This is a typical reverse voltage protection circuit used on many guitar effects pedals. I have used it at times myself and it is effective within limits.
Under normal conditions when a power supply with the properly polarity is being used, the D1 diode is not conducting. A few milliamps are being drawn through the series resistor and the power dissipation is less than the max rating of the resistor, which is typically 1/4 watt.
However, if a reverse polarity voltage supply is plugged into this circuit, the D1 diode will conduct and current passes through it and back to the power source through the 100 ohm resistor. This is when the problems start.
The diode has about 0.7v drop across it but the rest of the voltage will be across the resistor and Ohm’s law can tell us how much current will flow: I = V/R
The AMZ LED calculator is a quick way to get the answers that are needed. The Vf of the diode is 0.7v and it is entered in the LED forward voltage box (an LED is a diode after all). So if you plug in the voltage supply, forward voltage and resistor value, the calculator will reveal how much current is used and also how many watts have to be dissipated by the resistor.
In this case, with 9v supply, 0.7v diode and a 100 ohm resistor, the current is 83 milliamps and almost 0.70 watts. This is clearly in excess of the rating for the 0.25 watt resistor and it will get quite hot after a few seconds. Usually the resistor melts down and becomes open shortly thereafter. No current will flow then since the resistor has acted like a fuse.
The diode doesn’t melt down because it is usually rated for 1.0A or more and this circuit is only pulling 0.083A through it.
A very well-known commercial pedal uses 47 ohms for the resistor and if you substitute that value in the calculator, it shows that the resistor will have to stand almost 1.5 watts! I have replaced the fried 47R power supply resistor in many of these pedals.
One solution is to use a 2w rated resistor for the series input from the power supply. This works but the resistor will get hot if the power is reversed but at least it won’t go up in flames.
A better solution is to add a new diode in series before the resistor. This protects the circuit and in reversed power conditions, no current flows. The downside is that there will be a voltage drop across the diode and the voltage that makes it to the circuit will be less; about 8.0 – 8.3 volts, depending on the diode and how much current the circuit draws.
The voltage drop can be minimized by using a low Vf diode such as the 1N5817 Schottky part. This is the best solution and the diode can typically be added where the positive voltage wire connects to the pcb.
If you have a pedal that is not working and there does not seem to be any voltage getting to the transistors or chips, check the series resistor on the plus power line.
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Curious – what is the value of D1 (assuming it’s not an LED).
Confused: can’t D1 just be put before the series resistor?
D1 is usually a 1N4001 or similar diode.
if you add the new series 1N5817 diode then D1 is not necessary but it does not have to be removed from a pedal where it is already in place. This lessens the chance that the pcb will be damaged when desoldering and allows you to easily return the pedal to stock condition if desired.
regards, Jack
You can put the diode before the 100R resistor, which is part of the R-C power supply noise filter. However, with it before, the diode will carry all of the current in a reverse voltage situation. If the power supply can provide more than 1A, the diode will get hot very quickly and burn out. Usually it opens up. In that case, the diode is open and the reverse voltage then flows on to the circuit that it was intended to protect, with possibly damage to the circuit.
The series diode, as shown in the article, is a better idea. It’s what I use in the pre-assembled effect modules.
That’s what happened to my Rotovibe when I lent it to a friend of mine
I have been designing and building pedal effects for a while and also tube amplifiers. In the pedal end in my builds I use 2 diodes most pedal designs use a Neg.pin supply for the DC inlet barrel connector, the barrel is Positive, the positive goes to the first diode ( usually 1N4007 ) Cathode ( stripe line on diode ) and anode to ground before the Filter Cap. ( diode is not conducting ). I don’t use a series resistor because I don’t like possible flame out when the supply polarity gets reversed. The other diode is wired from the Neg. pin to cathode side to ground, ( forward bias in a way ) usually to the input jack TRS for battery operation or mono for no battery, which ever the case may be. It all depends how you design the grounding system in your pedal. I use star grounding from input jack then to the P.C.B. The jacks are grounded to the box. This means the second diode lifts the ground slightly ( because of internal resistance ) but shorted to all round box ground. When the polarity gets reversed ( which I have tested ) from the DC inlet the current has no where to go and it just humss, there is little current return flow because of internal resistance and no flame out. This scheme seems to work out for me and I have incorporated this in most of my builds.
An even better solution might be to use a mosfet wired as ‘active diode’.
See this application note from TI for more information.
http://www.ti.com/lit/an/slva139/slva139.pdf