5A Dummy Load - Page 1
Forum section: electronics / projects

author: geekbot
author_rank: Contributor
timestamp: June 05, 2014, 09:44:31 am
content: I was inspired to make a dummy load after watching Dave's video and knew it was essential to build one before making a power supply. It's the bare bones circuit with an opamp and a mosfet. I connected the Mosfet in the feedback path of the opamp and fed the 0-6v to the non inverting terminal of the opamp. The resistor is a 1.2 ohm 50W. The problem is that I get only about 0-3.5 amperes through the resistor! I tried increasing the reference voltage to about 8v but still the voltage across the resistor was less than 6v. Then I went ahead and measured the reference voltage which should be 6v was the same value as the voltage across the resistor. I know it's the opamp action but it should have worked the other way around. The voltage at the non inverting terminal should be across the resistor right?! When I remove the potential divider from the non inverting terminal and measure the voltage it's back to 6v. I tried using IRFp250 and CSD18502kcs(logic level) mosfets and LM6482 opamp(rail to rail). The opamp rails were from 0 to 12v

author: madires
author_rank: Super Contributor
timestamp: June 05, 2014, 12:22:43 pm
content: At first glance I'd assume a V_GS issue, i.e. lowering the value of the shunt resistor should fix that. But please post the circuit diagram.

author: geekbot
author_rank: Contributor
timestamp: June 05, 2014, 04:17:34 pm
content: Quote from: madires on June 05, 2014, 12:22:43 pmCan you explain the VGS Issue? I am having problem of not getting higher drain current.

author: madires
author_rank: Super Contributor
timestamp: June 05, 2014, 08:42:48 pm
content: Quote from: geekbot on June 05, 2014, 04:17:34 pmLet's assume the MOSFET needs a V_GS of 5V to pass 1A. 1A through the shunt resistor creates a voltage drop of 1.2V, i.e. the MOSFET's source is at 1.2V. So the voltage at the gate needs to be 5V + 1.2V = 6.2V (referenced to ground) for passing 1A. If you increase the reference voltage to raise the current, the voltage drop of R1 will increase too. And since that voltage is also the source voltage, the gate voltage has to be inceased further to have the MOSFET pass the higher current. The supply and reference voltage would need to be higher than the maximum voltage drop of the shunt resistor by V_GS (for the specified maximum current). A load of 5A would cause a voltage drop of 6V. Please check the datasheet of your MOSFET for the required V_GS at 5A. Add that voltage to 6V and you'll get the required gate voltage which is also the maximum reference voltage.

author: geekbot
author_rank: Contributor
timestamp: June 06, 2014, 06:24:55 am
content: Thanks! It makes sense now, but what if I place the resistor in the drain of the the mosfet? And connect the inverting terminal above it? RDs On is very less, so it doesnt make much change right?

author: madires
author_rank: Super Contributor
timestamp: June 06, 2014, 10:47:38 am
content: Quote from: geekbot on June 06, 2014, 06:24:55 amFor a dummy load the MOSFET is driven to be a resistive element and not used as a switch. Therefore RDS_on doesn't apply here. Basically R1 is just the current sense resistor for the control loop. Of course, it's also a small load, but the real load is the MOSFET. The control loop drives the MOSFET to pass the current you've selected by the reference voltage while R1 provides the feedback. When you increase the voltage of the power source ,the control loop will lower V_GS to increase the resistance of the MOSFET. It's all about Ohm's law. If the resistance of the MOSFET increases, the voltage drop across the MOSFET increases too. If you put R1 top of the MOSFET and also move the feedback to that point, you would change the constant current load into a constant voltage load (and R1 isn't required anymore).

author: rob77
author_rank: Super Contributor
timestamp: June 06, 2014, 11:40:50 am
content: raise the supply voltage to at least 15V and will be ok. as it was mentioned - your MOSFET has no sufficient gate voltage left because of the voltage drop on the resistor.another option would be to use smaller resistor 0R47 or even less and keep the 12V supply rail.actually i'm designing and building a similar device , but controlled with atmega. i'm using a 1R resistor and 4 paralleled mosfets (IRFZ34N - cheap ones) to make sure those will be running within the SOA. and the rail is 15V while i'm designing it for only 5A max current, just to have the approx 10V Ugs available which is needed to fully open the mosfets.other features will be: 80W power dissipation max. , it will have a constant current and constant power modes and thermal shutdown/throttling.btw.. be aware of the thermal de-rating of the mosfets. for example a 80W max heat dissipation mosfet will easily become a 30W one at 100 degree Celsius junction temperature.here is mine - the power module only - fully operational (temperature sensors not wired yet). but the whole device is still a work in progress.

author: geekbot
author_rank: Contributor
timestamp: June 06, 2014, 04:17:40 pm
content: Quote from: rob77 on June 06, 2014, 11:40:50 amLooks Great! I paralleled 24 12ohm 2watt resistors to get a 0.5ohm resistor.What are the functions you are going to implement using your atmega micro? I am gonna use one in mine to display the current by means of a difference amplifier or a current sense amplifier. I am aiming for a 100W dissipation, I think I will need to parallel another IRF250 for that. I dont know how to read SOA curves yet.

author: rob77
author_rank: Super Contributor
timestamp: June 06, 2014, 04:47:25 pm
content: Quote from: geekbot on June 06, 2014, 04:17:40 pmactually it's still an open topicbut i'm definitelly going to implement:1. constant current load (set voltage generated by the atmega's PWM out with a low-pass filter).2. constant power mode (measure the input voltage with the Atmega's ADC and adjusting the current to have constant power sink)3. thermal throttling (sink as much as it can without overheating).4. thermal shutdown (for const current & const. power modes)5. over-voltage protection (small relay in the path of the positive input - open by default and the MCU will have to switch it on - voltage measurement will be before the ralay)6. temperature controlled cooling fan (on/off only - handled with comparators, not the MCU).i'm using Atmega328 (cheap Chinese clone of arduino nano 3.0) + LCD display (16char 2 rows). display will be used to show current, voltage, temperature, estimated time remaining till thermal shutdown...etc..currently i'm wiring up the thing on a protoboard and will write the software during the weekend. once done, i'll release the whole design including software as an open source project.

author: jaxbird
author_rank: Frequent Contributor
timestamp: June 06, 2014, 04:58:39 pm
content: Lots of Electronic Load build logs here on the forum.Here my solution, I think similar specs to what you want to achieve:https://www.eevblog.com/forum/projects/electronic-load-build-log/It's a great device, I use it often.

author: rob77
author_rank: Super Contributor
timestamp: June 06, 2014, 05:10:15 pm
content: Quote from: jaxbird on June 06, 2014, 04:58:39 pmexcellent job !it's a candy for the eyesvery nice project and very well done !actually i'm going to borrow the mAh counter idea from you and add that one into my specs

author: geekbot
author_rank: Contributor
timestamp: June 06, 2014, 05:23:47 pm
content: Quote from: rob77 on June 06, 2014, 05:10:15 pmRob, it worked with the 0.5 ohm resistor. Gives me near 0 to 8.7 amps. Now to implement the instrumentation amplifier for the current readout.

author: jaxbird
author_rank: Frequent Contributor
timestamp: June 06, 2014, 05:34:27 pm
content: Quote from: rob77 on June 06, 2014, 05:10:15 pmThanks a lotI am happy if my build can inspire others, it's quite simple compact, but it will do continuous 50W+ and report accurately within ~10mV/mA/mW/R over the full range.It is really a very useful device to have.