- Joined Jan 7, 2020
Moderation: Link to old thread. @crutschowAfter some additional thought I believe the circuit below can give you a nearly full range current limit without any added power dissipation.
The circuit senses the voltage across the 10mΩ shunt resistor in the ground side of the load.
The Trim connection is the Trim input on the DC-DC converter.
(There is also a resistor to ground from the Trim input to determine the normal output voltage).
The load current across the 10mΩ shunt resistor is amplified by U1.
When this voltage (buffered/isolated by U2 and D1) exceeds the Trim reference voltage of 0.591V it will override that voltage, causing the DC-DC converter to reduce its voltage, thus reducing the current.
The adjust range is from ≈100mA to the current limit of the DC-DC (no added limit).
The one disadvantage of this circuit is that the output limit current is a very non-linear function of the pot position.
A logarithmic (audio taper) pot would thus be better for U3.
V1 is the input power to the DC-DC.
Note that the ground side of R_Shunt should be connected directly to the ground terminal on the DC-DC converter.
View attachment 158354
I bumped into this post the other day. Seem to be almost exactly what i'm looking for.
I have a DC power supply (Turnigy 540W) that have a variable resistor to vary the output voltage from 12 to 17 volts.
It can deliver 30 Amps
The problem is that its not current limited.
The power supply have a display that shows me the output voltage and the current.
I would like to make a current limiting function to this power supply.
I found out that the display have an op amp OP07 with single supply +5V.
The output is a function of the current.
0 V = 0 A, 1 V = 16 A, 1,55 V = 24 A
I also found out that the variable resistor for setting output voltage is 10K Ohm
When I measure the terminals of the resistor in different setting I got this:
Resistance, Voltage at pot pin, Output voltage
9150 Ohm, 1.1 V, 12 V
335 Ohm, 0,18 V, 16.8 V
0 Ohm, 0 V, 17.8 V
If I disconnect the potmeter I got 1.33 V at pin and 10.8 Volt power supply output.
I also tried to remove the potmeter completely and apply voltage directly to the pin and got the same function at the table over.
From my little knowledge of op amps I think it would be possible to make a small circuit that will lower the output voltage (increes "pin voltage") when the current goes over the limit that I have chosen.
Where to start ?
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