# Halfwave rectifier simulation

• 21-05-13, 10:02
udhay_cit
Halfwave rectifier simulation
I am using Ltspice for my simulations. I have simulated a halfwave rectifier with two diodes as shonw in the circuit diagram. While measuring output waveform at node 1 with reference to ground i am getting a negative peak of 6v. I am so confused, where the negative peak come from..! The two diodes are acting like a diode potential divider.

Anybody explain me.... Is it the simulation tool problem? can anybody simulate the same schematic with some other tool?

Regards
Udhay

http://obrazki.elektroda.pl/16529856...3143_thumb.jpg
http://obrazki.elektroda.pl/77764511...3194_thumb.jpg
• 21-05-13, 10:33
ZekeR
Re: Halfwave rectifier simulation
Nope, it's not LTspice's fault; LTspice is actually a pretty good program.

Consider the equivalent circuit when the diodes turn off (which is the period under question). When they turn off, they become open circuits (with some parallel capacitance). Thus, the two diodes have roughly zero current flowing through them, but substantial voltage across them. Since you have two diodes in series, the voltage they block (12V) will be divided evenly between them. And that's why you see 6V across each diode during the off period.
• 21-05-13, 10:40
udhay_cit
Re: Halfwave rectifier simulation
So you mean that the two diodes are acting like a capacitor while in OFF state?
• 21-05-13, 11:28
pplus
Re: Halfwave rectifier simulation
• 21-05-13, 13:04
Re: Halfwave rectifier simulation
Falstad's simulator handles it the same way.

The ground icon is defined as 0V. Therefore the simulator divides the negative supply voltage across all components in the branch, in proportion to each one's resistance at that moment.

http://obrazki.elektroda.pl/78720522...4209_thumb.jpg
• 21-05-13, 13:51
pplus
Re: Halfwave rectifier simulation
Quote:

Falstad's simulator handles it the same way.

Try please to simulate my circuit. Clear where LTspice takes this capacitance -
. MODEL 1N5819HW D (IS = 191U RS = 42.0M BV = 40.0 IBV = 1.00M CJO = 239P M = 0.333 N = 1.70 TT = 7.20N VPK = 40.0V IAVE = 1.00A MFG = DIODESINC TYPE = SCHOTTKY)
http://obrazki.elektroda.pl/2274119700_1369136822.png
It is not clear where it takes your simulator.
• 23-05-13, 14:01
albbg
Re: Halfwave rectifier simulation
I suspect there is something wrong with the model (and not the simulator itsel) of udhay_cit. Try the behaviour of different diodes.

Try to add some small capacitance (shoud be negligible with respect to the frequency) in parallel to one of the two diodes ore use two diodes in parallel; if it's matter of C of the diodes you should see the -6V changing.

It could be you can find different models for 1N5819 searching the network.
• 23-05-13, 14:21
pplus
Re: Halfwave rectifier simulation
Quote:

Originally Posted by albbg
I suspect there is something wrong with the model (and not the simulator itsel) of udhay_cit. Try the behaviour of different diodes.

Try to add some small capacitance (shoud be negligible with respect to the frequency) in parallel to one of the two diodes ore use two diodes in parallel; if it's matter of C of the diodes you should see the -6V changing.

It could be you can find different models for 1N5819 searching the network.

The diodes produce a capacitive divider, as their model is the same voltage drop is Uin/2. Oscilloscope probe has a finite resistance in contrast to the probe LTspice and the effect is reduced.
• 24-05-13, 09:16
shamanthdn
Re: Halfwave rectifier simulation
Hi,

Actually there is no problem with the output. The out put you have got is correct.
In the -ve half cycle v1=-12V and the diodes block this voltage. So there is 6V drop across each voltage. Since Vi=-12V and drop across diode D1 is 6V node1 is at -6V. Since there is no current node2 is also at -6V.