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My power supply is filtered regulated 5 volt 500 mA . I am using a 555 in monostable mode to switch the motor on for a certain amount of time after the 555 is trigerred. (The 555 is being trigerred by a counter circuit). But the output current from the 555 is too low. How can I use a transistor like 2N3055 to get full 500mA current? What other ways are there to achieve the same thing?

Will This Work? Will this work?

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up vote 7 down vote accepted

An emitter follower will have a voltage in the emitter about 0.6v lower than the voltage at the base. It will work if you don't have a problem with the reduced voltage level. Note that a 555 with 5v supply can have an output as low as 3v depending on the output current.

The alternative is to use a transistor or mosfet as a low side switch (switching the ground side of the load)


simulate this circuit – Schematic created using CircuitLab

The mosfet will be a better solution (more efficient) because it has a lower voltage drop across it (drain-source), and it doesn't need a constant current in the gate (for static operation) as the transistor does. Just select a logic level Nmosfet so that it can turn fully on with low resistance

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Values of the resistors, transistors mosfets...? – Soumadeep Saha Aug 5 '14 at 12:54
You can use any logic level mosfet, just visit any vendor or eshop and use the filters to select a device that suits your needs making sure it can be fully turned on with 3-4v Vgs. Regarding R2, it should be calculated for a base current of about 1/10 to 1/20 of the collector current so that the transistor is saturated and has a low Vce drop. – alexan_e Aug 5 '14 at 13:12
Yeah, I can confirm the mosfet solution works well. I used a similar approach to driving a set of CNC stepper motors, and used mosfets as switches driven by a low current device like a 555 (it was a microprocessor, but the issue is the same). I used a generic mosfet I got at radio shack (IRF510 perhaps). The current ratings on mosfets go up into the whole amps range. – Calphool Aug 5 '14 at 16:29
Per ti.com/lit/ds/symlink/ne555.pdf the high-level output can be as low as 2 volts. – WhatRoughBeast Aug 5 '14 at 17:12
@alexan_e I am sorry ... The problem wasn't with anything else it was a loose wire... Thanks for the solution. – Soumadeep Saha Aug 6 '14 at 11:06

As a high side driver, the 555 is only rated for something like 100 mA, and it takes a good sized output voltage hit at that current, so it'll never work directly as a motor driver.

The emitter follower you've shown is better, but you can't drive it into saturation so you'll still lose about a volt across the transistor when it's on, starving your motor somewhat.

If you use a P channel MOSFET with a reasonable Rds(on) for the high side driver then you'll get nearly all of the 5V supply across the motor when the MOSFET turns on, plus it'll dissipate very little power.

The circuit shown below assumes you're stuck with a high side driver, and it'll work with a bipolar 555. The LTspice circuit list is here if you want to play with the circuit, and if you want to switch to a low side driver get rid of Q1B and replace R6 with the motor and D2.

Just as an aside, the FDS4559 is a pair of about 3 amp MOSFETs in the same package, one P channel and one N channel for less than a dollar in onesies at DigiKey.

enter image description here

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+1 A simple output booster with a lot of 'potential', EMF. – Spehro Pefhany Aug 5 '14 at 14:16
Thanks. :-) It seems like it'll work nicely for the 'current' application. – EM Fields Aug 5 '14 at 14:40

Using the transistor as an emitter follower will work, but you'll only get about 4.3 V at the output, because of the 0.7V drop of the transistor's B-E junction.

If possible, it would be better to put the load on the "high side" of the transistor — between the +5V supply and the collector, with the emitter tied directly to ground. You'll also need a resistor in series with the transistor's base to limit the current. 100Ω would give you a base current of 50 mA, which should be plenty.

For even lower loss, substitute an N-channel MOSFET for the NPN transistor.

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That was my thought, too, but the NE555 minimum Vout with a 5V supply is about 2.75V – Scott Seidman Aug 5 '14 at 12:50
@ScottSeidman: I don't understand what point you are trying to make. The minimum Vout is at the maximum Iout (which is about 200 mA for a regular, non-CMOS device). The 555 should be capable of driving either the BJT or the FET to very close to 5V. – Dave Tweed Aug 5 '14 at 13:24

Will this work?

Nope. Not even close. Take a look at the NE555 data sheet p.5 "High level output voltage". Although you are using an extra transistor to reduce the load on the 555, it demonstrates that the output is not good at sourcing current. Add the base-emitter drop in the transistor (~.7 volts), and you are not certain at all of getting a good motor drive. This, on the other hand, will:


simulate this circuit – Schematic created using CircuitLab

Depending on how much the motor draws, R6 may need to be decreased. This will certainly work to 150 mA.

Note that you can simplify the circuit if you use a PNP for the output stage. Then Q3 goes away, and the collector of Q2 becomes the motor +, with motor - at ground. R4 then controls the amount of drive current and would definitely need reduction.

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Since the output of the 555 is a totem pole, R2 can go away, and since Q2 is being pulled up to +5 by R3 and R4 when Q1 is cut off, R5 can also go away. – EM Fields Aug 5 '14 at 15:51
You're right about R2 - I was just being reflexively careful. And the schematic has an error which I had intended to correct but forgot - I intended to eliminate R3, which will work better than eliminating R5. – WhatRoughBeast Aug 5 '14 at 17:09

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