I am working on making a drink mixer inspired by BAR2D2, I am using a gravity fed pump that is capable of running on a 9v battery. I am using a circuit with a 555 timer to control how long the pump will run. When I hook the pump directly to the battery the pump runs perfect. When I plug it onto my circuit, I get nothing. It turns out I am not getting enough amperage to run the pump.

How can I get more amperage to my pump but still have the timing controlled by my 555 timer?

This is the circuit diagram I used to hook up the 555 timer circuit:

enter image description here

When I put power to the circuit I want the pump to run for 4 seconds which will fill up a cup. the green LED is where I am putting the power leads that go to the pump. When I use the green LED in the circuit the circuit works perfectly. When I replace it with the pump the pump doesn't come on. When I hook the pump to the battery the pump works. My best guess is the pump needs 2 amps to run and it is not getting that.

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    \$\begingroup\$ Without a circuit diagram, any answer or comment is merely a blindfolded dart toss. So, here's my toss: have you considered using a relay? \$\endgroup\$ Commented Apr 20, 2013 at 23:28
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    \$\begingroup\$ @TrishaWrightBuck slow down and read what folk are saying and examine what you are telling us... For a start read what you said: "When I hook the pump directly to the pump the pump runs perfect" - this does not make sense and, trying to fathom out what you have tried without a scheme of some sort doesn't make much sense either. Help us help you. BTW I'm not a councillor but because I like drink I'm prepared to help!! \$\endgroup\$
    – Andy aka
    Commented Apr 20, 2013 at 23:44
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    \$\begingroup\$ Arrgh apparently I have forgotten how to speak english, when I hook the pump directly to the BATTERY it works. \$\endgroup\$ Commented Apr 20, 2013 at 23:51
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    \$\begingroup\$ @Andyaka: The OP has been unable to have a drink, hence slowing down may be difficult. We need to organize an emergency airlift of motorized drinks to the OP's current location. \$\endgroup\$ Commented Apr 21, 2013 at 0:01
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    \$\begingroup\$ Did you use a power NPN transistor which can handle the collector current, with an adequate heat sink? Did you give it enough base drive so that it is fully saturated? Those three things will lead to a "one use only" transistor: inadequate rating; inadequate heat sinking for the device, current, and ambient temperature; inadequate turn-on, allowing a significant voltage drop across the transistor at a current that is still large. I would not use a relay here because the gravity fed pump (valve, really?) runs on a small voltage. There is no need for the isolation. \$\endgroup\$
    – Kaz
    Commented Apr 21, 2013 at 3:11

2 Answers 2


You need to buffer that output with a power stage. You can do that in many ways, using a relay is one of them, but I would not advise that.

A MOSFET power stage will be faster and more durable because there are no mechanical components.

As Alfred Centauri points out it is not starightforward to know why your npn design didnt work without an schematic, but Ill suggest a solution.

Ill assume is that you have an unregulated supply, and you are powering everything at 9V. This is enough to correctly bias any N channel MOSFET. What youll do is connect the output of your 555 directly to the mosfet's gate, the mosfet's source to gnd, and the mosfet's drain to the pump's negative terminal. And you will need two extra components, first and most important a 1n4007 diode in parallel with the pump, with the cathode on the pump's positive terminal. And also, a 1Meg resistor between the mosfet's gate and source.

enter image description here

The only thing you need to make sure is that the mosfet can handle the current needed by the pump. To do that, you need to measure the pump's current while connected to the battery directly. Now, you need a mosfet whose Rdson times that current squared (this will be the mosfet's on state power dissipation in your circuit) is less than the maximum power that the device can dissipate. For a TO220 Mosfet, that will be in the order of 0.5-1W depending on how hot you want it to run.

EDIT: The 1M pull-down is used to ensure that the MOSFET is properly switched off, draining the charge that may remain in the gate. Some might say that in this case it is not needed because the 555's output is push-pull, and can properly drive the gate to gnd. But I view this as decoupling caps, you always have to use them or you'll regret it. Besides, in this case there is a reason to use it; when power is removed, I wouldn't trust the 555 to drive the output low, and you want that MOSFET to turn off in that case. A G-S resistor is almost always a good idea.

  • \$\begingroup\$ Thank you. I am about to try this idea and I will let you know how it turns out. \$\endgroup\$ Commented Apr 21, 2013 at 0:20
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    \$\begingroup\$ Why a 1M gate pull down? \$\endgroup\$
    – Matt Young
    Commented Apr 21, 2013 at 0:54
  • \$\begingroup\$ I didn't get a chance to try the diode and mosfet yet, I had to leave the lab. I did stop and get the mosfet and the correct diode so maybe tomorrow will bring new hope.Before I left though my partner was trying to run the circuit with a relay. I am not sure on his set up but I know he was using a push button to start and stop the circuit, and the problem he was having was powering the relay and the 555 timer was getting hot. \$\endgroup\$ Commented Apr 21, 2013 at 3:01
  • \$\begingroup\$ @payala Let me rephrase my previous comment. 1M is a super weak pulldown. Anything over 100k in that capacity, I start to wonder if noise would be able to turn the MOSFET on. Why not 10k? \$\endgroup\$
    – Matt Young
    Commented Apr 21, 2013 at 16:55
  • \$\begingroup\$ @Matt 1M is not that high for a MOSFET gate, I have designed many industrial equipment with this type of circuit without problems. Of course, you have to know the context where the circuit will be. And in low power equipment this is very common. Besides, the MOSFET gate is low impedance because the 555's output is driving it. One thing you learn when you design circuits professionally, is that inductions are not so strong generally. Many junior engineers when a circuit doesn't work as expected blame some form of induction just 30 minutes after working on the problem. \$\endgroup\$
    – payala
    Commented Apr 21, 2013 at 22:33

A generic 555 timer IC can only source (or sink) about 200 mA. The pump is going to use a lot more current than that.

A few ways to boost the output current that can be used are a relay, a power transistor, or a MOSFET.


Even though a relay would work in this situation, I really wouldn't recommend it. A relay draws a lot of current and puts noise on the power rails.

Power Transistor

You can hook up a transistor to the output of the 555 to amplify the current that you need. A 2N2222 NPN transistor can output 800mA. If you need more, 2N3055 NPN transistor can give you up to 15A of current. Remember, though, proper heatsinking is required!


See payala's answer -

A MOSFET is a lot like a bipolar junction transistor, except that it is controlled by voltage instead of current. If you want to use a MOSFET, you would connect the 555 to the gate of the MOSFET.

  • \$\begingroup\$ Oh, actually, never mind! Something glitched I think. I have way more than enough to post images. \$\endgroup\$
    – fuzzyhair2
    Commented Apr 21, 2013 at 15:13

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