DIY electronic load troubleshooting - Page 1
Forum section: electronics / projects

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 22, 2019, 08:24:38 pm
content: Greetings,I followed Dave's video and built myself an electronic load. Main difference is the supply voltage, which is 12V, so I can use a standard MOSFET and have a bit more current capability.Problem is that I can only get about 5A through it (I should be able to get about 10A with a 1R resistor). The maximum gate voltage is 10V, which according to the datasheet is enough to drive the transistor so that it sinks a drain current of 11A.I suspect a fake transistor, but it should've blown up by now.I have attached the schematic below.

author: Dave
author_rank: Super Contributor
timestamp: July 22, 2019, 08:43:52 pm
content: The maximum output voltage of the opamp with respect to ground is 10V, but that's not what VGSwill be with 5A flowing through a 1R resistor.Why did you use the buffer opamp?

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 22, 2019, 08:51:52 pm
content: That's true, and I completely missed that. The actual gate voltage is 5V.So the first obvious solution is to lower the value of the resistor?I used both op-amps because it's a dual op-amp package, and one was unused.

author: duak
author_rank: Super Contributor
timestamp: July 23, 2019, 12:35:44 am
content: Yes, you will have to reduce the value of the R5.After you reduce the value I think you will find that Q1 is undersized for 10 amps and load voltages much above 5 to 10 V. The IRF640 data sheet shows that it is possible, but it is difficult to achieve. The 1 ohm resistor is probably dissipating much of the power.This circuit will not work well at low load currents. The TL072 is not specified to operate reliably with its inputs within 3 V of Vcc-. You will have to provide a -3 V or greater negative supply voltage or use an opamp that can operate with its inputs at 0 V relative to Vcc-.The first opamp connected as a buffer is redundant. I would consider redesigning the circuit to use two IRF640 with their drain terminals connected together and driven by the two opamp sections with two sense resistors (R5).

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 23, 2019, 07:10:41 am
content: I noticed that it goes down to around 10mA minimum. However, I'm not really interested in low currents, it's designed to test power supplies and their overcurrent capabilities.Any recommendation for a suitable op-amp?Sent from my POCO F1 using Tapatalk

author: Kleinstein
author_rank: Super Contributor
timestamp: July 23, 2019, 07:23:06 am
content: The Tl072 may not go low enough to make the MOSFET turn off all the way. Even more trouble some is the common mode range for teh input. So one should use single supply OPs. Alternative would be a +- supply.A few possible candidates are LM358, TLE2021, OPA197, OPA170, LT1013, OPA145Driving the MOSFET directly from the OP can cause oscillations. The usual way is to add a capacitor and resistor.The IRF640 is not a bad choice, but with the relatively small case the power is limited. Usually TO220 gets tricky with more than 50 W. So for higher power it is more like using 2 sections in parallel.

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 23, 2019, 07:31:45 am
content: What about LM324? The datasheet says it can swing to GND even with single supply. And since it's a quad package, I can parallel 3-4 transistors.Sent from my POCO F1 using Tapatalk

author: magic
author_rank: Super Contributor
timestamp: July 23, 2019, 10:27:49 am
content: It's same thing as 358 but in quad.

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 23, 2019, 06:47:33 pm
content: This is the new version. Please comment if there is anything wrong.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 23, 2019, 08:57:22 pm
content: The added caps are at the wrong position - this way they make things even worse.The resistor in series to the gate is Ok, but most OPs want more like >= 100 Ohms to isolate a capacitive load.The capacitor would be right as direct feedback from the OPs output to the - input. In addition the feedback from the shunt would go through a resistor (e.g. some 2-10 K).So the high frequency feedback is through the capacitor and only the slower part (e.g. < some 20 kHz) is through the shunt.The Lm358/324 are cheap, but due to the cross over distortions they don't like capacitive loads very much. Another common OP I missed in the list is the TLC272 - not true single supply, but sufficient.

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 24, 2019, 04:02:56 pm
content: Done. What exactly is the purpose of the resistor from the shunt to the - input?

author: Kleinstein
author_rank: Super Contributor
timestamp: July 24, 2019, 04:45:11 pm
content: The principle circuit looks OK, but the capacitor can be smaller, more like 1-3 nF.The resistors form the shunt together with the capacitor in feedback set the cross over between where the feedback is directly from the OP and where the feedback is from the shunt. The cross over should be at a frequency considerably lower than the time constant of gate capacitance and 100 Ohms resistor at the Gate. It should usually also be lower than the OPs GBW.

author: bogdan2014
author_rank: Regular Contributor
timestamp: July 24, 2019, 06:37:18 pm
content: Great, thanks! I'll come back with the whole thing done.