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I designed a boost converter using a 555 timer in order to power a Peltier device. This is mostly just for fun and to understand circuits a bit better.

I ordered all the parts so that they are the same as the simulated circuit in the pictures. The only component that might not be the same is the MOSFET. When I designed the physical circuit I used my voltmeter and found that the boost converter output stabilized around 60V using a 9V battery. This is without a load attached.

Before I connected the Peltier device, which requires 12V, I added a 100k resistor in series between it and the boost converter. Once connected I read 25V across the Peltier device, but the device didn't heat up. When I get a chance I'm going to review my design, but I wanted to see if any of you can see something I can not.

First part of circuit

second part of circuit

final part of circuit

Entire Circuit

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    \$\begingroup\$ I designed a boost converter using a 555 timer: Please don't. That thing is a terrible choice. There's many, and many cheap, boost converter ICs that have actual voltage feedback, loop compensation, protection, integrated switches, way better switch drives, higher switching frequency leading to cheaper inductors being feasible... \$\endgroup\$
    – Marcus Müller
    Commented Jun 24, 2021 at 11:04
  • \$\begingroup\$ Welcome to stackexchange! Please draw a schematic of how you connected everything. \$\endgroup\$
    – winny
    Commented Jun 24, 2021 at 11:04
  • \$\begingroup\$ I'd say, it's nice that you have three partial screenshots, better than nothing, but if I want to answer your question and the first thing I have to do is get an image editing program to piece your three images together to one, then I wonder why you haven't done that, especially seeing your phone program seems to have a screenshot functionality. \$\endgroup\$
    – Marcus Müller
    Commented Jun 24, 2021 at 11:06
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    \$\begingroup\$ This is mostly just for fun and to understand circuits a bit better. That's a good goal. However, since this circuit isn't going to behave as you expect it (because a 555 isn't good for a DCDC converter and a Peltier element needs a lot of power) so you will not have "fun", instead you will have frustrations and you will not learn anything because you do not understand yet why this doesn't work like you want it to. I strongly advise you to build circuits that are known to work. Like flash an LED using a 555. Then experiment with that. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 24, 2021 at 11:17

2 Answers 2

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  1. Don't use a 9V battery for this kind of thing. Peltier devices use a lot of current. 9V batteries are made for something less than 200mA. A 9V battery will not last long - minutes at the most.
  2. Boost converters aren't magic. You get (more or less) the same power out as power in. If you need 12W out (12V at 1 ampere) then your booster will take in at least 1.3A from the 9V battery (12W / 9V.)
  3. Peltier devices need lots of current. That 100k resistor you put in there means your Peltier device is getting at most one milliampere of current. Even if it is 100 ohms rather than 100k, you still won't be able to operate your Peltier device on it.
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The only component that might not be the same is the MOSFET.

The MOSFET in everycircuit is not modelled based on any real world MOSFET. In everycircuit, you need to model the MOSFET yourself by setting the channel width, length, kp, lambda etc. This is true for the diode as well.

A former user called @kiani has created a MOSFET library in EC, which is very popular. You can use one of those models to get better results.

When I designed the physical circuit I used my voltmeter and found that the boost converter output stabilized around 60V using a 9V battery. This is without a load attached.

This is due to the fact that the boost converter that you have designed is an open loop one with no load. Therefore, the output voltage of such a converter depends highly on the input voltage and the type of components used.

The everycircuit diode has a default cut in voltage of around 0.6V and an ohmic resistance of 1ohm. In addition to this, the on resistance of the MOSFET (according to your configuration) is high. All these factors result in the simulated circuit outputting a lower voltage than the breadboard-built circuit. The MOSFET and diode used in your breadboard must have better specifications than the EC model.

My suggestion: use an off-the-shelf boost converter like the LM2577-ADJ to boost 9V to 12V and use LTspice to simulate your circuit. LTspice is a much more robust simulator and is much more accurate than EC because it uses proper models. Everycircuit only uses ideal components.

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