a minimum regulated output of 0.1 A, a 65 ohm shunt would yield a min reg output of 10 mA, etc. And while 10 mA is a nice min value for a low energy circuit, for the same circuit adjusted to 100 mA the shunt would have to have 6.5 V across it, which is a large headroom requirement for a relatively low output current.
In post #1 here:
http://cr4.globalspec.com/thread/57232/need-high-dc-voltage-low-constant-current-diagram
the top right hand circuit is closest to what I'm describing. The 3K shunt resistor sets the minimum current, in this case only 200 microamps. Whatever the shunt is, if you bridge it with a higher-value pot and connect the wiper to the upper transistor's base, you now have an adjustable output current without all of the current going through the pot, which is the original question here.
Another way is to use a regulated voltage and a resistor to create a regulated current that drives a current mirror:
http://circuit-diagram.hqew.net/Current-sources-and-mirrors_14545.html
Circuits 6 and 7 will work for this thread. You can turn them from sinks to sources by changing to PNP's.
Commandeered a nurse's station, Quickie circuit, not tweaked for power levels, component selection, etc.; just a concept drawing with parts in my design library.
ak
Without an opamp, differential pair, or some other zero-headroom (zero as in way less than Vbe) gain device, a constant current regulator has limitations. When using a transistor's Vbe as both the reference voltage and the setpoint, first to go is the idea of regulating down to zero or very low output current. For example, if you use the standard 2-transistor CC circuit, you can put a low power pot across the shunt resistor for the kind of adjustability I think the OP wants. But the minimum regulated output current always will be Vbe / Rshunt. For example, a 6.5 ohm shunt would yield a minimum regulated output of 0.1 A, a 65 ohm shunt would yield a min reg output of 10 mA, etc. And while 10 mA is a nice min value for a low energy circuit, for the same circuit adjusted to 100 mA the shunt would have to have 6.5 V across it, which is a large headroom requirement for a relatively low output current.
In post #1 here:
http://cr4.globalspec.com/thread/57232/need-high-dc-voltage-low-constant-current-diagram
the top right hand circuit is closest to what I'm describing. The 3K shunt resistor sets the minimum current, in this case only 200 microamps. Whatever the shunt is, if you bridge it with a higher-value pot and connect the wiper to the upper transistor's base, you now have an adjustable output current without all of the current going through the pot, which is the original question here.
Another way is to use a regulated voltage and a resistor to create a regulated current that drives a current mirror:
http://circuit-diagram.hqew.net/Current-sources-and-mirrors_14545.html
Circuits 6 and 7 will work for this thread. You can turn them from sinks to sources by changing to PNP's.
Commandeered a nurse's station, Quickie circuit, not tweaked for power levels, component selection, etc.; just a concept drawing with parts in my design library.
ak
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