Using a software circuit designer - Page 1
Forum section: electronics / projects

author: lion
author_rank: Contributor
timestamp: April 22, 2017, 03:27:59 pm
content: Hello folks,I made a thread (link below) talking about a load sensor circuit.https://www.eevblog.com/forum/projects/schematic-interpretation/msg1026428/#msg1026428I tested the circuitin real life and it works(I didn't use the D1, D2 or D3)The circuit that was posted isn't mine andI'm having trouble simulating it usingMultisim.I'm not very experienced with MultisimThe circuitIdesigned in Multisim (Updated with values).The circuit on the previous thread (the one I didn't built but that's working).Anyone call help me by telling whats the problem with my Multisim design?

author: Buriedcode
author_rank: Super Contributor
timestamp: April 22, 2017, 05:53:57 pm
content: No sure you'll get many responses without a question. In what way ar eyou having trouble simulating it? Simply saying 'I'm having trouble' doesn't really tell us whats going on.. does it generate errors? does it crash your computer?The first picture looks like C1 is wired backwards - the right side of the diode bridge is + and the left is -I believe D1,D2 and D3 clamp the voltage across C1 to ~ 2.1V. These may be require in a real-world application. Also if you're no that experienced in electronics, you probably shouldn't be dealing with mains.I haven't used multisim in years so I'm afraid I can't help you in terms of using the software.

author: lion
author_rank: Contributor
timestamp: April 22, 2017, 06:52:21 pm
content: I updated the picture with values. I guess something is not well connected.The C1 sounds fine according to the schematic. I tested the circuit in real life (without D1,D2, D3) and it worked.

author: Zero999
author_rank: Super Contributor
timestamp: April 22, 2017, 06:57:18 pm
content: Quote from: lion on April 22, 2017, 06:52:21 pmYou have posted two schematics.In the first one, C1 has the incorrect polarity and in the second one, C2 has the correct polarity.As you've mentioned D1 to D3, I presume you build the second, not the first schematic.

author: Ian.M
author_rank: Super Contributor
timestamp: April 22, 2017, 07:08:47 pm
content: Its probably having a snit-fit because nothing on the live side of the circuit being simmed is grounded. If you ground the - terminal of source V1, it will correspond to the reality of North American power distribution which almost invariably has the 120V neutral grounded at either the main distribution panel. or the pole pig or both.If you try to build this, the capacitor will explosively vent and the optocoupler and probably the 100R resistor will die. The three diodes you removed are essential to clamp the voltage across the capacitor to under its voltage rating and prevent excessive voltage causing excessive current through the series combo of the 100R resistor and the optocoupler LED.

author: lion
author_rank: Contributor
timestamp: April 22, 2017, 07:23:25 pm
content: Quote from: Hero999 on April 22, 2017, 06:57:18 pmIs the C1 on first schematic wrong? It seams good to me..And yes, the first schematic is mine. The second is from the web.QuoteLike this? Still doesn't work

author: Zero999
author_rank: Super Contributor
timestamp: April 22, 2017, 07:44:06 pm
content: Quote from: lion on April 22, 2017, 07:23:25 pmNo, look at the current path through the rectifier. As drawn the capacitor will charge up backwards and explodehttp://www.reuk.co.uk/wordpress/electric-circuit/bridge-rectifier/The absence of D1 to D3 will also cause an excessive voltage to develop across both the capacitor and R1, causing the destruction of them both.QuoteThe second schematic is correct.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 23, 2017, 12:49:21 am
content: Green symbols mean there's no simulation model attached.Substitute with parts that have models. Don't mind if you can't get exact part numbers. You're going for functional, not exact reproduction.Don't forget to include ground references for floating loops: namely, the 5V battery will generate a singular matrix error. Connect one of the nodes in that loop to ground, either directly or with a resistor of any value.Note that all measurements are made with respect to ground (e.g., the measurement boxes), unless specifically changed (you can place two measure boxes and set one for reference, or use the oscilloscope object and measure between two net connections). You may find it's more natural to ground the DC- side of the rectifier.Tim