[SOLVED] 555 circuit problem

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Veketti

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Dear All,

I'm building a circuit using 555 chip to open and close injector. My problem is that if resistor R5 is any smaller value than ~700ohm, it gets very warm. If it's like 100ohm, it gets so hot that you cant touch it. Is it normal or is there something wrong with my circuit? I'm driving this circuit by 12V from J2. If I leave Q1 mosfet unconnected R5 wont heat up. With current values R5 820ohm and R6 4000ohm temps are bearable. I tried to increase R8 value if it has any effect on R5 heating but it didn't.

Sorry that it looks bit stupid as eg. switch is illustrated as two pins (J4 & J5) as I'm installing leads from there to the actual switch. And layout is awful as I'm not yet good on arranging components.. :lol: Is there other big no no's or errors?



Thank you in advance.
 

Re: Newbie 555 circuit problem

hi, why do you want R5 smaller than 700 ohm? Discharge pin in 555 has a current limit that you can find in Datasheet? may you can modify your circuit by changing other parts.
let me why do you want R5 smaller than 700 ohm?
 

Re: Newbie 555 circuit problem

The reason for your heat in resistor is the R5 will directly come into Vcc and Ground when the circuit is powered and the output is low (The Discharge pin will also low at that time).

Two ways to solve your problem..
1, calculate a new resistor value for R5
2, calculate the power dessipation in R5 and put the proper wattage

for your circuit Ton = 0.693 (R5 + R6) * C7
Toff = 0.693 (R6) * C7
With this you can calculate the desired values

power dessipation in R5 is
(Vcc * Vcc / R5) * Toff / (Ton + Toff) in w
 
Re: Newbie 555 circuit problem

Here is a link to a very useful application note: http://www.sophphx.caltech.edu/Physics_5/Data_sheets/555appnote.pdf

Under selecting components, you will find the following:



This is not different than advice already give. Its main purpose is to encourage you to study that application note. Unfortunately, the note was no longer available on the Philips site when I last looked.

John
 
Re: Newbie 555 circuit problem

Thank you all for the replies. Now I understand why R5 gets hot. Why I originally wanted to use small value on that is that I could get the R6 as close to R5 as possible so that duty cycle would be as close to 50% as possible. When using kohm resistors next value is always few hundred ohm higher. And I didn't want to put two resistors in parallel. Anyway, I'm happy with these values. One more thing to ask, is it ok to dump that R8 between 555 and mosfet? That mosfet gate consumes only microamps and according to datasheet can handle +-20V so I assume it's ok?
 

Re: Newbie 555 circuit problem

555 is spec'd to +/-200mA output. With max rated at +/-225mA.
Meanwhile, your R8=10 draws 1.2A spikes. Bad form.

FET needs high gate current only to switch internal charge then low current.

To improve you need to buffer the 555 such as complementary source or emitter followers, then R8 can be reduced to 5 Ohms or whatever max Gate current allows such that P=12^2/5 * % duty cycle is being dissipated.
 

Re: Newbie 555 circuit problem

R8 helps to isolate the capacitive load of the MOSFET gate from the 555. Although the MOSFET has a very high input resistance, it also has a relatively high capacitance so when fed with anthing but DC (such as the oscillating output of the 555) it can draw significant current. However, in your low frequency application and at low voltages, you should be able to remove it quite safely.

As long as you keep the ratio of the resistors the same you wil get the same mark/space ratio but you can use much higher values to reduce the current consumption. As you make the values larger the frequency will drop but you can restore it by dropping the value of the timing capacitor.

Brian
 
Re: Newbie 555 circuit problem


Something I don't quite get as I removed the R8 from the circuit and measured current between 555 pin 3 and mosfet gate and it was 55uA. If it went directly to ground it should have produced much higher currents? What am I missing.

- - - Updated - - -

Just played around and checked what component gets hottest and it's the diode D1 37.5°C as 280mA goes through it. Everything else few degrees less.

555 is bit hidden under the wires.
 

Re: Newbie 555 circuit problem

Veketti said:
Something I don't quite get as I removed the R8 from the circuit and measured current between 555 pin 3 and mosfet gate and it was 55uA. If it went directly to ground it should have produced much higher currents? What am I missing.
Click to expand...

This is what you are missing from understanding how FETs work
Imagine a switch that changes capacitance quickly during transition.

You ought to know Q=CV thus C=Q/V an the slope indicates Capacitance and zero slope in the middle indicates a rapid change in capacitance such that the charge stays constant.

We know that Ic=C*dv/dt but also additional current comes from Ic=V*dC/dt.


You can't measure this with a DC ammeter.
 
Re: Newbie 555 circuit problem

Just assume the Gate of Mosfet as a capacitor, and it doesnt have any resistor in series, so for the AC you are giving its nearly like a short circuit, but if you put a resistor It will limit the maximum current value.

also you cant measure this gate current with a DC ammeter as it is a AC spikes of current, what you measuring as 55uA is error value..
 
Re: Newbie 555 circuit problem

Lets not forget this oscillator is only running at 35Hz so although the issues of gate charge are absolutely valid, it is unlikely that the gate current will cause any problems. If the component count is really critical, the resistor from the gate to ground could be removed as the 555 itself is more than capable of ensuring the gate is pulled low.

For better efficiency, the timing resistor should be made higher values though. Changing them to 8.2K/39K and the timing capacitor to 0.47uF will give exactly the same output waveform with much less supply current.

Nobody has commented on the other glaring error yet - C1 is reversed !

Brian.
 
Re: Newbie 555 circuit problem

Yeah, then C7 is also reversed. I think the symbol is wrongly considered in the circuit.

Surely if the resistor values increased then the duty cycle will be better,
From the equation in #3
for 50% duty cycle Ton = Toff
expected R5 = 0 !!
instead Duty cycle = R6 / (2R6+R5)
if you put 400 k in R6 8.2 k in R5 and 47nf in C7 still you will get same result with more perfect duty cycle.

Also the Gate parallel resistor may be brought near the Mosfet if any problem with the signaling then it will be useful..
 
Re: Newbie 555 circuit problem

Alright Gents, spent some time with this again. I checked by scope is there spike between pin 3 and gate like you can see from coils when disconnected. Indeed there is but it's only ~2V higher than the top of wave. This was done without the R8.

Also measured with AC side on ammeter and it registered ~60uA. Probably it's not responding fast enough to get real picture of the peak amps.? What kind of instrument should be used to see the true peak amps?

Those electrolytic capacitor symbols are missing + sign and somehow I managed to put them reverse on that drawing. On actual breadboard they are the right way.

What comes to the duty cycle I didn't thought possibility that you don't need to get those two resistor values very close to each other, same can be done if R6 is very high value compared to R5. Good point, thanks. I only got 100nf ceramic capacitor so ended up using R6 272kohm and R5 2.2kohm. This gave me 50.4% +duty cycle.
 

put 1 ohm instead of R8 and measure the voltage across its two ends using oscilloscope.

Now you will see the current in oscilloscope, but it will be very very narrow for your 35 Hz..
 
We don' really know the injector impedance, so current sensing R on both gate and drain ought to show the real problem with the layout and that is spurious resonance. Use very low value <<1 R on drain.
 
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