http://www.thefatherjackband.com/amps/D ... 0Power.pdf
This schematic has the positive leads of the caps going to ground... is this correct in this application?
Thanks,
Andy
Question on 70's ODS Relay PS in this schematic...
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v00d00blues79
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Re: Question on 70's ODS Relay PS in this schematic...
To strictly answer your question....no, the caps are backwards. However, there are a bunch of problems with the circuit you have there.
If I read this correctly, you have a dedicated 12V transformer for the relay circuit and a 12V relay. So...
1. The diode/capacitor configuration is a voltage doubler. This vould give you 24V DC. I believe the circuit that you are modifying was meant to drive a 12V realy from the 6.3V filament supply. I would just put a full wave rectifier bridge after the transformer with 100 uf capacitor for filtering to get about 11 VDC. Plenty for the relay.
2. Im not sure what the 56K resistor and the 100 uf cpacitor are for. The 56k resistor means that, with the 250 ohm resistance of the relay coil, the entire volage drop is across the resistor. The coil will never pull in.
3. You need a snubber diode around the resistor coil
4. I suppose I can see a purpose to isolate the relay coil from ground with the .1 uf capacitor, but (to me) it would just be simpler to just leave that part open with no connection.
5. The 1k resistor is, I assume, for an LED supply in the foot pedal. That value may be too low, depending on the LED you are using. Assuming a 12V supply and the LED forward voltage of 2V, that gives 10V across the 1K resistor and a current of 10 mA. I don't have LED data here, but that seems really high to me.
A simpler circuit for what I think you are attempting can be found at http://www.hoffmanamps.com/projects/relay_switch3.htm. Bear in mind that circuit is coming off the 6.3 V supply and is driving a 5V relay. Use a 12V transformer and a 12V and omit the 180 ohm resistor (straight wire). Should work fine.
If I read this correctly, you have a dedicated 12V transformer for the relay circuit and a 12V relay. So...
1. The diode/capacitor configuration is a voltage doubler. This vould give you 24V DC. I believe the circuit that you are modifying was meant to drive a 12V realy from the 6.3V filament supply. I would just put a full wave rectifier bridge after the transformer with 100 uf capacitor for filtering to get about 11 VDC. Plenty for the relay.
2. Im not sure what the 56K resistor and the 100 uf cpacitor are for. The 56k resistor means that, with the 250 ohm resistance of the relay coil, the entire volage drop is across the resistor. The coil will never pull in.
3. You need a snubber diode around the resistor coil
4. I suppose I can see a purpose to isolate the relay coil from ground with the .1 uf capacitor, but (to me) it would just be simpler to just leave that part open with no connection.
5. The 1k resistor is, I assume, for an LED supply in the foot pedal. That value may be too low, depending on the LED you are using. Assuming a 12V supply and the LED forward voltage of 2V, that gives 10V across the 1K resistor and a current of 10 mA. I don't have LED data here, but that seems really high to me.
A simpler circuit for what I think you are attempting can be found at http://www.hoffmanamps.com/projects/relay_switch3.htm. Bear in mind that circuit is coming off the 6.3 V supply and is driving a 5V relay. Use a 12V transformer and a 12V and omit the 180 ohm resistor (straight wire). Should work fine.