Can't make sense of this circuit on a controller loop - Page 1
Forum section: electronics / projects

author: osflores@qdusa.com
author_rank: Newbie
timestamp: May 02, 2024, 04:31:52 pm
content: Hi All,The circuit in the attachment is used on a voltage control loop. As you can see the integrator stage U301B is followed by a non inverting configuration with positive feedback U301A (which adds hysterisis to the circuit I assume). The tricky part is the MOSFET used in the positive feedback path Q300. Is someone familiar with this type of circuit and where I can learn about it?

author: mikerj
author_rank: Super Contributor
timestamp: May 02, 2024, 08:36:47 pm
content: Quote from: osflores@qdusa.com on May 02, 2024, 04:31:52 pmThe common source connected transistor Q300 inverts the signal being fed back to the non-inverting input so it's negative feedback, not positive. This is an inverting amplifier with the gain set by R302/R303, in this case a gain of -1 which corrects the inverted output of the integrator.

author: moffy
author_rank: Super Contributor
timestamp: May 02, 2024, 09:48:54 pm
content: The schematic segment shown is a bit too brief, what is Q300 driving? What is the 'gate' signal connected to? Is the output of Q300 a binary 'on'/'off' level or a more analogue value?

author: Cuchi
author_rank: Newbie
timestamp: May 07, 2024, 10:29:02 am
content: Hello,I am a new member of this site and a beginner in the subject. I don't know how to navigate it yet but I came across your post while doing a search and I'm telling you my problem thinking that you might be able to help me find the solution or tell me where to ask my question more effectivelyI purchased a controller as specified belowan PWM type single channel H-Bridge DC brushed motor driver for automotive useThe pin assignment is as follow:an Output high-side current monitoring function (OCM pin):+VIN; GND and M+;M-A Microcontroller(inputs):PWM1: Driver control signal input pin 1PWM2: Driver control signal input pin 2ENGBD+12V OUTI plan to use it to vary the speed of a 12 volt electric motor (a motorcycle starter) connected to a 12 volt motorcycle battery as well.The problem is that when I connect the throttle to the controller inputs (+12v pin; Gnd pin and signal pin) the output signal of the throttle connected to the PWM1 input of the controller, I obtain an inverted effect, i.e. the motor spins when the throttle is closed and at rest and the motor stops when the throttle is opened with the lever pushed all the way. If I connect the throttle signal pin to the PWM2 of the controller nothing happensMy question is how to reverse things, i.e. make the motor spin as the throttle is applied towards more gas and not the other way around.Thanks

author: perieanuo
author_rank: Frequent Contributor
timestamp: May 07, 2024, 12:46:04 pm
content: hi, if by throttle you mead a potentiometer-like device, just reverse throttle wires Dc+ and GND.for me, you are not driving the bridge like it should, maybe pwm inputs need pwm signal, but i this case I cannot explain why pwm2 is different

author: osflores@qdusa.com
author_rank: Newbie
timestamp: May 07, 2024, 05:02:59 pm
content: 1. It is driving a HV Voltage source with a resitor dividor feeding back to the controller. We am trying to make the high voltage source as stable as possible with this controller.2. The gate signal is connected to a BJT arrangement that pulls the gate to ground (I assue that disables the circuit).3. That is what I am trying to find out, I am not sure what to make of this circuit. From a simulation that I did there is not much change adding or removing this last segment of the control loop, rather it seems like it is used to turn on and off the output throught Q300? This is just speculation but I hope someone can help me figure it out.

author: moffy
author_rank: Super Contributor
timestamp: May 07, 2024, 11:17:08 pm
content: Quote from: osflores@qdusa.com on May 07, 2024, 05:02:59 pmJust broad strokes as part of the input to U301B is missing (should be pulling current out, -ve voltage), U301B is the slow loop integrator with some feedforward provided by R300, stability/phase reasons. It is reset by a +5V pulse that sets U301B fully negative. It integrates driving Q300 which is effecting the voltage being controlled until some form of equilibrium is reached. Q300 has some local negative feedback, the resistor divider and Q302 again possibly for stability reasons. It's like looking at part of a puzzle and trying to construct the whole picture, so take what I say with that in mind.

author: moffy
author_rank: Super Contributor
timestamp: May 07, 2024, 11:19:32 pm
content: Quote from: Cuchi on May 07, 2024, 10:29:02 amYou should start your own post on the matter, this post is dedicated to the OP. If you do that more people will see your issue.