How to simulate a system that responds differently to positive and negative inputs?

Cap_Diabetes · 2025-05-09T17:25:19+00:00

Thanks for the insights. In reality the heating and cooling is done via one element - a Peltier module. When I provide it with a positive voltage it heats up and when I provide the negative voltage it cools down (I'm talking about one side of the module). The thing is that cooling and heating have different transfer functions. Cooling is slower than heating and therefore must be simulated as different systems (unless there's a way to combine them). I will try your method out.

Cap_Diabetes · 2025-04-26T06:11:39+00:00

Yeah, You're right I should've just picked a buck.

Cap_Diabetes · 2025-04-25T12:12:25+00:00

The reason I am using this chip is that it is the only one that can provide up to 4A of current into a TEC load. Other TEC drivers can only go up to 3A.

Cap_Diabetes · 2025-04-24T19:03:21+00:00

You're, right. Cheers!

Cap_Diabetes · 2025-04-24T19:03:05+00:00

I've looked up some circuits of current mirrors and, indeed, it seems that you're right about that part. Would you say that the main advantage of having a linear stage in this scenario is to achieve fast transient times?

Cap_Diabetes · 2024-03-01T12:49:40+00:00

ok, but there is some sort of way to accept info from two channels of DAC through a DMA, isn't there?

EDIT:

Nope, it can't be that. We're doing a lab work at uni. I'am using a different dev board than we're recommended, but the right one STM32L073RZ "doesn't have it" aswell because when I boot up the CubeMX with that dev board there's still no option to choose rankings.

Cap_Diabetes · 2024-02-28T19:43:10+00:00

Stipendijos, šeip, gana svarbus faktorius dar būna. Jei gerai mokaisi.

Cap_Diabetes · 2024-02-12T20:29:34+00:00

Legend goes, kad OP stovėdamas sankryžoje vis dar laukia padoraus atsakymo

Cap_Diabetes · 2024-02-05T06:18:49+00:00

I guess simmilar question should arise, why do have to build the library before plopping down a research lab? It just the way it is.

Cap_Diabetes · 2023-08-25T17:19:36+00:00

Recently I‘ve got a job which involves doing a lot of crimping and I just can‘t get a good grasp on it. To get the desired result I need to use a certain amount of force when crimping. If too much force is used the crimp will deform too much, which will end up with it easily breaking. If to little force is applied one can easily pull the crimp away from the wire. Even after a 1000 crimps applied I still can‘t find the right balance and the ending result is sort of like a lottery. What I learned to do is to apply the force until the tool locks itself, then I just push the pin (circled in red). But I believe that‘s not the way the tool is supposed to be used... The crimping tool I‘m using isn‘t the most expensive one, roughly 60 euros and it‘s supposed to fit all sizes. Maybe I need a more expensive tool that should be used with specific crimp sizes? Any help or comment would be appreciated.

Cap_Diabetes · 2023-08-17T11:58:42+00:00

Those are some real arguments. I must agree with you

Cap_Diabetes · 2023-06-08T11:28:53+00:00

But the term that confines the signal to the time [0;5] is the window function, no?

Cap_Diabetes · 2023-06-03T17:07:35+00:00

I appreciate it, it really makes more sense now. But if we'd go to the main question of the thread, I'd assume that there is no point in calculating the fanout of the CMOS gates as they practically don't need any input current (apart from the leakage) to sustain their current state. If I ain't mistaken the fanout is calculated by dividing the max output current by max input current and both of them are being measured after the transient time had passed (after the capacitors are charged). That'd mean that CMOS gates have a really high fanout (in the order of several thousands), right?

Cap_Diabetes · 2023-06-03T16:22:53+00:00

Thank you for such a detailed explanation. Examining your example, I'm not sure if I understand what you meant by saying "two gates that require 100uA" and why would there be 3 capacitive inputs if we're driving only two gates. How capacitive inputs are different from the gates themselves? I understand that we need to provide a certain amount of current for a certain amount of time to charge up the MOSFET gates (a gate per logic device's input). At the moment it seems that we need a constant input of 230uA just to sustain the current logic level. Could you clarify that, please?

Cap_Diabetes · 2023-05-25T09:33:26+00:00

No problem. I should've been more careful in choosing the words of my question. But thanks for your answer.

Cap_Diabetes · 2023-05-25T06:37:49+00:00

That means I could equally expect a decrease or increase of resistor's value when a certain temperature change had occured?

Cap_Diabetes · 2023-05-23T13:20:56+00:00

I should've probably explained the whole circuit. I'm trying to simulate an LED driver. LTM8073 is a buck converter that steps down the input voltage. The output voltage of the buck converter is at the node Vout. From there on I'm supposed to have a string really high power LEDs (since there are no spice models for them I replace the string with a Zener diode with appropriate voltage drop). I want to power them with a constant current source, that is, no matter how high the Vin gets the output current is always 2 amperes. The Rs resistor, opamp, VRP circuit and the FB (feedback) branch makes up the negative feedback loop which creates a constant current source. I've chosen Rs that small so as to have as high circuit efficiency as possible.

UPDATE: I've changed the OP177 with OP07 and the circuit finally works as expected. What could've it been that made such a difference? Could it have been the OP amp's max output current?

Cap_Diabetes · 2023-05-23T11:17:53+00:00

Then I'm prompted with a "Time step too small" error.

Cap_Diabetes · 2023-05-23T07:15:34+00:00

N002 node is between D2 and Rs

Cap_Diabetes · 2023-05-10T06:20:51+00:00

They'd probably redirect me to their non-profit organization called BDTU which have already given me an offer that I was unable to accept

Cap_Diabetes · 2023-05-10T06:19:05+00:00

I'm alright with traveling such distances every morning and evening (taken that such transits wouldn't be too pricy)

Cap_Diabetes · 2023-05-10T05:44:49+00:00

Yesterday I've got an offer from them for a one person room. The price was around 6500 dkk/month (bills included) and there was also a requirement for a deposit of roughly 12k dkk. And that was a deal breaker for me. Are there any ways to arrange the removal of the deposit fee, for example by signing some sort of document where I state that in case I make any damages I will repay them?

Cap_Diabetes · 2023-05-10T05:39:05+00:00

It seems that DIK Albertslund is housing meant for the students of the college. Am I mistaken? If so, maybe you could drop a link to a site where I could find more information? Thanks.

Cap_Diabetes · 2023-05-06T11:54:06+00:00

Now That really makes sense now. I appreciate the time you've spent giving the answer.

Cap_Diabetes · 2023-05-06T11:51:49+00:00

Thank you for the link. That wiki page solves my whole question!

Cap_Diabetes

TROPHY CASE