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I want to test a Constant Current LED Driver across its voltage range without using a LED for test purpose.

How can I clamp the output voltage of this driver to required voltage levels? For Example - LED Driver Ratings - 800mA Constant Current Voltage Range - 17V to 34V. Now how do I test this driver at 17V, 18V etc to make sure this driver works under different voltages.

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I also tried to clamp the voltage using an Op-amp but in vain. I am not sure if output voltage is stiff enough to clamp LED driver output voltage. Also, current is getting drawn from the output of Opamp whereas opamp is primarily used to just clamp the voltage.

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Any thoughts on how this can be achieved?

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Set your electronic load to resistance mode and adjust the synthetic resistance to get the desired voltage drop.

I think this is probably best general answer for almost all of those with commercial electronic loads. All I have used have constant current, constant resistance and constant power modes. That's the whole point of such devices- to act as flexible loads- most will also allow you to simulate changing loads etc. If yours does not for some reason, keep reading:


If yours does not, you can use a shunt regulator based on an op-amp with a PNP darlington (and a reference voltage) or use the ciruit in the TL431 data sheet, substituting a PNP darlington (eg. TIP125) for the PNP transistor and choosing the B-E resistor so that the TL431 always conducts at least 1mA.

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The resistor in series with Vi is not required in this application- the series impedance of the current source takes its place (and should be quite high in dynamic resistance if it is a good constant current source). Vref for the TL431 is about 2.5V (2.495V nominally). Depending on the discrepancy between the load and source the transistor could see a lot of dissipation (18V * 0.8A = 14.4W if the load was completely disconnected). That would require a large heatsink. If the load is set to a higher current than the source, the output voltage will collapse which may cause your source to misbehave, or worse.

If your "load" is just a current sink (CC mode only) it is not very appropriate for this purpose). You might as well just use the above circuit alone with a suitable heatsink.

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  • \$\begingroup\$ What do you mean by synthetic resistance? Is there some kind of variable resistance which I vary to achieve this voltage drop? \$\endgroup\$ – Mithun N Aug 23 '17 at 3:09
  • \$\begingroup\$ Also my electronic Load does not have a LED mode. Also, how do I realize this in circuit? \$\endgroup\$ – Mithun N Aug 23 '17 at 3:10
  • \$\begingroup\$ Most electronic loads have a CR (constant resistance) mode, as used for testing (for example) primary batteries to data sheet specifications. If you set 20 ohms and your driver supplies 800mA the voltage drop should be 16V. \$\endgroup\$ – Spehro Pefhany Aug 23 '17 at 3:28
  • \$\begingroup\$ But most Electronic Loads use some kind of MOSFET for their operation. Here current is controlled and in Constant Resistance mode, only current is varied with applied voltage to maintain constant resistance. MOSFET's used in E'load tend to have very less Rds because of which voltage drop across MOSFET due to Rds will be small. So there is no real voltage drop happening in this case. Can you please comment and correct on this point? \$\endgroup\$ – Mithun N Aug 23 '17 at 3:52
  • \$\begingroup\$ In CR mode, the current is controlled, but the voltage drop is also considered, in order to maintain (as much as possible) a constant ratio of V/I, which represents the synthetic resistance. Rds(on) of the MOSFET(s) plus the shunt resistor represent the lower limit to the set resistance. Similarly, in constant power mode, the current is controlled to maintain a constant V*I product, which simulates the load represented by a switching power supply. \$\endgroup\$ – Spehro Pefhany Aug 23 '17 at 3:58
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The fancy way would be to use a 2-quadrant or a 4-quadrant power supply. These will sink current as well as source current. Although they're not exactly like your LED (because they source), they would still test your LED driver.

But a real hands-on engineer would roll his own. Here's the simplest circuit I could think of:

schematic

simulate this circuit – Schematic created using CircuitLab

Basically, the transistors start to pull down on "In" when the voltage exceeds about 17.2 Volts. Q1 is an old giant transistor in a TO-3 case. Bolt Q1 to a huge heatsink and aim a fan at it. You're specs mean a lot of power. 27 Watts is a lot. Q2 increases the lousy gain of Q1 and greatly reduces the power zener D1 has to handle.

You would have to change the zener diode D1 to get different voltages.

You could also skip Q2 and just use a zener rated for several Watts.

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You start a design validation test or DVT plan with specs then realize these variables for Vin, Vout, Iout & Pd while changing each variable.

To simulate LEDs one can use a fixed voltage sink with adequate power such as an adjustable negative LDO for a +ve CC source or a positive LO for a -be CC sink.

But then you dont really need an LDO, just use a suitable biased FET to with a dummy load and adjust it to achieve 17,18,... V while monitoring current with a 80mV drop @800mA or 0.10 Ohm Rsense. Use Ohms law to anticipate power dissipation and select suitable loads for up to 20W.

You should not be surprised to see current stay well regulated <2% but may be offset by >2%, but always consult the specs. (if they exist) and test it for input and thermal variations

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  • \$\begingroup\$ Dear Tony, You said to use an FET and bias it suitaly. But how do you make voltage drop to set value across the FET? \$\endgroup\$ – Mithun N Aug 26 '17 at 9:36
  • \$\begingroup\$ CC Driver should source a constant across a wide voltage range. So we will have to vary 2 things here.1. Current to limit or set the current. 2. To set voltage drop across the driver. How do you do the latter? \$\endgroup\$ – Mithun N Aug 26 '17 at 9:59
  • \$\begingroup\$ The DUT being tested regulates both CC and V. THe external FET must have a current sense R to gnd so you can measure it. When you bias the FET either manually with a pot or with feedback using a comparator, it will limit the current until setpoint, then above this, the DUT will limit current and regulate voltage down as you increase bias on FET. Simple? THus you can manually adjust load RdsOn and observe Vout vs Iout to safely test. If FET Pd is too high add an equivalent load in series for about 80% so FET only draws 20% of total Pd. Tungsten bulbs work well with a FET . \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 26 '17 at 22:24
  • \$\begingroup\$ I might consider a suitable size halogen bulb to take most of the heat of 15~20W \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 26 '17 at 22:26

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